Important Questions

Punjab State Board PSEB 11th Class Physics Important Questions Chapter 4 Motion in a Plane Important Questions and Answers.

PSEB 11th Class Physics Important Questions Chapter 4 Motion in a Plane

Very short answer type questions

Question 1.
When do we say two vectors are orthogonal?
Solution:
If the dot product of two vectors is zero, then the vectors are orthogonal.

Question 2.
What is the property of two vectors \(\overrightarrow{\boldsymbol{A}}\) and \(\overrightarrow{\boldsymbol{B}}\) such that \(\overrightarrow{\boldsymbol{B}}+\overrightarrow{\boldsymbol{A}}=\overrightarrow{\boldsymbol{C}}\) and \(\overrightarrow{\boldsymbol{A}}+\overrightarrow{\boldsymbol{B}}=\overrightarrow{\boldsymbol{C}}\)?
Solution:
The two vectors are parallel and acting in the same direction i. e., θ = 0 °.

Question 3.
What are the minimum number of forces which are numerically equal whose vector sum can be zero?
Answer:
Two only, provided that they are acting in opposite directions.

Question 4.
Under what condition the three vectors cannot give zero resultant?
Answer:
When the three vectors are not lying in one plane, they cannot produce zero resultant.

Question 5.
Can the scalar product of two vectors be negative?
Solution:
Yes, it will be negative if the angle between the two vectors lies between 90° to 270°.

Question 6.
Can the walking on a road be an example of resolution of vectors?
Answer:
Yes, when a man walks on the road, he presses the road along an oblique direction. The horizontal component of the reaction helps the man to walk on the road.

Question 7.
A particle cannot accelerate if its velocity is constant, why?
Answer:
When the particle is moving with a constant velocity, there is no change – in velocity with time and hence, its acceleration is zero.

Question 8.
A football is kicked into the air vertically upwards. What is its (i) acceleration and (ii) velocity at the highest point?
(NCERT Exemplar)
Answer:

(i) Acceleration at the highest point = -g
(ii) Velocity at the highest point = 0.

Question 9.
Why does a tennis ball bounce higher on bills than in plains?
Answer:
Maximum height attained by a projectile ∝ 1/ g. As the value of g is less on hills than on plains, so a tennis ball bounces higher on hills than on plains.

Short answer type quetions

Question 1.
Explain the property of two vectors \(\overrightarrow{\boldsymbol{A}}\) and \(\overrightarrow{\boldsymbol{B}}\) if \(|\overrightarrow{\boldsymbol{A}}+\overrightarrow{\boldsymbol{B}}|=|\overrightarrow{\boldsymbol{A}}-\overrightarrow{\boldsymbol{B}}|\).
Solution:
As we know that
\(|\vec{A}+\vec{B}|=\sqrt{A^{2}+B^{2}+2 A B \cos \theta}\)
and \(|\vec{A}-\vec{B}|=\sqrt{A^{2}+B^{2}-2 A B \cos \theta}\)
But as per question, we have
\(\sqrt{A^{2}+B^{2}+2 A B \cos \theta}=\sqrt{A^{2}+B^{2}-2 A B \cos \theta}\)
Squaring both sides, we have (4 AB cos θ) = 0
⇒ cosθ = 0 or θ = 90°
Hence, the two vectors \(\vec{A}\) and \(\vec{B}\) are perpendicular to each other.

Question 2.
The sum and difference of two vectors are perpendicular to each other. Prove that the vectors are equal in magnitude.
Solution:
As the vectors \(\vec{A}+\vec{B}\) and \(\vec{A}-\vec{B}\) are perpendicular to each other, therefore
\((\vec{A}+\vec{B}) \cdot(\vec{A}-\vec{B})\) = 0
\(\vec{A} \cdot \vec{A}-\vec{A} \cdot \vec{B}+\vec{B} \cdot \vec{A}-\vec{B} \cdot \vec{B}\) = 0
or A² – B² = 0     [∵ \(\vec{A} \cdot \vec{B}=\vec{B} \cdot \vec{A}\)]
⇒ A = B

Question 3.
The dot product of two vectors vanishes when vectors are orthogonal and has maximum value when vectors are parallel to each other. Explain.
Solution:
We know that \(\vec{A} \cdot \vec{B}\) = AB cos θ, when vectors are orthogonal, then, θ = 90°.
So, \(\vec{A} \cdot \vec{B}\) = AB cos 90 ° = 0, when vectors are parallel, then, θ = 0°
So, \(\vec{A} \cdot \vec{B}\) = AB cos ° = AB (maximum)

Question 4.
Can a flight of a bird, an example of composition of vectors. Why?
Answer:

Yes, the flight of a bird is an example of composition of vectors as the bird flies, it strikes the air with its wings W, W along WO. According to Newton’s third law of motion, air strikes the wings in opposite directions with the same force in reaction. The reactions are \(\overrightarrow{O A}\) and \(\overrightarrow{O B}\). From law of parallelogram vectors, \(\overrightarrow{O C}\) is the resultant of \(\overrightarrow{O A}\) and \(\overrightarrow{O B}\). This resultant upwards force \(\overrightarrow{O C}\) is responsible for the flight of the bird.

Question 5.
How does the knowledge of projectile help, a player in the baseball game?
Answer:
In the baseball game, a player has to throw a ball so that it goes a certain distance in the minimum time. The time would depend on velocity of ball and angle of throw with the horizontal. Thus, while playing a baseball game, die speed and angle of projection have to be adjusted suitable so that the ball covers the desired distance in minimum time. So, a player has to see the distance and air resistance while playing with a baseball game.

Question 6.
A skilled gun man always keeps his gun slightly tilted above the line of sight while shooting. Why?
Answer:

When a bullet is fired from a gun with its barrel directed towards the target, it starts falling downwards on account of acceleration due to gravity.
Due to which the bullet hits below the target. Just to avoid it, the barrel of the gun is lined up little above the target, so that the bullet after traveling in parabolic path hits the distant target.

Question 7.
Establish a relation between angular velocity and time period.
Answer:
We know that angular velocity A0
ω = \(\frac{\Delta \theta}{\Delta t}\)
For motion with uniform angular velocity, in one complete revolution A0 = 2JI radian and At = T s, hence
ω = \(\frac{2 \pi}{T}\) or T = \(\frac{2 \pi}{\omega}\).

Question 8.
A fighter plane is flying horizontally at an altitude of 1.5 km with speed 720 km/h. At what angle of sight (w.r.t. horizontal) when the target is seen, should the pilot drop the bomb in order to attack the target? (NCERT Exemplar)
Answer:
Consider the adjacent diagram. Let a fighter plane, when it be at position P, drops a bomb to hit a target T.

Let < P’PT= θ
Speed of the plane = 720 km/h
= 720 × \(\frac{5}{18}\) m/s = 200m/s
Altitude of the plane (PT) = 1.5km = 1500 m
If bomb hits the target after time t, then horizontal distance travelled by the bomb.
PP’ = u × t = 200t
Vertical distance travelled by the bomb,
P’T = \(\frac{1}{2}\)gt² ⇒ 1500 = \(\frac{1}{2}\) × 9.8t²
⇒ t² = \(\frac{1500}{49}\) ⇒ t = \($\sqrt{\frac{1500}{49}}$\) = 17.49s
Using value oft in Eq. (i),
PP’ = 200 × 17.49 m
Now,
tanθ = \($\frac{P^{\prime} T}{P^{\prime} P}=\frac{1500}{200 \times 17.49}$\) 0.49287 = tan23°12′
θ = 23°12′
Note Angle is with respect to target. As seen by observer in the plane motion of the bomb will be vertically downward below tbe plane.

Long answer type questions

Question 1.
An airline passenger late for a flight walks on an airport moving sidewalk at a speed of 5.00 km/h relative to the sidewalk, in the direction of its motion. The sidewalk is moving at 3.00 km/h relative to the ground and has a total length of 135 m.
(i) What is the passenger’s speed relative to the ground?
(ii) How long does it take him to reach the end of the sidewalk?
(iii) How much of the sidewalk has he covered by the time he reaches Hie end?
Solution:
The situation is sketched in figure. We assign a letter to each body in relative motion, P passenger, S sidewalk, G ground. The relative velocities υ ps and υ SG are given

υ_PS = 5.00 km/h, to the right
υ_SG = 3.00 km/h, to the right

(i) Here, we must find the magnitude of the vector υ_PG, given the magnitude and direction of two other vectors. We find the velocity υ_PG by using the relation
υ_PG = υ_PS + υ_SG
Here, the vectors are parallel, and so the vector addition is quite simple (see figure). We add vectors by adding magnitudes.
υ_PG = υ_PS +υ_SG
= 5.00 km/h + 3.00 km/h
= 8.00 km/h
= 8 × \(\frac{5}{18}\) m/s = \(\frac{40}{18}\) = 2.22 m/s

(ii) The length of the sidewalk is 135 m, and so this is the distance Δ x_G the passenger travels relative to the ground. So, our problem is to find Δt when Δx_G =135 m. The rate at which this distance along the ground is covered by the passenger is υ_PG, where
υ_PG = \(\frac{\Delta x_{G}}{\Delta t}\)
Therefore, Δ t = \(\frac{\Delta x_{G}}{v_{P G}}\) = \(\frac{135 \mathrm{~m}}{2.22 \mathrm{~m} / \mathrm{s}}\) = 60.8 s

(iii) The problem here is to determine how much of the sidewalk’s surface the passenger moves over. If he was standing still and not walking along the surface, he would cover none of it. Because he is moving relative to the surface at velocity υ_PS, he does move some distance Δ xs relative to the surface. The problem is to find Δ X_S when Δt = 60.8 s, since we found in part (ii) that this is the time interval during which he is on the moving sidewalk. His velocity relative to the sidewalk is υ_PS = Δx_S / Δt, and so
ΔX_S = υ_PS = Δt = (5.00 km/h) × (60.8s)
= \(\frac{25}{18}\) × 60.8 (∵ 1 km/h = \(\frac{5}{18}\) m/s)
= 84.4 m

Question 2.
A hunter aims his gun and fires a bullet directlyiafoi monkey in a tree. At the instant, the bullet leaves the barrebdi,;the gun, the monkey drops. Will the bullet hit the monkey? Substantiate your answer with proper reasoning.
Solution:
Let the monkey stationed at A, be fired with a gun fromO with a velocityu at an angle 0 with the horizontal direction OX.
Draw AC, perpendicular to OX. Let the bullet cross the vertical line AC at B after time t and coordinates of B (x, y) be w.r.t. origin O as shown in figure.
∴ t = \(\frac{O C}{u \cos \theta}=\frac{x}{u \cos \theta}\) ………….. (i)
In ∆ OAC, AC = OC tanO = x tanθ ……………. (ii)
Clearly, CB = y = the vertical distance travelled by the bullet in time t. Taking motion of the bullet from O to B along Y-axis, we have y₀ = 0, y = y,U_y = usin0, a_y = -g,t = t

It means the bullet will pass through the point B on vertical line AC at a vertical distance \(\frac {1}{2}\)gt² below point A.
The distance through which the monkey falls vertically in time t = \(\frac {1}{2}\)gt²
= AB. It means the bullet and monkey will pass through the point B simultaneously.
Therefore, the bullet will hit the monkey.

Punjab State Board PSEB 11th Class Physics Important Questions Chapter 10 Mechanical Properties of Fluids Important Questions and Answers.

PSEB 11th Class Physics Important Questions Chapter 10 Mechanical Properties of Fluids

Very Short Answer Type Questions

Question 1.
Three vessels have same base area and different neck area. Equal volume of liquid is poured into them, which will possess more pressure at the base?
Answer:
If the volumes are same, then height of the liquid will be highest in which the cross-section area is least at the top. So, the vessel having least cross-section area at the top possess more pressure at the base (∵ P = ρgh).

Question 2.
What is the use of barometer?
Answer:
Barometer is used to measure the atmospheric pressure.

Question 3.
What is the use of open tube manometer?
Answer:
Open tube manometer is used for measuring pressure difference.

Question 4.
‘What is the gauge pressure?
Solution:
The difference between absolute pressure and atmospheric pressure is known as gauge pressure.
As, P_absolute = P_a+ ρgh
So, P_absolute – P_a = ρgh
i.e., P_gauge = ρgh
Here ρ is the density of a fluid of depth h.

Question 5.
If a wet piece of wood bums, then water droplets appear on the other end, why?
Answer:
When a piece of the wet wood bums, then steam formed and water appear in the form of droplets due to surface tension on the other end.

Question 6.
Why soap bubble bursts after some time?
Answer:
Soap bubble bursts after some time because the pressure inside it become more than the outside pressure.

Question 7.
Can two streamlines cross each other? Why?
Answer:
Two streamlines can never cross each other because if they cross them at the point of intersection there will be two possible direction of flow of fluid which is impossible for streamlines.

Question 8.
A hot liquid moves faster than a cold liquid. Why?
Answer:
The viscosity of liquid decreases with the increase in temperature. Therefore, viscosity of hot liquid is less than that of cold liquid. Due to this hot liquid moves faster than the cold liquid.

Question 9.
Is viscosity a vector? (NCERT Exemplar)
Answer:
Viscosity is a property of liquid it does not have any direction, hence it is a scalar quantity.

Question 10.
Is surface tension a vector? (NCERT Exemplar)
Answer:
No, surface tension is a scalar quantity.
Surface tension = \(\frac{\text { Work done }}{\text { Surface area }} \) , where work done and surface area both Surface area are scalar quantities.

Short Answer Type Questions

Question 1.
A large force is needed to normally separate two glass plates having a thin layer of water between them. Why?
Answer:
The thin layer of water between the glass plates forms a concave surface all around. This decreases the pressure on the inner side of the liquid film. Thus, a large amount of force is required to pull them apart against the atmospheric pressure.

Question 2.
Two soap bubbles in vacuum having radii 3 cm and 4 cm respectively coalesce under isothermal conditions to form a single bubble. What is the radius of the new bubble?
Solution:
Surface energy of first bubble = Surface area x Surface tension
= 2 x 4 πr²₁T = 8πr²₁T
Surface energy of second bubble = 8πr²₂T
Let r be the radius of the coalesced bubbles.
∴ Surface energy of new bubble = 8πr² T
According to the law of conservation of energy,

∴ r = 5 cm

Question 3.
A balloon with hydrogen in it rises up but a balloon with air comes down. Why?
Answer:
The density of hydrogen is less than air. So, the buoyant force on the balloon will be more than its weight in case of the hydrogen. So, in this case the balloon rises up. In case of air, the weight of balloon is more than the buoyant force acting on it, so balloon will come down.

Question 4.
It is easier to spray water in which some soap is dissolved. Explain why?
Answer:
When the liquid is sprayed, it is broken into small drops. The surface area increases and hence the surface energy is also increased. Therefore, work has to be done to supply the additional energy. Since surface energy is numerically equal to the surface tension, so when soap is dissolved in water, the surface tension of the solution decreases and hence less energy is spent to spray it.

Question 5.
Why are the wings of an aeroplane rounded outwards while flattened inwards?
Answer:
The special design of the wings increases velocity at the upper surface and decreases velocity at the lower surface. So, according to Bernoulli’s theorem, the pressure on the upper side is less than the pressure on the lower side. This difference of pressure provides lift.

Question 6.
The surface tension and vapour pressure of water at 20°C is 7.28 x 10^-2 Nm^-1 and 233x 10³ Pa, respectively. What is the radius of the smallest spherical water droplet which can form without evaporating at 20°C?
Answer:
Given, surface tension of water (S) = 7.28 x 10^-2 N/m
Vapour pressure (p) = 2.33 x 10³ Pa
The drop will evaporate if the water pressure is greater than the vapour pressure.
Let a water droplet or radius R can be formed without évaporating.
Vapour pressure = Excess pressure in drop.
∴ p = \(\frac{2 S}{R}\) or R= \(\frac{2 S}{p}=\frac{2 \times 7.28 \times 10^{-2}}{2.33 \times 10^{3}}\)
= 6.25 x 10^-5 m

Long Answer Type Questions

Question 1.
if a sphere of radius r falls under gravity through a liquid of viscosity q, its average acceleration is half that of in starting of the motion. Then, show that the time taken by it to attain the term mal velocity is independent of the liquid density.
Solution:
Let the density of sphere’s material is ρ and that of liquid is σ.
When the sphere just enters in the liquid.
Downward force on the sphere, F = weight of the sphere – weight of the fluid displaced by it.
F= \(\frac{4}{3} \pi r^{3}\) ρg – \(\frac{4}{3} \pi r^{3}\)σg
∵ Mass = Volume xDensity = \(\frac{4}{3} \pi r^{3}\) (ρ-σ)g
∴ Acceleration of the sphere at this instant.

When the sphere approches to terminal velocity, its acceleration becomes zero.
∴ Average acceleration of the sphere = \(\frac{a+0}{2}\)
= \(\frac{\left(1-\frac{\sigma}{\rho}\right) g}{2}=\left(1-\frac{\sigma}{\rho}\right) \frac{g}{2}\)

If time t taken by the sphere to attain the terminal velocity As we know that,
Terminal velocity, ν = \(\frac{2}{3} \frac{r^{2}}{\eta}(\rho-\sigma) g\)
∵ The sphere falls from rest,
∴ u=O
Using ν=u+at
Putting values in above eqdation, we get

Thus, t is independent of the liquid density.

Question 2.
(a) Derive the expression for excess of pressure inside:
(i) a liquid drop.
(ii) a liquid bubble.
(iii) an air bubble.
(b) Derive the relation between the surface tension and the surface energy
Solution:
(a) (i) Let r = radius of a spherical liquid drop of centre O.
T = surface teñsion of the liquid.
Let p_i and p₀ be the values of pressure inside and outside the drop.

∴ Excess of pressure inside the liquid drop = p_i -p₀
Let Δr be the increase in its radius due to excess of pressure. It has one free surface outside it.
∴ increase in surface area of the liquid drop

= 8πr Δr …………………………… (i)

(∵ Δr is small ∴ Δr² is neglected.)
∴ increase in surface energy of the drop is
W = surface tension x increase in area
=T x8πr Δr …………………………………… (ii)

Also W = Force due to excess of pressure x displacement
W = Excess of pressure x area of drop x increase in radius
= (p_i -p₀ )4πr² Δr ………………………………… (iii)
From eqs. (ii) and (iii), we get
(p_i -p₀ ) x 4 πr² Δr = T x8πr A r Δr
or p_i -p₀ = \(\frac{2 T}{r}\)

(ii) In a liquid bubble : A liquid bubble has air both inside and outside it and therefore it has two free surfaces.
r,Δr, T = ? as above
Thus increase in its surface area
= 2 [ 4 π(r+Δr)² – 4 πr²]
= 2 x 8 πrΔr
= 16πrΔr
∴ W = T x 16πrΔr, …………………… (iv)

Also W= (p_i -p₀ ) x 4πr² x Δr ………………………. (v)
∴ From (iv) and (v), we get
(p_i -p₀ ) x 4πr² x Δr = T. 16πrΔr
or p_i -p₀ = \(\frac{4 T}{r}\)

(iii) Inside an air bubble: Air bubble is formed inside liquid, thus air bubble has one free surface inside it and liquid is outside.
If r = radius of the air bubble.
Δr = increase in its radius due to excess of pressure (p_i -p₀ ) inside it.
T = surface tension of the liquid in which bubble is formed, increase in surface area = 8 πrΔr
∴ W = T x 8 πrΔr
Also W = (p_i -p₀)x 4 πr²Δr
∴ (p_i -p₀) x 4 πr²Δr = T x 8 πrΔr
or p_i -p₀ = \(\frac{2 T}{r}\)

(b) Let ABCD be a rectangular frame of wire. Let LM be a slidable cross-piece. Now dip the wireframe in the soap solution so that a film is formed over the frame. Due to surface tension, the film has a tendency to shrink and thereby, the cross-piece LM will be pulled in inward direction which can be kept in its position by applying an equal and opposite force F on it.
∴ F = T × 2l
where T = surface tension and l = length of LM.
It has been taken 21 as the film has two free surfaces.
Let x = small distance by which LM moves to L’M’.
∴ 2l × x = increase in the area of the film
if W = work done in increasing the area by 2l × x,
then W = F × x = (T × 2l) × x

If U be the surface energy, then by definition
U = \(\frac{\text { Work done in increasing the surface area }}{\text { increase in surface area }} \)
= \(\frac{T \times 2 l \times x}{2 l \times x}\)
U = T
Thus, U is numerically equal to the surface energy.

Punjab State Board PSEB 11th Class Physics Important Questions Chapter 5 Laws of Motion Important Questions and Answers.

PSEB 11th Class Physics Important Questions Chapter 5 Laws of Motion

very short answer type questions

Question 1.
Bodies of larger mass need greater initial effort to put them in motion. Why?
Answer:
According to the Newton’s second law of motion, F = ma, for given acceleration a, if m is large, F should be more i. e., greater force will be required to put a larger mass in motion.

Question 2.
The distance travelled by a moving body is directly proportional to time. Is any external force acting on it?
Solution:
When S ∝ t, so acceleration = 0. Therefore, no external force is acting on the body.

Question 3.
A body is acted upon by a number of external forces. Can it remain at rest?
Answer:
Yes, if the external forces acting on the body can be represented in magnitude and direction by the sides of a closed polygon taken in the same order.

Question 4.
If force is acting on a moving body perpendicular to the direction of motion, then what will be its effect on the speed and direction of the body?
Answer:
No change in speed, but change in direction is possible. Forces acting on a body in circular motion is an example.

Question 5.
An impulse is applied to a moving object with a force at an angle of 20° w.r.t. velocity vector, what is the angle between the impulse vector and change in momentum vector?
Answer:
Impulse and change in momentum are along the same direction. Therefore, angle between these two vectors is zero degree.

Question 6.
A body is moving in a circular path such that its speed always remains constant. Should there be a force acting on the body?
Answer:
When a body is moving along a circular path, speed always remains constant and a centripetal force is acting on the body.

Question 7.
The mountain road is generally made winding upwards rather than going straight up. Why?
Solution:
When we go up a mountain, the opposing force of friction
F = μR = μ mg cosθ.
where θ is angle of slope with horizontal. To avoid skidding, F should be large.
∴ cosθ should be large and hence, θ must be small.
Therefore, mountain roads are generally made winding upwards. The road straight up would have large slope.

Short answer type questions

Question 1.
A body of mass 500 g tied to a string of length 1 m is revolved in the vertical circle with a constant speed. Find the minimum speed at which there will not be any slack on the string. Take g = 10ms^-2
Solution:
The tension T in the string will provide the necessary centripetäl force
\(\frac{m v^{2}}{r}\) i.e., T = \(\frac{m v^{2}}{r}\)
Here, m = 500g = \(\frac{1}{2}\)kg; r = 1m
T = \(\frac{1}{2}\)υ²N ……………. (i)
There will not be slack 1f T ≥ weight of the body
i.e., T ≥ mg or \(\frac{1}{2}\)υ² ≥\(\frac{1}{2}\) × 10
υ² ≥ 10 or υ ≥ \(\sqrt{10}\) ms^-1
So the minimum speed = \(\sqrt{10}\) ms^-1 = 3.162 ms^-1

Question 2.
A light, inextensible string as shown in figure connects two blocks of mass M₁ and M₂. A force F as shown acts upon M₁. Find acceleration of the system and tension in string.

Solution:
Here as the string is inextensible, acceleration of two blocks will be same. Also, string is massless so tension throughout the string will be same. Contact force will be normal force only. Let acceleration of each block is a, tension in string is T and contact force between M₁ and surface is N₁ and contact force between M₂ and surface is N₂
Applying Newton’s second law for the blocks;
For M₁, F – T = M₁ a ……………. (i)
M₁ g – N₁ = 0 …………….. (ii)
For M₂, T = M₂ ……………… (iii)
M₂g – N = 0 ……………… (iv)
Solving equations (i) and (iii), we get
a = \(\frac{F}{M_{1}+M_{2}}\)
and T = \(\frac{M_{2} F}{M_{1}+M_{2}}\)

Question 3.
A block of mass m is held against a rough vertical wall by pressing it with a finger. If the coefficient of friction between the block and the wall is μ and the acceleration due to gravity is g, calculate the minimum force required to be applied by finger to hold the block against the wall? (NCERT Exemplar)
Solution:
Given, mass of the block = m
Coefficient of friction between the block and the wall = μ
Let a force F be applied on the block to hold the block against the wall.
The normal reaction of mass be N and force of friction acting upward be f.
In equilibrium, vertical and horizontal forces should be balanced separately.
f = mg …………….. (i)
∴ and F = N …………… (ii)

But force of friction (f) = μN
= μF [using eq. (ii) ] ………….. (iii)
From eqs. (i) and (iii), we get
μF = mg
or F = \(\frac{m g}{\mu}\)

Question 4.
A bird is sitting on the floor of a closed glass cage and the cage is in the hand of a girl. Will the girl experience any change in the weight of the cage when the bird (i) starts flying in the cage with a constant velocity, (ii) flies upwards with acceleration and (iii) flies downwards with acceleration?
Solution:
In a closed glass cage, air inside is bound with the cage. Therefore,
(i) there would be no change in weight of the cage if the bird flies with a constant velocity.
(ii) the cage becomes heavier, when bird flies upwards with an acceleration.
(iii) the cage appears lighter, when bird flies downwards with an acceleration.

Question 5.
When walking on ice, one should take short steps rather than long steps. Why?
Solution:
Let R represent the reaction offered by the ground. The vertical component R cosθ will balance the weight of the person and the horizontal component R sinθ will help the person to walk forward.

Now, normal reaction = R cosθ
Friction force = R sinθ
Coefficient of friction, μ = \(\frac{R \sin \theta}{R \cos \theta}\) = tanθ
In a long step, θ is more. So tanθ is more. But μ has a fixed value. So, there is danger of slipping in a long step.

Question 6.
A body of mass m is suspended by two strings making angles α and β with the horizontal as shown in fig. Calculate the tensions in the two strings.

Solution:
Considering components of tensions T₁ and T₂ along the horizontal and vertical directions,
We have
-T₁cosα + T₂cosβ = 0
or T₁cosα = T₂cosβ …………… (i)
and T₁ sinα + T₂ sinβ = mg
From eq. (i) T₂ = \(\frac{T_{1} \cos \alpha}{\cos \beta}\) and substituting it in eq. (ii), we get

Question 7.
State the law of conservation of momentum. Establish the same for a ‘n’ body system.
Solution:
When no external force acts on a system the momentum will remain conserved. Consider a system of a n bodies of masses m₁ ,m₂ ,m₃ , ………… ,m_n. If p₁ , p₂ , P₃ , ………. ,P_n are the momentum associated then the rate of change of momentum with the system,
\(\frac{d p}{d t}=\frac{d p_{1}}{d t}+\frac{d p_{2}}{d t}+\frac{d p_{3}}{d t}\) + ………. + \(\frac{d p_{n}}{d t}=\frac{d}{d t}\) = (p+₁ +p₂ +p₃+ ………. +p_n )
If no external force acts, \(\frac{d p}{d t}\) = 0
∴ p = constant, i.e., P₁ + p₂ + P₃ +………… +P_n = constant.

Question 8.
A block slides down from top of a smooth inclined plane of elevation θ fixed in an elevator going up with an acceleration a0. The base of incline has length L. Find the time taken by the block to reach the bottom.

Solution:
The free body force diagram is shown. The forces are
(i) N normal to the plane (ii) mg acting vertically down (iii) ma₀ (pseudo-force).

If a is the acceleration of the body with respect to incline, taking components of forces parallel to the incline mg sinθ + ma0 sinθ = ma
a = (g + a₀)sinθ
This is the acceleration with respect to elevator.
The distance travelled is \(\frac{L}{\cos \theta}\) If t is the time for reaching the bottom of
incline, using equation of motion, s = ut + \(\frac{1}{2}\)at2, we get
\(\frac{L}{\cos \theta}\) = 0 + \(\frac{1}{2}\)(g + a₀)sinθ.t²
t = [latex]\frac{2 L}{\left(g+a_{0}\right) \sin \theta \cos \theta}[/latex]^1/2

Long answer type questions

Question 1.
Figure shows (x – t), (y – t) diagram of a particle moving in 2-dimensions.

If the particle has a mass of 500 g, find the force (direction and magnitude) acting on the particle. (NCERT Exemplar)
Given, mass of the particle (m) = 500 g = 0.5 kg
x – t graph of the particle is a straight line.
Hence, particle is moving with a uniform velocity along x-axis, i. e., its acceleration along x-axis is zero and hence, force acting along x-axis is zero.
y – t graph of particle is a parabola. Therefore, particle is in accelerated motion along y – axis.
At t = 0, u_y = 0
Along y – axis, at t = 2s, y = 4m
Using equation of motion, y = u_yt + \(\frac{1}{2}\) a_yt²
4 = 0 × 2 + \(\frac{1}{2}\) × a_y × (2)²
or a_y = 2 m/s²
∴ Force acting along y – axis (f_y) = ma_y = 0.5 × 2 = 1.0 N (along y – axis)

Question 2.
When a body slides down from rest along a smooth inclined plane making an angle of 45° with the horizontal, it takes time T. When the same body slides down from rest along a rough inclined plane making the same angle and through the same distance, it is seen to take time pT, where p is some number greater than 1. Calculate the coefficient of friction between the body and the rough plane. (NCERT Exemplar)
Solution:
On smooth inclined plane Acceleration of a body sliding down a smooth inclined plane, a = g sinθ
Here, θ = 45°
a = gsin45°= \(\frac{g}{\sqrt{2}}\)
Let the travelled distance be s.
Using the equation of motion, s = ut + \(\frac{1}{2}\) at² ,
We get
s = 0 .t + \(\frac{1}{2} \frac{g}{\sqrt{2}}\)T²
or s = \(\frac{g T^{2}}{2 \sqrt{2}}\) ………… (i)

On rough inclined plane
Acceleration of the body,
a = g (sinθ – μ cosθ)
= g (sin 45° – μ cos 45°)
= \(\frac{g(1-\mu)}{\sqrt{2}}\) [as sin 45°= cos 45° = \(\frac{1}{\sqrt{2}}\)]
Again using equation of motion,

Punjab State Board PSEB 11th Class Physics Important Questions Chapter 6 Work, Energy and Power Important Questions and Answers.

PSEB 11th Class Physics Important Questions Chapter 6 Work, Energy and Powere

Very short answer type questions

Question 1.
Under what condition is the work done by a force inspite of displacement being taking place?
Answer:
Work done by a force is zero inspite of displacement being taking place, if displacement is in a direction perpendicular to that of force applied.

Question 2.
Can acceleration be produced without doing any work? Give example.
Answer:
Yes, for uniform circular motion, no work done but a centripetal acceleration is present.

Question 3.
Does the amount of work done depend upon the fact that how fast is a load raised or moved in the direction of force?
Answer:
The amount of work does not depend upon the fact that how fast is a load raised or moved in the direction of force.

Question 4.
A body is moving along a circular path. How much work is done by the centripetal force?
Answer:
For a body moving along a circular path, the centripetal force acts along the radius while the displacement is tangential, i. e., θ = 90 °, therefore,
W = Fscos90° = 0.

Question 5.
What is the source of kinetic energy of the bulelt coming out of a rifle?
Answer:
The source of kinetic energy of bullet is the potential energy of the compressed spring in the loaded rifle.

Question 6.
A spring is cut into two equal halves. How is the spring constant of each half affected?
Answer:
Spring constant of each half becomes twice the spring constant of the original spring.

Question 7.
Is collision between two particles possible even without any physical contact between them?
Answer:
Yes, in atomic and subatomic particles collision without any physical contact between the colliding particles is taking place e. g., Rutherford’s alpha particles scattering.

Question 8.
Why is electrical power required at all when the elevator is descending? Why should there be a limit on the number of passengers in this case? (NCERT Exemplar)
Answer:
When the elevator is descending, then electric power is required to prevent it from falling freely under gravity.
Also, as the weight inside the elevator increases, its speed of descending – increases, therefore, there should be a limit on the number of passengers in the elevator to prevent the elevator from descending with large velocity.

Short answer type questions

Question 1.
A block of mass M is pulled along a horizontal surface by applying a force at an angle θ with horizontal. Coefficient of friction between block and surface is μ. If the block travels with uniform velocity, find the work done by this applied force during a displacement d of the block.

Solution:
The forces acting on the block are shown in figure. As the block moves with uniform velocity the forces add up to zero.
∴ Fcosθ = μN ………….. (i)
Fsinθ + N = Mg ……………. (ii)
Eliminating N from equations (i) and (ii)
F cosθ = μ(Mg – F sinθ)
F = [Latex]\frac{\mu M g}{\cos \theta+\mu \sin \theta}[/Latex]
Work done by this force during a displacement d
W = F. d cosθ = [Latex]\frac{\mu M g d \cos \theta}{\cos \theta+\mu \sin \theta}[/Latex]

Question 2.
Two springs have force constants K₂ and K₂ (K₁ > K₂ )• On which spring is more work done when they are stretched by the same force?
Solution:

As x₁ < x₂
∴ W₁ < W₂ or W₂ > W₁

Question 3.
A particle is moving in a circular path of radius r with constant speed. Due to change in the direction of motion of the particle continuously, the velocity of the particle is changing. But the kinetic energy of the particle remains the same. Explain why ?
Solution:
Kinetic energy is given by
E = \(\frac{1}{2}\) mυ² = \(\frac{1}{2}\) m(\(\vec{v} \cdot \vec{v}\))
Since \(\vec{v} \cdot \vec{v}\) – υ², a scalar quantity, so it is the speed which is taken into account while calculating the kinetic energy of the particle. As the speed is constant, so kinetic energy of the particle will also remain constant.

Question 4.
Can a body have energy without momentum? If yes, then explain how they are related with each other?
Solution:
Yes, when p = 0,
Then, K = \(\frac{p^{2}}{2 m}\) = 0
But E = K + U = U (potential energy), which may or may not be zero.

Question 5.
Two bodies A and B having masses m_A and m_B respectively have equal kinetic energies. If p_A and p_B are their respective momenta, then prove that the ratio of momenta is equal to the square root of ratio of respective masses.
Solution:
Let υ_A and υ_B be the velocities of A and B respectively.
Since their kinetic energies are equal,

Question 6.
Two ball bearings of mass m each, moving in opposite directions with equal speed υ, collide head on with each other. Predict the outcome of the collision, assuming it to be perfectly elastic.
Solution:
Here, m₁ = m₂ = m
u₁ = υ,u₂ = -υ
Velocities of two balls after perfectly elastic collision between them are

After collision, the two ball bearings will move with same speeds, but their direction of motion will be reversed.

Question 7.
An engine is attached to a wagon through a shock absorber of length 1.5 m. The system with a total mass of 50,000 kg is moving with a speed of 36 kmh^-1 when the brakes are applied to bring it to rest. In the process of the system being brought to rest, the spring of the shock absorber gets compressed by 1.0 m. If 90% of
energy of the wagon is lost due to friction, calculate the spring constant. (NCERT Exemplar)
Solution:
Given, mass of the system (m) = 50,000 kg
Speed of the system (υ) = 36 km/h
= \(\frac{36 \times 1000}{60 \times 60}\) = 10 m/s
Compression of the spring (x) = 1.0 m
KE of the system = \(\frac{1}{2}\) mυ² = \(\frac{1}{2}\) × 50000 × (10)²
= 25000 × 100 J = 2.5 × 10⁶J
Since, 90% of KE qf the system is lost due to friction, therefore, energy transferred to shock absorber, is given by
ΔE = \(\frac{1}{2}\)kx² = 10% of total KE of the system
= \(\frac{10}{100}\) × 2.5 × 10⁶ J or k = \(\frac{2 \times 2.5 \times 10^{6}}{10 \times(1)^{2}}\)
= 5.0 × 10⁶ N/m

Question 8.
An adult weighting 600 N raises the centre of gravity of his body by 0.25 m while taking each step of 1 m length in jogging. If he jogs for 6 km, calculate the energy utilised by him is jogging assuming that there is no energy loss due to friction of ground and air. Assuming that the body of the adult is capable of converting 10% of energy intake in the form of food, calculate the energy equivalents of food that would be required to compensate energy utilised for jogging. (NCERT Exemplar)
Solution:
Given, weight of the adult (w) = mg = 600 N
Height of each step = h = 0.25m
Length of each step = 1 m
Total distance travelled = 6 km = 6000 m
∴ Total number of steps = \(\frac{6000}{1}\) = 6000
Total energy utilised in jogging = n × mgh
= 6000 × 600 × 0.25J = 9 × 10⁵ J
Since, 10% of intake energy is utilised in jogging.
∴ Total intake energy = 10 × 9 × 10⁵J = 9 × 10⁶J

Long answer type questions

Question 1.
A body of mass 0.3 kg is taken up an inclined plane length 10 m and height 5 m, and then allowed to slide down the bottom again. The coefficient of friction between the body and the plane is 0.15. What is the
(i) work done by gravitational force over the round trip?
(ii) work done by the applied force over the upward journey?
(iii) work done by the frictional force over the round trip?
(iv) kinetic energy of the body at the end of trip? (Take g = 10 ms^-2)

Solution:
Upward journey

Let us calculate work done by different forces over upward joume Work done by gravitational force
Wi = (mg sinθ)s cos 180°
W ₁= 0.3 × 10 sin30° × 10 (-1)
W₁ =-15J
Work donp by force of friction
W₂ = (μ mg cosθ)s cos180°
W₂ = 0.15 × 0.3 × 10 cos30° × 10 (-1)
W₂ =-3.879 J
Work done by external force
W₃ = F_ext × s × cos0°
W₃ = [mg sinθ + μ mg cosθ] × 10 × 1
W₃ = 18.897 J

Downward journey

mg sin30°> μ mg cos30°
Work done by the gravitational force
W₄ = mg sin 30° × scos0°
W₄ = 0.3 × 10 × \(\frac{1}{2}\) × 10 = +15J
Work done by the frictional force
W₅ = μmg cos30° × s cos180°
= 0.15 × 0.3 × \(\frac{10 \sqrt{3}}{2}\) × 10 × (-1) = – 3.897 J
(i) Work done by gravitational force over the round trip
= W₁ + W₄ = 0J
(ii) Work done by applied force over upward journey
= W₃ = 18.897J
(iii) Work done by frictional force over the round trip
W₂ + W₅ = – 3.897 + (-3.897) = – 7.794 J
(iv) Kinetic energy of the body at the end of the trip
W₄ + W₅ = 11.103 J

Question 2.
Prove that when a particle suffers an oblique elastic collision with another particle of equal mass anil initially at rest, the two particles would move in mutually perpendicular directions after collisions.
Solution:
Let a particle A of mass m and having velocity u collides with particle B of equal mass but at rest. Let the collision be oblique elastic collision and after collision the particles A and B move with velocities υ₁ and υ₂ respectively inclined at an angle 0 from each other.

Applying principle of conservation of linear momentum, we get
mu = mυ₁ +mυ₂ or u = υ₁+ υ₂
or u² = (υ₁ + υ₂) – (υ₁ + υ₂)
= υ₁² + υ₂² + 2υ₁υ₂cos0 ………….. (i)
Again as total KE before collision = Total KE after collision
∴ \(\frac{1}{2}\) mu² = \(\frac{1}{2}\)mυ₁² + \(\frac{1}{2}\)mυ₂²
⇒ u² = υ₁² + υ₂² ……………. (ii)
Comparing eqs. (i) and (ii), we get 2υ₁υ₂ cosθ = 0
As in an oblique collision both υ₁ and υ₂ are finite, hence cos0 = 0
⇒ θ = cos^-1(0) = \(\frac{\pi}{2}\)
Thus, particles A and B are moving in mutually perpendicular directions after the collision.

Punjab State Board PSEB 11th Class Biology Important Questions Chapter 20 Locomotion and Movement Important Questions and Answers.

PSEB 11th Class Biology Important Questions Chapter 20 Locomotion and Movement

Very short answer type questions

Question 1.
Give the name of the cells/tissues in human body which:
(i) exhibit amoeboid movement.
(ii) exhibit ciliary movement. [NCERT Exemplar]
Answer:
(i) Macrophages,
(ii) Ciliated, epithelium of trachea.

Question 2.
Which property of muscles is used effectively in muscular movement?
Answer:
Contractile property of muscles.

Question 3.
Give the name of the oxygen-carrying pigment present in skeletal muscle.
Answer:
Myoglobin or muscle hemoglobin.

Question 4.
Label the different components of actin filament in the diagram given below: [NCERT Exemplar]
Answer:

Question 5.
What causes muscle fatigue?
Answer:
Accumulation of lactic acid.

Question 6.
The three tiny bones present in middle ear are called ear ossicles. Write them in correct sequence beginning from eardrum. [NCERT Exemplar]
Answer:
Malleus, incus and stapes.

Question 7.
What is the difference between the matrix of bone and cartilage? [NCERT Exemplar]
Answer:
The matrix of bone is hard due to calcium salts, whereas, the cartilage has slightly pliable matrix due to chondroitin salts.

Question 8.
How many total bones are there in human body? Name the largest and strongest bone.
Answer:
Human body contains 206 bones. Femur is the largest and strongest bone of human body.

Question 9.
Give the name of the cavity in the girdle to which head of femur articulates.
Answer:
Acetabulum.

Question 10.
Give the name of the funny bone.
Answer:
Olecranon process on top of the ulna is called the funny bone.

Question 11.
Give the location of ball and socket joint in a human body. [NCERT Exemplar]
Answer:
Shoulder joint (between pectoral girdle and head of humerus).

Question 12.
What substance is responsible for lubricating the freely movable joint at the shoulder? ,
Answer:
Synovial fluid.

Short answer type questions

Question 1.
Explain anaerobic breakdown of glycogen in muscles and its effect.
Answer:
Anaerobic Breakdown of Glycogen: The reaction time of the fibres can vary in different muscles. Repeated activation of the muscles can lead to the accumulation of lactic acid due to anaerobic breakdown of glycogen in them, causing fatigue.

Question 2.
Describe the structure of the human skull.
Answer:

• The skull is composed of two sets of bones-cranial and facial, that totals to 22 bones.
• Cranial bones are 8 in number. They form the hard protective outer covering, cranium for the brain.
• The facial region is made up of 14 skeletal elements which form thefront part of the skull.
• A single U-shaped bone called hyoid is present at the base of the buccal cavity and it is also included in the skull.
• Each middle ear contains three tiny bones-Malleus, Incus and Stapes, collectively called Ear Ossicles.
• The skull region articulates with the superior region of the vertebral column with the help of two occipital condyles (dicondylic skull).

Question 3.
Explain the structure of the vertebral column of human.
Answer:
Vertebral Column: Our vertebral column is formed by 26 serially arranged units called vertebrae and is dorsally placed. It extends from the base of the skull and constitutes the main framework of the trunk. Each vertebra has a central hollow portion (neural canal) through which the spinal cord passes.
First vertebra is the atlas and it articulates with the occipital condyles.

The vertebral column is differentiated into following regions starting from the skull:

1. cervical (7),
2. thoracic (12),
3. lumbar (5),
4. sacral (1-fused) and
5. coccygeal (1-fused) regions

The number of cervical vertebrae are seven in almost all mammals including human beings. The vertebral column protects the spinal cord, supports the head and serves as the point of attachment for the ribs and musculature of the back. Sternum is a flat bone on the ventral midline of thorax.

Question 4.
Describe the structure of the rib cage of human.
Answer:
Rib Cage: There are 12 pairs of ribs. Each rib is a thin flat bone connected dorsally to the vertebral column and ventrally to the sternum. It has two articulation surfaces on its dorsal end and is hence called bicephalic.
First seven pairs of ribs are called true ribs. Dorsally, they are attached to the thoracic vertebrae and ventrally connected to the sternum with the help of hyaline cartilage.

The 8th, 9th and 10th pairs of ribs do not articulate directly with the sternum but join the seventh rib with the help of hyaline cartilage. These are called verte brochondral (false) ribs. Last 2 pairs (11th and 12th) of ribs are not connected ventrally and are, therefore, called floating ribs. Thoracic vertebrae, ribs and sternum together form the rib cage.

Question 5.
Give a description of the appendicular skeleton in human.
Answer:
Appendicular Skeleton: The bones of the limbs alongwith their girdles constitute the appendicular skeleton. Each limb is made of 30 bones.

Bones of Limbs
Fore’ Limb	Hind Limb
Humerus,	Femur,
Radius,	Tibia,
Ulna	Fibula,
Carpals (8)	Tarsals (7)
Metacarpals (5)	Metatarsals (5)
Phalanges (14)	Phalanges (14)
	Patella

Question 6.
Write a short note on disorders of muscular and skeletal systems.
Answer:
Disorders of Muscular System

• Myasthenia gravis: It is an auto-immune disorder, affecting the neuromuscular junction leading to progressive weakening and paralysis of skeletal muscles.
• Muscular dystrophy: It is a genetic disorder resulting in progressive degeneration of skeletal muscles.
• Tetany: It refers to the continued state of contraction or wild contractions of muscles due to low Ca++ in body fluids.

Disorders of Skeletal System:

• Arthritis: Inflammation of joints.
• Osteoporosis: Age-related disorder characterized by decreased bone mass and increased chances of fractures. Decreased levels of estrogen is a common cause.
• Gout: Inflammation of joints due to accumulation of uric acid crystals.

Long answer type questions

Question 1.
Give answer for the following:
(i) Female pelvis is larger and has a broader front than male pelvis. Why?
(ii) Name the different curves of vertebral column.
(iii) What is a sesamoid bone? Name it.
(iv) Which bones have become modified to form ear ossicles?
Answer:
(i) Female pelvis is larger and has a broader front. This is an adaptation for childbirth.
(ii) Vertebral column forms four curves, i.e., cervical, thoracic, lumbar, and sacral located in the neck, thorax, abdomen, and pelvis respectively.
(iii) A bone embedded within a tendon is called a sesamoid bone, e.g., Patella which covers the knee ventrally.
(iv) Articular bone of lower jaw modifies to be malleus. Quadrate bone of upper jaw becomes incus and hyomandibular gets modified to become stapes.

Punjab State Board PSEB 9th Class Science Important Questions Chapter 15 Improvement in Food Resources Important Questions and Answers.

PSEB 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Long Answer Type Questions:

Question 1.
How to meet the needs of ever-increasing population of our country? List two main steps. Give in a tabulated form some of the crops grown in India.
Answer:
The present population of 1.21 billion people will reach about 1.343 billion people by the year 2020. This population will need about 241 million tonnes of grain production per year. Following two measures will help up to meet the demand.
1. Increase food production of both plants and animals.
2. Sustainable agriculture where by we should minimize using chemicals as fertilizers and insecticides. These can be replaced by biological resources.
Some crop plants grown in India:

Type	Some Examples
Cereals or grain crops Fibre crops Pulses Oil seeds Fodder crops Root crops Tuber crops Sugar crops Plantation crops Products from animals	Rice, Wheat, Barley, Ragi, Maize, Jowar, Bajra. Jute, Cotton, Hemp, Coir. Grams, Peas, Beans, Masoor, Mung. Mustard, Groundnut, Sunflower, Coconut, Taramira. Barseem, Oat, Sudan grass. Sweet potato, Carrot, Radish, Beet. Potato, Tapioca. Sugarcane, Beetroot. Coffee, Tea, Rubber, Coconut. Fish, Egg, Milk, and Meat.

Question 2.
How is manure prepared?
Answer:
Method of preparing compost manure:
Manure is a natural organic substance obtained by decomposition of animal wastes and plant residue through the action of microbes. It is of three types
1. Farmyard manure
2. Compost
3. Green manure.

1. A trench having the desirable size of 4-5 m long, 1.5 to 1.8m broad with a depth of 1.0 to 1.8 m is made.
2. A layer of about 30 cm in thickness containing well mixed refuse is spread in the trench.
3. This layer is sprayed with water containing slurry of cow dung.
4. Another layer of mixed refuse in trench up to the height of 45-60 cm (height of fish layer included).
5. Top of these two layers is covered by thin layer ot earth.
6. After a gap of about three months, material is taken out of trench, moistened with water and covered with earth.
7. Compost is ready for use after gap of 1-2 months.

Question 3.
Differentiate between manures and fertilizers.
Answer:
Differences between manures and fertilizers:

Manure	Fertilizers
1. Manures are partially decayed wastes and animal residues by microbes. 2. Organic substances. 3. Voluminous, bulky, difficult to store and transport. 4. Not very rich in minerals like N, P and K. 5. Contain all nutrients, although in small amount. 6. Slow absorption, being less soluble in water. 7. Plenty of humus is added to soil and improves the texture of soil.	1. Fertilizer is a salt or organic compound containing essential plant nutrients. 2. Inorganic salts or organic compounds. 3. Compact, can be easily stored and transported. 4. Rich m minerals like N, P, K. 5. Specific. Every fertilizer contains one or more nutrients. 6. Rapid absorption due to easy solubility in water. 7. Humus is not added to soil.

Question 4.
What is mixed cropping? Discuss the advantages of mixed cropping.
Answer:
Mixed cropping is growing of two or more crops simultaneously on the same piece of land.
Objectives of mixed cropping:

1. To minimise risk and insure against crop failure.
2. To reduce cultivation expenses.
3. To provide balanced nutrition to farmer and his family.

Advantages:

1. It acts as insurance against possible total crop failure in poor rainfall areas.
2. It saves time and labour of the farmer.
3. It provides different types of food materials.
4. Thus, farmer and his family can get balanced nutrition.

Some of the prominent mixed cropping practices:

1. Maize + urdbean
2. Cotton + mungbean
3. Groundnut + sunflower
4. Wheat + chickpea
5. Wheat + mustard

Question 5.
Define livestock. Classify the cattle on the basis of their utility and give one example of each.
Answer:
Livestock includes domesticated animals like cows, buffaloes, sheep, goats, pigs, horses, etc. Cattle and buffaloes are most important livestock animals. These are used in agricultural operations and transportation though also provide milk, meat, hides (for leather goods), dung manure and fuel (in biogas plants). There are 30 breeds of cows and 10 breeds of buffaloes in India.

1. Milch breeds: Milk-yielding varieties of cows but their males are not useful as working animals e.g. Gir, Sahiwal, etc.
2. Draught breeds: Their males are used as beast of burden and good work animals but their cows are poor milk-yielding e.g. Malvi, Hallikar, etc.
3. Dual-purpose breeds: Their cows are good milk-yielding while their males are good Work animals and help in agricultural operations e.g. Haryana, Tharparkar, etc.

Indian breeds of cows and buffaloes

1. Cows:

(a) Milch breeds:

1. Gir
2. Sahiwal
3. Red Sindhi

(b) Draught breeds:

1. Malvi
2. Nageri
3. Hallikar
4. Kangayam

2. Buffaloes:

1. Murrah
2. Mehsana
3. Surti
4. Nili Ravi

Buffalo milk is richer in fat, tocopherol, proteins, calcium, phosphorus and contains low sodium, potassium, cholesterol. Buffalo milk is ideal for making milk products like khoa, rabri, dahi and ghee.

Question 6.
What is the need of proper shelter to cattle? List the characters of a good animal shelter.
Answer:
Shelter to Cattle: A good animal shelter not only increases the milk-production but also improves the health of animals.
A good animal shelter has following characteristics:
(a) It should provide protection to the animals from unfavourable environmental factors and predators.
(b) It should be clear, dry, airy, spacious and well ventilated (proper sunlight).
(c) It should have arrangement for the hygienic disposal of animal excreta.
(d) It should have arrangement for clear drinking water for animals.
(e) It should have hygienic conditions to protect the animals from various diseases.

Question 7.
What are additive feeds for cattle? How can one detect that an animal is sick?
Answer:
The cattle feed consists of two main components i.e. roughage and concentrate. However high yielding cows require more feed. Dairy animals require minerals and antibiotics and they are commonly called additive. Additive feed performs following functions:
1. Protects the animals from diseases.
2. They enhance milk yield.
3. Promote growth of the animals.

Symptoms of detection of sick animals:

1. Laziness, tired and prefer to stay alone.
2. Stop feeding or take a very little feed.
3. Walks very slowly.
4. Fall in milk or egg production.
5. Waste passed out is dilute.
6. Rise in temperature with shivering and sneezing.
7. Secretion of excess of saliva.

Question 8.
Write a short note on variety improvement of poultry farming.
Answer:
Variety Improvement of Poultry Farming. It involves cross-breeding of indigenous varieties with exotic breeds. The improved varieties are developed for the following desirable traits:

1. Quality and quantity of chicks.
2. Dwarf broiler parent for commercial chick production.
3. Summer adaptation capacity/tolerance to high temperature.
4. Low maintenance requirements.
5. Reduction in the size of the egg-laying bird with ability to utilise more fibrous cheaper diets formulated using agricultural by-products.

Question 9.
What steps should be taken to improve production of food from animal sources in our country?
Answer:

1. Introduction of high-yielding varieties.
2. The output of research work carried out at various centres such as NDRI Kamal, Central Institute of Fresh Water Aquaculture (CIFA) Bhubaneshwar, should be made available to the public for their projects.
3. Protection of animals from diseases.
4. Providing proper shelter to animals.

Question 10.
What are the practices adopted to improve crop production?
Answer:
The practices adopted to improve crop production are as follows:

1. Fertilizers: These are the chemical compounds which are added to the soil to increase the fertility. They make up for the deficiency of the required nutrients and help in increasing the crop production.
2. Selective Breeding: Disease-resistant seeds are produced by selective breeding. Regular use of high yield variety results in better crop production.
3. Weed Control: The unwanted plants or weeds are controlled by using certain chemicals called weedicides.
4. Control of Plant Diseases: Crops should be protected from insects, fungi, animals and other diseases.

It is very useful for increasing crop production. Insects are very harmful to crops. So insecticides should be used to kill insects.

Question 11.
Discuss the preventive and control measures to check losses of grain during storage.
Answer:
Preventive and control measures
These are used before storing and the grains are stored for future use.
1. This can be done by proper drying of the produce in sun followed by drying in shade.

2. Maintenance of Hygiene. Godowns and stores should be properly cleaned. All sort of dust, dirt, rubbish, webbing or refuse of the previous grain should be swept away. Cracks and holes in the wall, floor or ceiling should be sealed. If old gunny bags are being used, clean them properly, turn inside out and expose to sun or fumigate. Earthen pots should also be cleaned and properly exposed to sun before using them for grain storage.

3. Fumigation. Chemicals which can exist in gaseous state in sufficient concentration to be lethal against the pest is known as fumigants. Aluminium phosphide tablets commonly known as black poison (3 g each) can be used at the rate of 2 tablets per ton grains.

4. Plant Products. The practice of adding small quantity of vegetable oil or mineral oil to grains or legumes to protect them from insect pests and mixing of neem kernel (seed) powder, crushed dried fruit of black pepper or cloves is also effective in controlling insects.

Question 12.
Write a brief account of procedure of Inland Fishery.
Answer:
Inland fishery involves the rearing of fishes in the specially designed breeding ponds near the rivers or other fresh-water natural sources.

It involves the following steps :
Breeding of good male and female culturable fishes in the breeding ponds either by natural breeding or induced breeding. In induced breeding, the male and female breeder fishes are injected with pituitary extract containing FSH and LH hormones (called Hypophysiation) which induce them to spawn within 24 hours. Fertilization occurs in the water so is external.

1. The fish seed (of fertilized eggs) are collected with the help of shooting net or benchi jal.
2. The fertilized eggs are kept in hatching pits, called hapas, and young ones are called hatchlings.
3. Hatchlings are allowed to grow in hapas for about 3-14 days to form fry.
4. Fries are allowed to feed and grow in the nursery ponds to form the fingerlings.
5. Fingerlings are allowed to grow in rearing ponds for about 3 months.
6. Fingerlings are allowed to attain full size in the still larger stocking ponds.
7. Harvesting or fishing involves the capturing of fully-grown fishes.

Question 13.
Briefly explain marine fisheries.
Answer:
Marine Fisheries: India’s marine fishery resources include 7500 km of coastline and the. deep seas beyond it. Popular marine fish varieties include pomphret, mackerel, tuna, sardines and Bombay duck. Marine fish are caught using many kinds of fishing nets from fishing boats.

The modern technologies for catching more fish include echosounders and use of satellite. Some marine fish of high economic value are also farmed in sea water. This includes finned fishes like mullets, bhetki and pearl spots, shellfish such as prawns, mussels and oysters.

Short Answer Type Questions:

Question 1.
How human beings depend upon plants and animals for nutrition?
Answer:

1. Man is omnivorous and takes plants and animals as food.
2. The plant sources of food are cereals (wheat, rice, maize), pulses, millets, fruits and vegetables.
3. The animal sources of food are meat, milk, fish, egg, milk products and liver oil.

Question 2.
Define nutrients. Give examples of macronutrients and micronutrients.
Answer:

• Nutrients: The elements needed for growth^of plants and animals are called nutrients. ^
• Macronutrients: The mineral elements needed by the plants in large amounts (more than 1 ppm) are called macronutrients.
• Examples: Nitrogen, Phosphorus, Carbon, Hydrogen, Oxygen, Potassium, Calcium, Sulphur.
• Micronutrients: The mineral elements needed by plants in small amount (or traces) are called micronutrients.
• Examples: Iron, Manganese, Copper, Zinc, Boron, Molybdenum, Chlorine.

Question 3.
What are Fertilizers?
Answer:
Fertilizers:

1. Fertilizers are commercially produced plant nutrients by using different chemicals.
2. Fertilizers supply Nitrogen, Phosphorus, Potassium (NPK), basically this is used for good vegetative growth (leaves, branches and flowers), giving rise to healthy plants.
3. Fertilizers are one of the major components for obtaining higher yields specially in high cost farming practices.
4. They are easy to store and transport.

Question 4.
What is the need of water to plants? Explain different types of irrigation system.
Answer:
Periodic Irrigation of Crops:
Crops are irrigated periodically due to the following reasons:
1. Plants absorb water from the soil by roots. The amount of water in the soil is not constant throughout the year. It is constantly lost by evaporation and percolation to lower depths of the ground.
2. The water is also lost by the aerial parts of the plants by transpiration.

Irrigation Systems:

1. Canal System: Here reservoirs or rivers supply water to canals. Canal is divided into sub-canals and distributaries. Further field channels may be made. Rotation system is usually followed to provide irrigation water to all the fields at the time of scarcity of water supply.

2. Tanks: Here water is stored, which is available due to run off. Small dams can be made at the base of higher elevation of catchment region. Outflow of water in tanks is kept in control. If it is not done, it may lead to:
(a) Uneven distribution of water.
(b) Shortage of water at tail end and excessive use at the top. This leads to uneven distribution of water.

3. Wells. Wells are made to exploit groundwater. They are of two types i.e. dug wells and tubewells. In dug well, water accumulates due to available groundwater table. From deeper areas of earth, tube wells can tap water. Here water is lifted by diesel or electricity-operated water pumps. Continuous supply can be ensured by this system.

4. River valley system: In Western Ghats of Karnataka and Kerala, many steep and narrow riverine valley are present. Rainfall in these areas is present only for shorter period i.e. 3-4 months. Extra water shifts in river in these months. In Rabi season, no rainfall is there. On slopes, plants like coconuts, arecanuts, coffee, rubber and tapioca are grown. These all are perennial plants. In basement areas, single crop like rice is grown.

5. River lift system: In this system, water is directly drawn from rivers for irrigation purpose. This all is done in areas nearby the rivers and canal flow is insufficient.

Question 5.
How do insect pests attack plants? Give examples.
Answer:
1. Cut root, stem and leaf – weevil attack wheat crops.
2. Suck cell sap from various parts of plants – aphids feed on mustard plants.
3. Bore into stem and fruits – top borer and shoot borer, larvae and caterpillars which bore into stem and fruits. The plant pathogens are transmitted to plants through water, soil, air and seeds.

Diseases caused by these pathogens include

1. Blast in paddy (rice).
2. Rust in wheat.
3. Stem rot in pigeon pea (mung).
4. Wilt in chickpea (gram).

Question 6.
What are weeds? What are the two types of weeds?
Answer:
Weeds: They are the small-sized unwanted plants which grow along with a cultivated crop in a field. Weeds are economically very important as they can severely reduce crop yields by competing for light, water and nutrients.

Based on the morphology of plants, weeds are classified into two types:

1. Narrow-leaf weeds, e.g., Wild sorghum, Wild oat.
2. Broadleaf weeds, e.g., Amaranthus viridis, Trianthima.

Question 7.
Discuss the three major methods of weed control.
Answer:
Methods of weed control:

1. Mechanical Methods: Uprooting, weeding with khurpi, hand hoeing, interculture, ploughing, burning and flooding.
2. Culture Methods: Proper seedbed preparation, timely sowing of crops, intercropping and crop rotation.
3. Chemical Methods: Spraying of some chemicals as herbicides or weedicides, is also done in case of heavy infestation.
4. Biological Control: Use of insects or some organisms which consume and destroy the weed plants.
5. Examples: Prickly-pear cactus (Opuntia) is controlled by cochineal insect and aquatic weeds are controlled by fish grass carp.

Question 8.
Briefly explain three types of manures on the basis of biological material used.
Answer:
Types of manure:

1. Farm Yard Manure (FYM): Livestock farm waste i.e. cattle excreta (cow dung and urine) is stored in a pit for decomposition. After 1-2 months this is used as FYM in farming practices.
2. Compost: The process in which waste material like vegetable waste, animals refuse, domestic waste, sewage waste, straw, eradicated weeds, etc. is decomposed in pits is known as composting. The compost is rich in organic matter and nutrients.
3. Green Manure: In cultivation field prior to the sowing of the seeds, some crops like sun hemp, guar, etc. are grown. After some time these plants are mulched by ploughing. These green plants thus turn into green manures which help in enriching the soil by N and P.

Question 9.
What are weeds? How does it affect the yield of crop?
Answer:
Weeds. They are unwanted plants which grow on their own along with crop plants. They are a harm to the crops.
Weeds damage the crops: They compete with the crop for nutrients and water in the field. They occupy space meant for crop plants. This leads to poor yield and quality of produce.

Methods of weed control:

1. Removal by hands: The weed can be uprooted and removed by hand.
2. Removal by instruments: The weeds can be removed by using a trowel (khurpa).
3. Removal by using chemicals: Weeds can also be destroyed by spraying special chemicals called weedicides.
4. These chemicals easily kill the broad-leaved weeds without affecting the crop.

Question 10.
List any five chemical fertilizers rich in nitrogen.
Answer:
Chemical fertilizers rich in nitrogen:

1. Urea
2. Ammonium sulphate
3. Ammonium nitrate
4. Sodium nitrate
5. Calcium ammonium nitrate

Out of these urea is organic while others are inorganic.

Question 11.
How has the excess use of pesticides and fertilizers proved harmful?
Answer:
If pesticides are not used with care, they lead to the disappearance of not only the undesirable insects but even the helpful ones. There is a danger that the pests may become resistant to pesticides and make them ineffective. Indiscriminate use of pesticides and fertilizers can lead to environmental degradation. Excess fertilizers is washed away into surrounding water bodies.

The resulting high concentration of nitrates and phosphates in ground and surface waters make them toxic and unfit for human and animal consumption.

Question 12.
What are the two main crop seasons? List the crops of respective season.
Answer:
1. Kharif Season. (June to October)
Crops grown: Paddy, Soyabean, Arhar, Maize, Cotton, Urad and Moong.
2. Rabi season. (November to April)
Crops grown: Wheat, Gram, Peas, Mustard, Linseed.

Question 13.
Write a note on organic farming.
Answer:
Organic Farming: It is a farming system with minimum or no use of chemicals as fertilisers, herbicides, pesticides, etc. and with maximum input of organic manures, recycled farm wastes, use of bio-agents such as culture of blue-green algae in preparation of biofertilisers, neem leaves or turmeric specifically in grain storage as bio-pesticides, with healthy cropping systems.

Question 14.
Define crop rotation. What is the role of leguminous plants in it?
Answer:
Crop Rotation: The process in which different types of crops are grown alternately in the same field is called crop rotation.

Leguminous crops save the nitrogenous fertilizers because such plants grown during crop rotation fix nitrogen from the air and enrich the soil with nitrate and nitrites. These nitrogen-containing compounds are used by plants.

Question 15.
What is the role of soil in better yield?
Answer:
Soil provides all the nutrients such as nitrogen, phosphorus, potassium etc. to the crop plants. Soil also serves as a source of water. If soil is deficient in one or more nutrients, the yield of crop will reduce.

Question 16.
What is mixed farming? Define it with suitable examples.
Answer:
Mixed farming can be defined as a system of farming on a particular farm to sustain and satisfy the essential needs of the farmers.
Examples: In mixed farming, crop production is combined with the rearing of livestock, poultry, fish and bees etc.

Question 17.
How can mixed farming sustain agricultural production? Answer With suitable examples.
Answer:
When earnings from one enterprise on small farms is not sufficient to sustain the family, farmer considers about the different possible combinations of enterprises.

In mixed farming from livestock, farmyard manure is made available to be used again in agricultural farms. With exact combination of mixed farming, a better money/ income is available. It provides the farmer work throughout the year. It provides the farmer with all the food needs of the family.

Question 18.
What are the uses of mixed farming?
Answer:

1.  From livestock, farmyard manure is available to be used again in agricultural farms.
2. Number of animals can be increased as per food available to them (as per crop availability) to provide milk and milk products.
3. By this method, straw, husks and chaffs of grains, refuse of household kitchen, shed grains in the field are converted into human food through the agency of cattle, sheep, poultry, pigs etc. as per choice of farmer.
4. With exact combination in mixed farming, a better income is available.
5. It provides work to all the members of a family throughout the year. Thus it provides subsidiary occupation to all the members of a household, without the need of employing special labour.

Question 19.
Write a note on crop protection management.
Answer:
Crop Protection Management: In fields, crops have to be protected from weeds, insect-pests and disease-causing organisms like fungi.

All these cause damage to crop plants so much so that most of the crop is lost. Thus, crops can be protected by the following methods:

1. Use of pesticides.
2. Use of resistant varieties.
3. Crop rotation and cropping system.
4. Summer ploughing.

Question 20.
Discuss the requirements for a good crop rotation.
Answer:
Requirements for a good Crop Rotation

1. The area of each crop should be nearly the same year after year.
2. The rotation should provide roughage and pastures for domestic animals.
3. Most profitable cash crops with wisdom should be selected for rotation.
4. Rotated crops can be cared for.
5. The rotation should include one tilled crop for elimination of weeds.
6. Organic matter should increase in soil due to rotation and feeding system. By growth of leguminous plant in rotation, nitrogen contents of soil increase.

Question 21.
List group of plants for crop rotation on the basis of one year, two year and three-year rotation.
Answer:

Duration	Rotation of Crops
1. One-year rotation 2. Two-year rotation 3. Three-year rotation	1. Maize-mustard 2. Rice-wheat1. Maize-mustard-sugarcane – fenugreek (Methi) 2. Maize-potato-sugarcane-peas1. Rice-wheat-moong-mustard – sugarcane-berseem 2. Cotton-sugarcane-peas-maize-wheat

Question 22.
List five groups of plants according to their soil needs for crop rotation.
Answer:
For crop rotation, plants are put into five groups according to their soil:

Roots	Legumes	Brassica	Others	Permanent
Potatoes Carrots Beet root	Peas Runner beans Broad beans French beans	Cabbages Cauliflowers Broccoli Turnips Radishes	Lettuces Onions Cucumber Spinach tomatoes	Asparagus Herbs Fruits

Question 23.
Describe biological method of pest control.
Answer:
Biological pest control methods
They can be further classified into three categories:
(a) Breeding disease resistant varieties. It is the development of varieties having resistance to pathogenic infection.
(b) Hyperparasitism. It involves the control of one pathogenic organism with the help of another organism, which parasitizes the pathogen.
(c) Trap crops. Many plants secrete some substances after infestation by some pathogens. These substances are toxic to pathogens. Such hosts are called trap or antagonistic plants.

Question 24.
What is inter-cropping? How does it differ from mixed cropping?
Answer:
Inter-cropping is the growing of two or more crops simultaneously in the same field in definite rows.
Differences between mixed cropping and inter-cropping

Characters	Mixed cropping	Intercropping
1. Aim 2. Pattern 3. Mixing of seeds 4. Application of fertilizers 5. Harvesting and threshing 6. Application of pesticides	To reduce the chances of crop failure. No definite pattern of rows. Seeds are mixed up before sowing. Fertilizers cannot be applied easily to different crops. Cannot be done separately for crops. Spraying of pesticides for separate crops not possible.	To enhance the production of crops per unit area. Grown in definite pattern of rows like 1 : 1, 1 : 2, 1 : 3. Seeds are not mixed before sowing. Can be applied as per need of individual crops. Crops can be harvested and threshed separately. Can be done easily.

Question 25.
Make a table showing the nutritional values of animal products.
Answer:
Nutritional values of animal products:

Animal products	Percent (%) Nutrients
	Fat	Protein	Sugar	Minerals	Water
Milk (Cow)	3.60	4.00	4.50	0.70	87.20
Egg	12.00	13.00	*	1.00	74.00
Meat	3.60	21.10	*	1.10	74.20
Fish	2.50	19.00	*	1.30	77.20

Present in very little amount

Question 26.
What does cattle feed consist of?
Answer:
Feed of Cattle: Supply of uncontaminated and balanced diet containing sufficient quantities of required nutrients is an essential need of animal husbandry. Cattle feed is formed of two main components: roughage and concentrate.

Question 27.
Differentiate roughage and concentrate.
Answer:
Roughage contains large amount of fibre but have low nutrients and include hay, fodder, silage and legumes like barseem, lucerene and cowpea. It also includes Common fodder grasses like Napier grass, Guinea grass and Elephant grass.

The concentrate is a mixture of cereals like maize, oat, barley, jowar, broken grams, rice polish, cotton seeds, gram barn and oilseed cake etc. moistened in water. These are rich in proteins and other nutrients, highly palatable and easily digestible.

Question 28.
Discuss the average daily feed of a cow.
Answer:

1. Green fodder = 15 – 20 kg
2. Grain mixture = 4 – 5 kg
3. Water = 30 – 35 litres
4. Additive feeds like antibiotics, hormones, minerals, etc.

Additive feed increases the milk yield and also protects the animals from diseases.

Question 29.
Name four animals which provide us food.
Answer:

1. Cows (milk and meat)
2. Buffaloes (milk)
3. Goats (milk and meat)
4. Pigs (provide pork)
5. Poultry birds (eggs and meat)
6. fishes (meat)

Question 30.
Mention the names of animal products which are used as food.
Answer:
Milk, beef (cow’s meat), pork (pig meat), mutton (sheep and goat meat), eggs (poultry birds) and fish meat and by-products of fishery such as fish meal, fish-protein concentrate, oil etc.

Question 31.
Name any two Indian breeds of:
1. Cows
2. Buffaloes
Answer:
1. Cow-breeds : Sahiwal and Gir.
2. Buffalo breeds : Murrah and Mehsana.

Question 32.
Name two exotic breeds of cows.
Answer:
1. Jersey of USA.
2. Brown-Swiss of Switzerland.

Question 33.
Mention the improved crossbreeds of cows.
Answer:

1. Karan-Swiss
2. Karan-Fries
3. Frieswal

Question 34.
List some Indian breeds of cows and buffaloes.
Answer:
1. Cows:

(a) Milch breeds

1. Gir
2. Sahiwal
3. Red Sindhi

(b) Draught breeds

1. Malvi
2. Nageri
3. Flallikar
4. Kangayam

(c) General utility breeds

1. Haryana
2. Ongole
3. Tharparkar

2. Buffaloes:

1. Murrah
2. Bhadawari
3. Jaffarabadi
4. Surti
5. Nagpuri
6. Nili Ravi
7. Mehsana

Question 35.
Briefly explain contribution of Dr. V. Kurien.
Answer:
Contribution of Dr. V. Kurien:
Dr. V. Kurein born on 26th November 1921, is the founder Chairman of the National Dairy Development Board (NDDB), which designed and implemented the world’s largest dairy development programme – the “Operation Flood”. Dr. Kurien is called the architect of India’s modem dairy industry and the father of white revolution.

Question 36.
Write a note on foot and mouth disease of cattle.
Answer:
It is a viral disease of cows and buffaloes. It is characterized by excessive salivation and reddish granules on their feet and mouth. It can be prevented by vaccinating the cows and by keeping their shelter clean and hygienic.

Question 37.
Write two infectious diseases of each of cows, poultry and fishes.
Answer:

Animal	Diseases
1. Cows 2. Poultry 3. Fishes	Anthrax (bacterial) and Foot and mouth disease (viral). Ranikhet (viral) and Salmonellosis (bacterial). Viral Haemorrhagic Septicemia (VHS) and Infectious Pancreatic Necrosis (IPN).

Question 38.
Mention a few measures for prevention of diseases in the animals.
Answer:
Preventive measures of diseases in animal

1. Compulsory vaccination of animals.
2. Proper disposal of dead animals and animal wastes.
3. Hygienic handling of all animal products and by-products.
4. Periodical screening of animals for diseases.
5. Providing a clean, dry, airy and well-ventilated good animal shelter with hygienic conditions.

Question 39.
List the main reasons for low milk yield of cattle in India.
Answer:
Preventive measures of diseases in animals:

1. The poor quality of feed.
2. The shortage of feed.
3. Low milk-yielding varieties of cattle.

Question 40.
Discuss the importance of poultry as a source of food.
Answer:
Poultry birds supply eggs and meat both being good sources of food. Whole egg contains 36% yolk, 64% proteins and vitamins like A and D. Poultry meat contains proteins (like myosin, globulins, actinomyosin, etc.), fats, vitamins, minerals, etc.

Question 41.
Name some indigenous and exotic breeds of fowls.
Answer:
1. Indigenous breeds: e.g. Aseel, Ghagus, Basara, Brahma and Chittagong,
2. Exotic breeds:
(a) American breeds e.g. Plymouth rock and Rhode Island Red.
(b) English breeds e.g. Sussex and Australorp.
(c) Mediterranean breeds e.g. Minorcha and White Leghorn.

Question 42.
Make a list of diseases of poultry.
Answer:

1. Poultry Diseases: The poultry birds suffer from various diseases caused by:
2. Virus – Fowl pox
3. Bacteria – Tuberculosis, Cholera, Diarrhoea
4. Fungi – Aspergillosis
5. Parasites – Worms, mites, lice, etc.

They also suffer from nutritional deficiency diseases. The most common disease of poultry is ‘Bird Flu’. This disease is damaging the poultry on vast scale.

Question 43.
Write the names of two exotic breeds of poultry.
Answer:

1. American breeds like Plymouth rock and Rhode Island Red.
2. English breeds like Australorp and Sussex.
3. Mediterranean breeds like White Leghorn and Minorcha.

Question 44.
What are important points to remember in poultry farming?
Answer:
Important points in poultry farming:

1. Maintenance of temperature and hygienic conditions in housing.
2. Proper poultry feed.
3. Prevention and control of diseases and pests.
4. Isolation of diseased birds.
5. Proper vaccination of birds.
6. Spraying of disinfectants at regular intervals.

Question 45.
List various measures of prevention of poultry diseases.
Answer:
Prevention of Poultry Diseases. To prevent the poultry from disease the following measures should be taken:

1. The poultry birds should be kept in good spacious, airy and ventilated shelter.
2. The shelter should be cleaned properly and regularly. Quick and hygienic disposal of excreta should be ensured.
3. External parasites should be controlled by applying insecticide solution.
4. Disinfectant should be sprayed to kill mosquitoes and other external parasites,
5. Every animal should be vaccinated at regular interval to minimise it against common infections and diseases.

Question 46.
Name three common fresh water and three marine food fishes.
Answer:
Fresh-water food fishes

1. Labeo rohita – Rahu
2. Catla catla – Katla or Theila
3. Wallago attu – Mullee

Marine food fishes
1. Harpodon – Bombay duck
2. Hilsa – Hilsa
3. Sardinella – Salmon

• Total fish production in India – 7th position in the world.
• Marine fish production – 10th position in the world.
• Aquaculture production – 2nd in south East Asian countries.
• Fish industry contribution – Rs. 400 crores annually as foreign exchange.

Question 47.
Why is fish meat considered advantageous than meat of other animals?
Answer:

1. Fish meat contains more proteins (13 – 20%) but less fats.
2. It has good amount of vitamins A and D and is rich in iodine (for thyroxine formation).
3. It is more easily digestible than other proteins.
4. So, fish meat is considered next to the mother’s milk as baby food.

Question 48.
Why are the major carps considered best culturable fishes?
Answer:

1. These survive even at high temperature and at low oxygen.
2. These have fast growth rate.
3. These have easily digestible and nutritive flesh.

Question 49.
Give the economy of fishes.
Answer:

1. Fish meat is rich in proteins (13 – 20%), vitamins and iodine but has less fats.
2. Their liver oil is rich source of vitamins A and D.
3. Fish meal is very rich source of proteins (55 – 70%), so is good food for domesticated animals.
4. Fish wastes can be used as manure for coffee, tea and tobacco plants.
5. Fish skin of sharks is used to form hand bags, shoes, tobacco pouches, etc.

Question 50.
Give the functions of followings in fishery:
1. Hapas
2. Nursery ponds
3. Traps
Answer:
1. Hapas: These are hatching pits. These are formed of cloth and supported by bamboo sticks. In these pits, hatching occurs and young ones called hatchlings, emerge and grow in hapas to form fries.
2. Nursery ponds: These are small-sized ponds located near the hapas. In these ponds, fries feed upon the planktons and grow into young ones called fingerlings.
3. Traps: These are used to harvest the fish from the stocking ponds.

Question 51.
Differentiate capture fishery and culture fishery.
Answer:
Capture fishery involves the catching of fish by various methods while culture fishery involves the rearing of fish in artificial freshwater bodies.

Question 52.
List certain common diseases of fishes.
Answer:
1. Main infectious diseases of fishes are:
(a) Viral Haemorrhagic Septicemia (VHS) and
(b) Infectious Pancreatic Necrosis (IPN).

2. Water-pollution caused diseases of fishes are:
(a) Gill rot (blackening of gills).
(b) Fin rot (cutting down of fins).
(c) Dropsy (swollen belly).

3. Fish-ectoparasites – e.g. fish lice – Argulus.

Question 53.
List some measures to control fish diseases.
Answer:

1. Pollution of fish farm should be avoided.
2. Regular monitoring of the level of oxygen, carbon dioxide and pH of the water of fish farm.
3. Argulus – fish lice can be controlled by adding 2.5 ml/litre of malathion in the pond water.

Question 54.
List the steps of fish seed production by induced breeding techniques.
Answer:
Fish Seed Production by induced breeding technique consists of the following steps:

1. Use of inducing agents.
2. Selection of healthy brooders.
3. Administration of hormones by injection.
4. Releasing sets of brooders in breeding pool.
5. Spawning.
6. Collection of Eggs.
7. Hatching.
8. Post care of fish seeds in nursery and rearing ponds.
9. Transfer of fingerlings in stocking ponds.

Question 55.
What are the uses of honey?
Answer:
Utility of honey:

1. Honey has great importance for its medicinal value specially in disorders related to digestion, dysentery, vomiting and stomach and liver ailments.
2. It also helps in growth of our body as it contains iron and calcium.
3. It is also used as a source of sugar in confectionary items.

Question 56.
Write the components of honey.
Answer:
Honey is a dense sweet liquid, containing sugars 20-40%, moisture 60-80%, minerals 0.22-0.3%, vitamins 0.2-0.5%, enzymes and pollen.

Question 57.
What are the advantages of beekeeping?
Answer:
Advantages of beekeeping:

1. Beekeeping is undertaken on commercial basis as an enterprise. Besides honey, other products obtained from beekeeping include wax, royal jelly and bee venom.
2. Beekeeping requires low investments, therefore, farmers along with agriculture also do beekeeping as an additional income-generating activity.
3. It also helps in cross-pollination as pollens are transferred from one flower to another by bees while collection of nectar.

Question 58.
List the varieties of honeybee used for beekeeping.
Answer:
Honeybee varieties used for beekeeping
The indigenous varieties of bee used for commercial production of honey are Apis cema indica F., commonly known as Indian bee, Apis dorsata, the rock bee and A-florae little bee. An exotic variety from Italy has been domesticated in India to increase yield of honey. This variety is called Italian bee i.e. Apis mellifera, commonly known as Italian bee.

Question 59.
How are honeybees affected by different factors?
Answer:
1. Honeybees generally get bacterial and viral diseases.
2. Common pests of bees are wasps, wax moths and mites. Wasps are controlled manually, by exposing bees in bee hive to sun or by increasing temperature. King crow and greenbee eater prey upon bees. They can be scared away by some devices.

Question 60.
Explain compound fish culture. List the factors.
Answer:
Composite Fish Culture:
Combination of 6 species is used in the culture system. This combination is highly
advantageous because these fishes do not compete for food among them having different types of food habits. Another advantage is that food available in all the parts/zones of the pond is utilized due to their food habits.

The food habits of six species are catla, a surface feeder, rohu feed in middle zone of the pond i.e. column feeder and mrigal and common carp feed at the bottom, where as grass carp feed on aquatic weeds in the pond. Amongst them three are foreign or exotic, i.e. transplanted from China and 3 species are of Indian origin.

Factors: Important factors to be taken into consideration for fish culture include:

1. Topography or location of pond.
2. Water resources and quality.
3. Soil quality i.e. composition particle size as well as nutrients.

Question 61.
What are the uses of chemical fertilizers?
Answer:
Uses of fertilizers:

1. They cover up the mineral deficiency of soil due to excessive and repeated cropping in a field.
2. They are required in smaller bulk.
3. They are easy to transport.
4. They are quickly available to their plant food constituents.
5. The chemical fertilizers are available both in soil form as well as in solution form.
6. Chemical fertilizers are soluble in water, hence are easily absorbed by plants.

Very Short Answer Type Questions:

Question 1.
What are the major sources of food for us?
Answer:
Plants and animals.

Question 2.
List the nutrients supplied by food.
Answer:
Proteins, carbohydrates, fats, vitamins and minerals.

Question 3.
Coin the terms for extensive production of :
1. fish
2. milk
3. oil
Answer:
1. Blue revolution
2. White revolution
3. Yellow revolution

Question 4.
What term is used for extensive production of pulses?
Answer:
Golden revolution.

Question 5.
List cereal crops which provide carbohydrates for energy.
Answer:
Wheat, rice, maize, minor millets, and sorghum.

Question 6.
Name a few pulses which provide proteins.
Answer:
Gram, Pea, Black gram, green gram, pigeon pea, lentil.

Question 7.
List any six oil seed crops which provide fatty acids.
Answer:
Soyabean, groundnut, seasame, castor, mustard, linseed, niger and sunflower.

Question 8.
From where we get minerals and vitamins?
Answer:
Vegetables, spices and fruit crops.

Question 9.
Name three fodder crops.
Answer:
Berseem, oat and Sudan grass.

Question 10.
Name the crops grown in rabi season (November to April).
Answer:
Wheat, gram, peas, mustard, linseed.

Question 11.
Name kharif season crops.
Answer:
Paddy, soyabean, arhar, maize, cotton, urad and moong.

Question 12.
How many nutrients are required by plants?
Answer:
16 nutrients.

Question 13.
Name four macronutrients.
Answer:
Nitrogen, phosphorus, potassium and sulphur.

Question 14.
Name nine micronutrients.
Answer:
Iron, Manganese, Boron, Zinc, Copper, Molybdenum, Chlorine, Calcium, Magnesium.

Question 15.
What is farmyard manure (FYM)?
Answer:
FYM is the decomposed mixture of cattle excreta (dung) and urine along with litter (i.e. bedding material used in night under cattle) and left over organic matter such as roughage or fodder.

Question 16.
What is composting?
Answer:
Composting is a biological process in which both aerobic and anaerobic micro-organisms decompose the organic matter.

Question 17.
What is vermicomposting?
Answer:
The degradation of organic wastes through the consumption by the earthworms is called vermicomposting.

Question 18.
What are nitrogenous fertilizers? Give one example.
Answer:
These fertilizers supply the macronutrient nitrogen. Example. Urea.

Question 19.
What are complex fertilizers?
Answer:
When a fertilizer contains atleast two or more nutrients (N, P₂O₅ and K₂O), it is called complex fertilizer.

Question 20.
Define irrigation.
Answer:
The process of supplying water to crop plants by means of canals, reservoirs, wells etc. is called irrigation.

Question 21.
Name two potassium fertilizers.
Answer:
Potassium sulphate and potassium chloride.

Question 22.
When even excessive application of manure does not cause pollution?
Answer:
Manures are biodegradable, so they do not cause any damage.

Question 23.
List three important nitrogen-containing fertilizers.
Answer:
Urea, Ammonium Sulphate and Ammonium nitrate.

Question 24.
List three phosphorus-containing fertilizers.
Answer:
Ammonium phosphate, Ammonium hydrogen phosphate, Calcium Superphosphate.

Question 25.
Name two potassium-containing fertilizers.
Answer:
Potassium sulphate, Potassium chloride.

Question 26.
What is being traditionally used as manure in our country?
Answer:
Cow-dung.

Question 27.
Define herbicide.
Answer:
Herbicide is the chemical agent that destroys or inhibits plant growth ; used to destroy weeds in the cultivated patch of land.

Question 28.
List any three manures.
Answer:
Farmyard manure, composted manure and green manure.

Question 29.
What is monoculture?
Answer:
Growing the same crop in a field year after year is called monoculture.

Question 30.
A fanner grows gram crop between two cereal crops. Which agricultural practice is being followed by him?
Answer:
Crop rotation.

Question 31.
Suppose you are incharge of a grain store. How will you find out the presence of pests? Mention any two indications.
Answer:
1. By presence of living or dead insects.
2. By noticing white powdery materials on the bags or on the floor.

Question 32.
What are weeds?
Answer:
Unwanted plants growing alongwith main crops are called weeds.

Question 33.
Define fungicide.
Answer:
The pesticides that kill fungi are called fungicides.

Question 34.
What is the basic objective of mixed cropping?
Answer:
Minimize risk and insurance against crop failure due to abnormal weather conditions.

Question 35.
List any two prominent mixed cropping practices.
Answer:
1. Maize and urd bean.
2. Cotton and mung bean.

Question 36.
Which crops can be grown along with wheat during mixed cropping practices?
Answer:
Chickpea and mustard.

Question 37.
What is intercropping?
Answer:
Intercropping is growing of two or more crops simultaneously in the same field in definite rows.

Question 38.
Expand HYV.
Answer:
High Yielding Varieties.

Question 39.
Name any two subsidiary occupations which are part of mixed farming.
Answer:
Daily farming and poultry farming.

Question 40.
What is row inter-cropping?
Answer:
Growing two more crops at the same time with atleast one crop planted in rows.

Question 41.
What is inter-cropping?
Answer:
Growing two or more crops together in strips wide enough to permit separate crop production using machines but close enough for the crops to interact.

Question 42.
Name any two improved varieties of wheat.
Answer:
Sonara, PPW 154.

Question 43.
Name any two improved varieties of rice.
Answer:
Kasturi, PNR-591-18.

Question 44.
What is green revolution?
Answer:
The tremendous increase in food grains (especially wheat) during the last three decades due to the use of HYV, high dose of fertilizers and irrigation is known as green revolution.

Question 45.
Define selection.
Answer:
Selection is the sorting out of individual plants or groups of plants from mixed population.

Question 46.
Name the oldest method of crop improvement.
Answer:
Introduction.

Question 47.
Give one example of crop combination used in mixed cropping.
Answer:
Soyabean and pigeon pea.

Question 48.
Define plant breeding.
Answer:
Plant breeding means production of new varieties or strains by a programme of artificial selection spanning several generations of the organism concerned.

Question 49.
Name the branch which deals with feeding, caring and breeding of domestic animals.
Answer:
Animal husbandry.

Question 50.
What are milch animals?
Answer:
Animals providing milk are called milch animals.

Question 51.
Define the livestock.
Answer:
Domesticated animals reared to provide milk, meat, etc. e.g. cows, buffaloes, sheep, goats, etc.

Question 52.
Give two uses of cattle.
Answer:
1. Cattles provide hide to prepare leather goods.
2. They are useful in agricultural operations like ploughing, harrowing, levelling, etc.

Question 53.
Give two examples of each of indigenous breeds and exotic milch breeds of cows.
Answer:
1. Indigenous breeds. Sahiwal and Red Sindhi.
2. Exotic breeds. Jersey and Brown Swiss.

Question 54.
List three categories of cattle on the basis of their utility. Give one example of each.
Answer:

1. Milch breed e.g. Sahiwal.
2. Draught breed e.g. Hallikar.
3. General utility breed e.g. Haryana.

Question 55.
Name two high milk-yielding crossbreeds of cows.
Answer:
Karan-Swiss and Karan-Fries.

Question 56.
Name two exotic breeds of cows.
Answer:
Jersey (USA) and Brown-Swiss (Switzerland).

Question 57.
Name two breeds of buffaloes.
Answer:
Murrah and Mehsana.

Question 58.
What are the two components of cattle feed?
Answer:
Roughage and Concentrate.

Question 59.
How roughage and concentrate differ from each other?
Answer:
Roughage contains fibres but less nutrients e.g. fodder, while concentrate is rich in proteins e.g. cereal grains.

Question 60.
What is ration of cow?
Answer:
Ration is the amount of food, which is given to animal during 24 hours period. For cow it is about 15 to 20 kg of green fodder and 4-5 kg of grain mixture.

Question 61.
What are additives?
Answer:
Dairy animals require certain additive feeds, which contain antibiotics, minerals and hormones, they promote growth of animals, facilitate good yield of milk and protect them from diseases.

Question 62.
Name the progeny of cross breed of Brown Swiss and Sahiwal.
Answer:
Karan Swiss.

Question 63.
What is artificial insemination?
Answer:
Introduction of semen of a high quality pedigree bull into the vagina of a healthy female cow by artificial means is called artificial insemination.

Question 64.
Give the term for the process by which a female cow of good breed is stimulated by the hormones to release more ova from its ovaries.
Answer:
Superovulation.

Question 65.
Give the significance of superovulation.
Answer:
It increases the chances of transmission of good characters to progeny.

Question 66.
Who is called “Father of White Revolution”?
Answer:
Dr. V. Kurien.

Question 67.
What is NDRI? Where is it located?
Answer:
National Dairy Research Institute. It is located at Kamal (Haryana).

Question 68.
Name two bacterial diseases of cattle.
Answer:
Anthrax and Brucellosis.

Question 69.
Name two viral diseases of cattle.
Answer:
Foot and mouth disease and cowpox.

Question 70.
Name two animal products in which carbohydrate is totally absent.
Answer:
Eggs and meat.

Question 71.
Name two non-leguminous dry fodders.
Answer:
Pounded straw of wheat and dry grass.

Question 72.
Give one term for the science dealing with rearing of birds.
Answer:
Poultry.

Question 73.
Name two indigenous breeds of fowls.
Answer:
Aseel and Brahma.

Question 74.
Name two exotic breeds of fowls.
Answer:
Rhode Island Red and White Leghorn.

Question 75.
Name two high-yielding crossbreeds of fowls.
Answer:
“B-77” and “HH-260”.

Question 76.
What are broilers?
Answer:
Meat-providing birds are called broilers.

Question 77.
What are layers?
Answer:
Egg-laying hens are called layers.

Question 78.
Name one viral and one bacterial disease of fowls.
Answer:
1. Ranikhet (Viral disease)
2. Salmonellosis or Pullorum (Bacterial disease)

Question 79.
What are vegetarian eggs?
Answer:
Infertile eggs are called vegetarian eggs.

Question 80.
Define pisciculture.
Answer:
Pisciculture is rearing and management of fishes.

Question 81.
Give one term for composite fish-farming.
Answer:
Polyculture.

Question 82.
What is economic importance of fish as food?
Answer:
Fish meat contains more proteins (13 – 22%), less fats, vitamins (A and D) and iodine.

Question 83.
Name two by-products of fishery.
Answer:
Liver oil and fish meal.

Question 84.
Which two vitamins are present in liver oil of certain fishes?
Answer:
Vitamins A and D.

Question 85.
Name two indigenous breeds of carps used as food fishes.
Answer:
1. Catla (Theila)
2. Labeo (Rahu)

Question 86.
Name two exotic breeds of carps used as food for fishes.
Answer:
Silver carp and Grass carp.

Question 87.
What is fish meal? Give its significance.
Answer:
It is prepared from non-oil type fishes and is rich in proteins (55 – 70%).

Question 88.
Define inland culture fishery.
Answer:
Rearing of fishes in the artificially prepared fresh water ponds.

Question 89.
Give two examples of marine food fishes.
Answer:
Sardinella (Salmon) and Harpodon (Bbmbay duck).

Question 90.
What is hypophysation?
Answer:
Process by which female and male fishes are injected the hormones of pituitary extract to induce spawning.

Question 91.
Give the term for the newly-hatched young ones of fishes.
Answer:
Hatchlings.

Question 92.
Define fingerlings.
Answer:
Fingerlings are about 4″ – 6″ sized fishes formed from fries in nursery ponds and rearing ponds.

Question 93.
What is harvesting?
Answer:
Capturing of fully grown fishes is called harvesting or fishing.

Question 94.
Name two methods of fishing.
Answer:
Angling and trapping.

Question 95.
What are hapas?
Answer:
These are rectangular boxes of mosquito net cloth in which hatching of fish eggs occur.

Question 96.
Name two infectious diseases of fishes.
Answer:
1. Viral Haemorrhagic Septicemia (VHS)
2. Infectious Pancreatic Necrosis (IPN)

Question 97.
Name two varieties of Indian fishes.
Answer:
Carps (e.g. Catla) and Cat-fishes (e.g. Mystus).

Question 98.
Mention seafood items other than fishes.
Answer:
Lobsters, prawns, oysters, mussels, etc.

Question 99.
Name the different types of ponds.
Answer:
Nursery ponds, rearing ponds, stocking ponds and broodstock ponds.

Question 100.
List the products obtained from beekeeping.
Answer:
Honey, wax, royal jelly and bee venom.

Question 101.
“Honey is a dense sweet liquid.” Write its composition.
Answer:
Sugar 20-40%, moisture 60-80%, minerals 0.22 to 0.3%, vitamins 0.2 – 5%, enzymes, pollen etc.

Question 102.
What is the utility of pollen for bees?
Answer:
Pollens serve as protein food for bees.

Question 103.
What is a beehive?
Answer:
A beehive is a wooden box of size 46 × 23 cm made up of wooden chambers for egg-laying, honey collection, as honey reserve.

Punjab State Board PSEB 9th Class Science Important Questions Chapter 14 Natural Resources Important Questions and Answers.

PSEB 9th Class Science Important Questions Chapter 14 Natural Resources

Long Answer Type Questions:

Question 1.
Write a note on formation of soil.
Answer:
Formation of Soil: The formation of soil depends on the parent rock material, the climate and topography of the area, the organisms present in the soil and the time over which the soil has been developing. Over long periods of time, thousands and millions of years, the rocks near the surface of the Earth are broken down by various physical, chemical and some biological processes. The end product of this breaking down is the fine particles of soil.

Processes for soil formation:
1. The Sun: The sun heats up rocks during the day so that they expand. At night, the rocks cool down and contract. The unequal expansion and contraction in different parts of the rock results in the formation of cracks and ultimately rocks break up into smaller pieces.

2. Water: Water helps in the formation of soil in two ways:

1. Water could get into the cracks in the rocks formed due to uneven heating by the sun. If this water freezes, it will widen the cracks.
2. Fast flowing water carries big and small particles of rock downstream, causing breakdown of rock particles into smaller, finer particles through their abrasive effects.

3. Wind: Strong winds also break down rocks. They also carry sand from one place to the other like the water does.

4. Living organisms: They also influence the formation of soil. While lichens grow on surface of rocks, they release certain substances that cause the rock surface to powder down and form a thin layer of soil. Likewise, small plants like moss and roots of big trees also break the rocks.

Question 2.
What are the effects of use of fertilizers and pesticides for long period on soil fertility? What are the causes of soil erosion?
Answer:
Effects of excessive use of fertilizers and pesticides:

1. Use of these substances over a long period of time can destroy the soil structure by killing the soil micro-organisms that recycle nutrients in the soil.
2. They also kill the earthworms which are instrumental in making the rich humus.
3. Fertile soil can quickly be turned barren if sustainable practices are not followed.
4. Major cause of soil pollution is removal of useful components from the soil and addition of other substances, which adversely affects the fertility of the soil and kills the diversity of organisms living in it, is called soil pollution.

Causes of Soil Erosion:

1. Wind causes soil erosion by carrying away the top loose soil particles.
2. Rain causes soil erosion on unprotected topsoil by washing it down.
3. mproper farming or tilling and leaving the field fallow for long time causes soil erosion.
4. Frequent flooding of rivers causes soil erosion by removing the fertile top soil of the fields near the river banks.
5. Deforestation also leads to soil erosion.

Question 3.
Why is soil as resource important for mankind? Mention the constituents of soil.
Answer:
Soil is a rich source of minerals and humus. It is important for growing crops. Soil water is used by plants for various functions.
Soil provides support to crops, grassland and forests thus it is an important natural resource.

Components of Soil. Soil is a mixture, it contains:

1. Small particles of rocks.
2. Bits of decayed living organisms which is called humus.
3. Soil also contains various forms of microscopic life.
4. It contains nutrients and availability of which depends on the rocks from which it was formed.
5. Soil water – 25% – 35%
6. Soil air – 15-25 %

Question 4.
Define soil fertility. How can it be maintained?
Answer:
Soil fertility. It is the ability of soil to provide minerals, water and other nutrients to the plants.
Conservation of Soil fertility:

1. Adding of manure to the soil.
2. Rotation of crops.
3. Keeping the land as such without growing any crop.
4. Addition of fertilizers.

Artificial methods to maintain soil fertility:

1. Nitrogenous and other fertilizers are added.
2. For natural restoration of nutrients, soil is kept uncultivated for certain period.

Role of humus:

1. Humus increases the soil fertility.
2. Humus has high retaining capacity for water.
3. It makes the soil porous and allows water and air to penetrate deep.

Question 5.
What is air pollution? Write the main sources and preventive measures.
Answer:
Air Pollution. Air pollution refers to the release into the atmosphere of materials that are harmful to man, other animals, plants and buildings or other objects.

Sources of Air pollution:
The major sources of air pollution are fossil fuels (coal and petroleum) and industries. Human Sources. Many activities done by man are the main sources of air pollution. These activities can be divided into following categories.

1. Combustion activities.
2. Industrial activities.
3. Agricultural works.
4. Use of solvents.
5. Activities concerned with atomic energy.

Preventive measures for air pollution:
To prevent and control air pollution two types of measures can be adopted.
1. Instead of releasing poisonous gases containing various pollutants into the atmosphere they could be destroyed or used by some other measures.
2. Converting harmful pollutants to harmless products and then releasing them into
the atmosphere.

Control measures for minimizing air pollution:
1. Simple combustible solid wastes should be burnt in incinerators.
2. Automobiles must be either made to eliminate the use of gasoline and diesel oil or complete combustion is obtained in the engine so that harmful products are omitted.

Question 6.
Write a few properties of water. Sketch water Cycle.
Answer:
Water is a liquid at room temperature. It is densiest (heaviest) at about 4°C. The dense water sinks and the lighter frozen water (ice) floats, ice also insulates the water below. This enables the aquatic life to survive under the ice in cold weather.

Question 7.
What is nitrogen cycle? Make a simple line sketch to show nitrogen cycle in the biosphere.
Answer:
The nitrogen cycle in the biosphere is regarded as a perfect cycle because the cycling process keeps the overall amount of nitrogen constant in the atmosphere and water bodies. The use of chemical nitrogeneous fertilizers like NPK and urea also help in the maintenance of soil nutrients and nitrogen cycle.

However, some of the nitrogen compounds present in soil get trapped within sedimentary rocks and therefore, they are not available to nitrogen cycle for circulation in the biosphere. However, this loss is compensated by volcanic eruptions and erosions and sedimentary rocks. Both these processes release nitrogen.

Micro-organisms involved in Nitrogen Cycle:

As already learnt, micro-organisms play a very important role in nitrogen cycle in nature. Different organisms are involved in different processes of nitrogen cycle. The main micro-organisms involved in nitrogen cycle are listed below:

Activity:

1. Nitrogen fixation
2. Ammonification
3. Ammonia to nitrites
4. Nitrification (Nitrites to nitrates)
5. Denitrification (Nitrates to free Nitrogen)

Organisms:

1. Rhizobium, blue-green algae
2. Decay bacteria, fungi
3. Nitrosomonas
4. Nitrobacter
5. Pseudomonas

The nitrogen cycle in the biosphere involves the following steps:

Question 8.
Briefly explain oxygen cycle.
Answer:
Oxygen cycle: Oxygen is very abundant element on earth. It is found in the elemental form in the atmosphere to the extent of 21%. It also occurs extensively in the combined form in the earth’s crust as well as also in the air in the form of carbon dioxide. In the crust, it is found as the oxides of most metals and silicon, and also as carbonate, sulphate, nitrate and other minerals. It is also an essential component of most biological molecules like carbohydrates, proteins, nucleic acids and fats (or lipids).

But when we talk of the oxygen cycle, we are mainly referring to the cycle that maintains the levels of oxygen in the atmosphere. Oxygen from the atmosphere is used up in three processes, namely combustion, respiration and in the formation of oxides of nitrogen. Oxygen is returned to the atmosphere in only one major process, that is, photosynthesis. And this forms the broad outline of the oxygen cycle in nature.

Question 9.
Briefly explain carbon cycle in nature.
Answer:
The Carbon Cycle. Carbon is found in various forms on the earth. It occurs in the elemental form as diamond and graphite. In abiotic environment, carbon is present in the following forms:

1. as carbon dioxide in the atmosphere.
2. as carbonate and hydrogen-carbonate salts in various minerals.
3. as dissolved carbonic acid and bicarbonates in water.
4. as fossil fuels like coal, petroleum and natural gas.
5. Plants utilise the atmospheric carbon dioxide in photosynthesis to produce carbohydrates, which are taken by herbivores and then pass through small and large carnivores.

Forms of Carbon: Carbon is found in various forms on the earth.

1. It occurs in the elemental form as diamonds and graphite.
2. In the combined state, it is found as carbon dioxide in the atmosphere, as carbonate and hydrogen-carbonate salts in various minerals.
3. All life forms are based on carbon-containing molecules like proteins, carbohydrates, fats, nucleic acids and vitamins.
4. Both plants and animals release carbon dioxide to the atmosphere as a product of respiration.
5. By decomposition of organic wastes and dead bodies by decomposers.
6. By burning of fossil fuels, like coal, and petroleum.

Question 10.
What are the causes of ozone depletion?
Answer:
Ozone depletion: Recently it was discovered that ozone layer was getting depleted. Various man-made compounds like CFCs (carbon compounds having both fluorine and chlorine which are very stable and not degraded by any biological process) were found to persist in the atmosphere.

Once they reached the ozone layer, they would react with the ozone molecules. This resulted in a reduction of the ozone layer and recently they have discovered a hole in the ozone layer above the Antarctica. It is difficult to imagine the consequences for life on earth if the ozone layer dwindles further, but many people think that it would be better not to take chances. Measures should be taken towards stopping all further damage to the ozone layer.

Question 11.
Write a note on freshwater resources.
Answer:
Fresh Water Resources:
Fresh water resources range from ponds to lakes and large rivers. It has the following characteristics :
(a) Freshwater is exhaustible, however, it is being made available again by oceans through hydrological cycle.

(b) Out of this three per cent, 77.2 per cent is stored in glaciers and ice caps. And 22.4 per cent is groundwater and soil moisture. Remaining 0.36 per cent is found in lakes, rivers, streams and swamps etc. Out of the total water evaporated from oceans 90 per cent falls on the oceans and remaining 10 per cent falls on the land. This water is utilised by various terrestrial ecosystems.

(c) Freshwater is essential for life on earth as well as for survival of human race.

(d) The total water in hydrosphere is 1.4 billion cubic kilometres (Km³). Total ocean
water is 97%. The ocean water cannot be consumed by human beings. Remaining three percent (freshwater) is available for human consumption. The water resources in India have an average run off in river system of 1,869 km² and 432 km³ groundwater.

Short Answer Type Questions:

Question 1.
What is atmosphere? Name its different layers.
Answer:
Atmosphere: Gaseous envelope surrounding the earth is called atmosphere. Several concentric layers can be identified in vertical profile of atmosphere. Density, temperature and composition differ in these layers. Near the earth’s surface, density is highest and with increase in latitude density decreases. Starting from earth’s surface five concentric layers can be distinguished:

1. Troposphere
2. Stratosphere
3. Mesosphere
4. Thermosphere
5. Ionosphere.

Exosphere which forms the outer fringe of atmosphere is highly rarefied and gradually get mixed with other space.

Composition of dry atmosphere:

Components	Volume
Nitrogen (N2)	78%
Oxygen (O2)	21%
Carbon dioxide (CO2)	0.03%
Argon	0.93%
Helium, Neon, Ozone, ammonia	0.04%

Helium, Neon and Argon are noble gases.

Question 2.
What is the role of atmosphere in climate control?
Answer:
Role of atmosphere in climate control. Atmosphere covers the earth like a blanket. Air is a bad conductor of heat. The atmosphere keeps the average temperature of the earth fairly steady during the day and even during the course of the whole year. The atmosphere prevents the sudden increase in temperature during the daylight hours.

During the night, it slows down the escape of heat into outer space. Moon, which is about the same distance from the sun that the earth is. Despite that, on the surface of the moon, with no atmosphere, the temperature ranges from -190° C to 110° C.

Question 3.
What is ozone layer? Write its importance.
Answer:
Ozone layer: It is the protective layer. Ozone in stratosphere is responsible for protecting the earth from high energy ultraviolet radiation. It forms a life-saving screen as it checks the entry of lethal UV- rays. Ozone found in troposphere has warming effect. Ozone is one gas which is harmful as well as useful for human beings.

16th September, 1996 was celebrated as “International Day for the Preservation of Ozone layer”. It was aimed to generate awareness about the dangers of ozone depletion in the stratosphere and it was drawn up by UNEP.

Question 4.
What is meant by ozone shield? How the CFC’s and ozone-depleting substances effect ozone shield?
Answer:
Ozone shield: An equilibrium is established between generation and destruction of O₃, leading to a steady-state concentration of ozone layer in the stratosphere between 20 and 26 km above the sea level. The thickness of the vertical column of stratospheric O₃ layer, condensed to standard temperature and pressure, average 0.29 cm above the equator and may exceed 0.40 cm above the poles at the end of the winter season. This layer acts as the ozone shield protecting the earth biota from harmful effects of strong UV radiations.

CFC’s produce active chlorine (Cl with CIO radical) in the presence of UV radiations. These radicals catalytically destroy ozone converting it into oxygen. CH₄ and N₂O also cause ozone destruction.

Question 5.
Write a note on the air pollution caused due to combustion.
Answer:
The mobile combustion sources are the main sources of air pollution especially in the cities. They include the locomotives, automobiles and aircrafts.
The air pollutants from these are:
1. (i) Carbon monoxide (ii) oxides of nitrogen and (in) a mixture of hydrocarbons.
2. The petroleum used as fuel in these sources contains lead as an impurity in the form of tetraethyl lead Pb (C₂H₅)₄, and tetramethyl lead Pb (CH₃)₄.

Question 6.
Discuss harmful effects of air pollution.
Answer:
Harmful Effects of Air Pollution
1. Air pollution affects the respiratory system causing breathing difficulties and diseases such as bronchitis, asthma, lung cancer, tuberculosis and pneumonia.

2. Burning of fossil fuels like coal and petroleum releases oxides of nitrogen and sulphur. Inhalation of these gases is dangerous. These gases also dissolve in rain to give rise to acid rain.

3. The combustion of fossil fuel also increases the amount of suspended particles in air. These suspended particles could be unbumt carbon particles or substances called hydrocarbons. The presence of high levels of all these pollutants, reduce visibility in cold weather where water also condenses out of air forming smog. Smog is an indication of air pollution.

4. Regular breathing in the polluted air increases allergies, cancer and heart diseases.

Question 7.
What is the role of biotic components in living organisms?
Answer:
The living or biotic components are plants and animals including us and non¬living or physical components are air, water, soil, light and temperature. All these components interact and effect each other, resulting in the establishment of a complex and complete balance in the environment. The environmental components like mountains, rivers, ponds, forests, minerals, coal and even petrol and other natural resources are of great importance to us.

CO₂ is fixed in two ways:
1. Green plants convert CO₂ into glucose in the presence of sunlight and chlorophyll pigments.
2. Many marine animals use carbonates dissolved in sea water to make their shells. Oxygen is required by eukaryotic and many prokaryotic cells. All the cells need oxygen to break down glucose molecules in order to release energy required to carry out this vital functions of life.

Question 8.
What are aerosols?
Answer:
Aerosols. Aerosols are certain chemicals released in the air with force in the form of mist or vapour. The important source of aerosols is the jet aeroplane emissions in the outer atmosphere. The aerosols contain fluorocarbons which deplete the ozone layer in the atmosphere.

Question 9.
What are acid rains?
Answer:
Acid Rain. It is the rain which contains small amount of acid in it that is formed from the gases like sulphur dioxide and nitrogen oxides present in polluted air. It causes damage to living and non-living things.

Question 10.
What is smog?
Answer:
Smog. Smoke and fog when combined together forms smog in the presence of sunlight. Various unbumt hydrocarbons produced from the automobile combustion react with oxides of nitrogen to form ozone, peroxyacyl nitrates and aldehydes. They are called photochemical oxidants. Together with smoke and fog they constitute smog which has a harmful effect on humans repiratory and nervous system; it also harm the plants and rubber goods.

Question 11.
Explain the role of sun in soil formation.
Answer:
Role of sun in soil formation:
The sun heats up rocks dtiring the day as a result that they expand. At night, these rocks cool down and contract. Since all parts of the rock do not expand and contract at the same rate, this results in the formation of cracks and ultimately the huge rocks break up into smaller pieces and small pieces further break up into still smaller pieces and fine particles.

Question 12.
How does water take part in soil formation?
Answer:
Role of water in soil formation:
1. Water enters into cracks formed as a result of uneven heating by sun. As this water freezes, it causes the cracks in the rocks to widen.
2. Flowing water wears away hard rocks over long period of time. The fast flowing water often carries big and small particles of rock downstream. These pieces of rock rub against other rock pieces and the abrasion causes the rocks to wear down into smaller and still smaller pieces.

Question 13.
Discuss the role of wind and living organisms in soil formation.
Answer:
Role of wind in the soil formation. Strong wind rubs against rocks and wear them down. The wind also carries soil particles from one place to another.

Role of living organisms in soil formation:

Living organisms also influence the formation of soil. The lichen that we read about earlier, also grows on the surface of rocks. While growing, they release certain substances that cause the rock surface to powder down and form a thin layer of soil.

Other small plants like moss, are able to grow on this surface now and they cause the rock to break up further. The roots of big trees sometimes go into cracks in the rocks and as the roots grow bigger, the crack further becomes bigger causing the rocks to break down to form soil.

Question 14.
What is the role of atmosphere in movement of air which causes winds?
Answer:
The movement of air causes winds:
1. The atmosphere gets heated from the radiation that is reflected back or re-radiated by the land or water bodies. As a result of heating convection currents are set up in the air. Since land gets heated faster than water, the air over land gets heated faster than the air above water bodies.

2. In coastal regions, during the day, the air above the land gets heated faster and starts rising. So a region of low pressure is created and air over sea moves into this area of low pressure. The movement of air from one region to the other region causes winds.

3. During the day, the direction of wind would be from the sea to the land and at night, both land and sea starts to cool. Since water cools down slower than the land, the air above water would be warmer than the air above land, thus the direction of wind would be from the land to the sea.

Question 15.
How are rainfall patterns decided?
Answer:

1. Rainfall patterns are decided by the prevailing wind patterns.
2. In large parts of India, rains are mostly brought by the southwest or northeast monsoons.
3. Depressions in the Bay of Bengal have caused rains in some areas, is the common comment during weather ireports.

Question16.
What is the role of atmosphere in causing rain?
Answer:
Role of atmosphere in causing rain:

1. When water bodies are heated during the day, a large amount of water evaporates and goes into the air.
2. The wind carries the water vapour to various places.
3. The air gets heated and rises up carrying the water vapour with it.
4. As this air rises, it expands and cools causing the water vapour in the air to condense in the form of tiny droplets.
5. Once the water droplets are formed, they grow bigger by the ‘condensation’ of these water droplets.
6. When the drops grow big and heavy, they fall down in the form of rain.

Question 17.
Comment “Water is one of the major resources which determine life on land”. List a few other factors also.
Answer:
The availability of water decides not only the number of individuals of each species that are able to survive in a particular area, but it also decides the diversity of life there. Of course, the availability of water is not the only factor that decides the sustainability of life in a region. Other factors like the temperature and nature of soil also matters. But water is one of the major resources which determine life on land.

Question 18.
What is water pollution?
Answer:
Water pollution: Addition of harmful materials to water is termed water pollution. The sources of inland water pollution are community wastewater (sewage) and wastes from industries and agricultural practices. Water pollutants include organic matter, pathogens, chemicals and minerals, solid particles, radioactive wastes and heat.

Question 19.
What are the sources of water pollution?
Answer:
Sources of Water Pollution:

1. Agriculture substances such as fertilisers and pesticides are used to increase crop yield, and some amount of these chemicals is washed into the water bodies that pollutes the water.
2. Sewage from homes and wastes from factories are dumped into rivers or lakes also cause water pollution.
3. Hot and cold water discharged from industries make a change in temperature, which is harmful for aquatic organisms.
4. All these affects the balance among various organisms that are found in water bodies.

Question 20.
What are the effects of water pollution?
Answer:
Effects of Water Pollution:

1. Water pollutants reach the sea directly from the coastal cities and ships, and indirectly with river water from distant places. Oil spilled in tanker accidents is a major threat to ocean life.
2. The substances like fertilisers and pesticides used in farming, mercury salts used by paper industries could be poisonous. There could also be disease-causing organisms, like the bacteria which causes cholera.
3. Industrial or household waste reduces the dissolved oxygen in water bodies, thereby affecting the aquatic life.
4. Aquatic organisms can stay alive in a certain range of temperature. Sudden change in temperature of water bodies is dangerous for aquatic organisms and affects their breeding.

Question 21.
Make a list of various diseases caused by polluted water.
Answer:
Diseases caused by polluted water

1. Bacterial diseases. Cholera, Typhoid, Diarrhoea, Dysentery.
2. Viral diseases. Jaundice, Polio etc.
3. Protozonal diseases. Diseases associated with stomach and intestines eg. Amoebic dysentery, Giardiasis etc.
4. Helminthic diseases. Infection of some intestinal parasites like Ascaris lumbricodies is through drinking water only.
5. Guinea worm diseases is through Cyclops present in the drinking water. Through contaminated water they reach to another host i.e. man.

Question 22.
What is greenhouse effect? Show the % age of gases that cause greenhouse effect.
Answer:
Greenhouse effect. Earth’s temperature is maintained by reradiated infra-red radiations by CO₂, CH₄, O₃, NO and NO₂ and slightly by water vapours in atmosphere. These gases prevent heat from escaping to outer space, so are functionally comparable to glass panels of a greenhouse and are called greenhouse gases (GHGs) and the process is called greenhouse effect. The CO₂ is added to atmosphere mainly by burning fossil fuels, volcanic activities and respiration.

Question 23.
List Various Measures for Soil Conservation
Answer:
Various Measures for Soil Conservation:

1. Stopping clear-cutting of forests and overgrazing checks soil erosion by streams and rivers.
2. Intensive cropping helps in checking soil erosion. A field always under a crop is protected against erosion.
3. Bunds around the fields contain rain water and check soil erosion besides washing away of minerals.
4. Irrigation channels in the fields should be so designed as to carry water at a slow speed.
5. Drainage canals to carry flood water will protect the fields against erosion.

Question 24.
How will you determine composition of soil?
Answer:

Determination of Soil Composition:

1. Materials required: 150 cc of sifted soil, a measuring glass cylinder of 1 litre capacity, water, glass rod.
2. Procedure: Take 150 cc of sifted soil in a glass cylinder. Pour about 750 cc of water over it. Stir the soil well with the help of a glass rod. Take out the rod. Allow the particles to settle. Observe after 30 minutes.
3. Results: The bottom of the cylinder has a layer of coarse sand. A layer of fine sand lies above it. Then there is a layer of silt. Clay lies above silt.

Turbid water occurs above the clay. It contains clay as well as mineral salts. Humus or organic matter floats over the top of turbid water.

Question 25.
Make an outline sketch of nitrogen cycle.
Answer:
Nitrogen Cycle

Very Short Answer Type Questions:

Question 1.
Life exists on which planet?
Answer:
Earth.

Question 2.
What are the materials required for life?
Answer:
Environment, Heat, Light, Water and food.

Question 3.
What is environment?
Answer:
Environment. The earth and everything which affects the living organisms constitute its environment.

Question 4.
Basic requirment of life are obtained from which sources.
Answer:
Energy from sun and resources present on earth.

Question 5.
What is atmosphere?
Answer:
Atmosphere: It is the multilayered gaseous envelope of air that covers the whole of the planet earth like a blanket.

Question 6.
How much surface of earth is covered by water?
Answer:
About 75 percent.

Question 7.
How atmosphere covers the earth?
Answer:
Atmosphere covers the earth as a blanket.

Question 8.
What is biosphere?
Answer:
It is the life-supporting zone of earth where the atmosphere, the hydrosphere and the lithosphere interact and make life possible.

Question 9.
What are biotic components of biosphere?
Answer:
All living organisms.

Question 10.
List the abiotic components of bioshphere?
Answer:
Air, water and soil.

Question 11.
Name the gaseous components of atmosphere.
Answer:
Nitrogen, Oxygen, CO₂ and Water vapour.

Question 12.
Name the gases present on Venus and Mars planet.
Answer:
95 to 97% CO₂.

Question 13.
What is respiration?
Answer:
A process in which O₂ is used and CO₂ is liberated.

Question 14.
How is CO₂ used so that balance is maintained?
Answer:
1. CO₂ is used during photosynthesis and carbohydrates are formed.
2. Used as carbonates by marine molluscs.

Question 16.
How is temperature regulated on earth?
Answer:
Atmosphere regulates the temperature on earth.

Question 17.
What is the minimum and maximum temperature on moon?
Answer:
Minimum temperature = – 190°C
Maximum temperature = 110°C.

Question 18.
What causes wind?
Answer:
Movement of air caused by uneven heating of the atmosphere in different regions of earth.

Question 19.
How are clouds formed?
Answer:
Clouds are formed by condensation of water droplets in the air.

Question 20.
Which wind gets hot : Water to earth surface or from surface of earth to upward.
Answer:
Land (surface of earth) to upward.

Question 21.
During night what is the direction of movement of air?
Answer:
From surface of earth (land) to the sea.

Question 22.
What is the cause of movement of wind?
Answer:
Interaction of atmospheric components.

Question 23.
List two factors which affect wind.
Answer:
1. Rotation of earth
2. Presence of mountain heights.

Question 24.
What is deforestation?
Answer:
Cutting of trees on large scale is called deforestation.

Question 25.
What is the effect of deforestation?
Answer:
Deterioration of atmosphere.

Question 26.
Is air a good or bad conductor of heat?
Answer:
Air is a bad conductor of heat.

Question 27.
What is the cause of rain on Indian Land.
Answer:
Rain in India is due to monsoon from south-west or east-west direction.

Question 28.
What is smog?
Answer:
Smoke mixed with fog is called smog.

Question 29.
What does the smog indicate?
Answer:
It indicates pollution of air.

Question 30.
Where is the purest form of water available.
Answer:
Snow/Ice caps.

Question 31.
Write one importance of water for living organisms.
Answer:
All cellular processes take place in water medium in living organisms.

Question 32.
Write one cause of pollution of water in town.
Answer:
Sewage.

Question 33.
What is soil?
Answer:
The top weathered part of earth’s surface is called soil.

Question 34.
What is the role of sun in the formation of soil?
Answer:
Heating of rocks causes cracking and ultimately breaking up into smaller pieces.

Question 35.
What is the role of wind in soil formation?
Answer:
Wind causes erosion of rocks.

Question 36.
What is soil erosion?
Answer:
Removal of useful components from the soil which affect the fertility of soil is called soil erosion.

Question 37.
What is the importance of soil?
Answer:
Soil supports terrestrial plants and animals and it decides the diversity of life in an area.

Question 38.
How is soil formed?
Answer:
Soil is formed by weathering of rocks.

Question 39.
What are three kinds of water sources?
Answer:
Rainwater, Groundwater and Subsoil water.

Question 40.
Write one advantage of seawater.
Answer:
It is a house of table salt (common salt).

Question 41.
What are the two types of water resources?
Answer:
1. Freshwater resources.
2. Saltwater (sea) resources.

Question 42.
What are the sources of freshwater?
Answer:

1. Rainwater
2. Surface water
3. Groundwater
4. Polar ice caps
5. Ponds and Pools

Question 43.
What do you understand by aquatic habitat?
Answer:
Organisms which are found in water possess aquatic habitat.

Question 44.
How much percent of nitrogen is present in atmosphere?
Answer:
78%.

Question 45.
How is nitrogen used in living organisms.
Answer:
Nitrogen is a component of proteins, nucleic acid (DNA and RNA).

Question 46.
Name two plants which are capable of fixing atmospheric nitrogen.
Answer:
Green pea and other leguminous plants.

Question 47.
Write two uses of mirco-organisms.
Answer:
Micro-organisms act as biofertilizers. They also produce antibiotics.

Question 48.
Name any two natural cycles operating in nature.
Answer:
1. Carbon cycle
2. Nitrogen cycle

Question 49.
Name the gaseous components of biosphere.
Answer:
CO₂, O₂ and Nitrogen.

Question 50.
Name the source of energy for the process of photosynthesis.
Answer:
Solar energy.

Question 51.
Define biomass.
Answer:
The total weight of a living organism.

Question 52.
Name two nitrifying bacteria.
Answer:
Nitrosomonas and Nitrobacter.

Question 53.
Define pollution.
Answer:
Pollution is an undesirable change in physical, chemical or biological characteristics of air, water or land caused by pollutants.

Question 54.
What are pollutants?
Answer:
The substances causing pollution are termed pollutants.

Question 55.
What are the three major types of pollution?
Answer:
Water, air and soil pollution.

Question 56.
What is soil pollution?
Answer:
Soil pollution is removal of useful components from soil and addition of other substances which adversely affect the soil is termed soil pollution.

Question 57.
Which part of solar radiation is absorbed by ozone layer?
Answer:
UV rays.

Question 58.
Name the major surface water pollutant from farm run off and bathroom water.
Answer:
Phosphorus.

Question 59.
Give the source of pathogens in the water.
Answer:
Domestic sewage.

Question 60.
What is the source of aerosols?
Answer:
Aeroplanes.

Question 61.
Which term is used for pollutants that are degraded by natural means?
Answer:
Biodegradable.

Question 62.
How can SO₂ pollution of air be checked?
Answer:
By using sulphur-free fuel in automobiles.

Question 63.
Mention the regions where rainfall is highest and lowest in India.
Answer:

• Minimum rainfall: Arid region having rainfall of 20 to 50 cm.
• Maximum rainfall: Wet region having rainfall of more than 200 cm.

Punjab State Board PSEB 9th Class Science Important Questions Chapter 13 Why Do we Fall Ill Important Questions and Answers.

PSEB 9th Class Science Important Questions Chapter 13 Why Do we Fall Ill

Long Answer Type Questions:

Question 1.
Discuss the modes of transmission of diseases.
Answer:
Transmission of Communicable Diseases: Diseases are transmitted from the reservoir of susceptible persons in tire following ways:
1. Direct transmission:
(a) Direct contact between infected and healthy person, e.g. chickenpox, leprosy.
(b) Droplet infection from sneezing, coughing, spitting and talking. e.g. T.B, Whooping cough.
(c) Contact with soil which contains saprophytic disease-causing agents.
(d) Bite of an animal, e.g. Rabies, by bite of rabid dog.
(e) Transplacental transmission (from mother to foetus).

2. Indirect transmission:
1. Air-borne diseases such as common cold, pneumonia and tuberculosis. Such disease causing microbes are spread through the air. This occurs through little droplets thrown out by an infected person who sneezes or coughs. Someone standing closeby can breathe in these droplets, and the microbes get a chance to start a new infection.

2. Water-borne diseases such as cholera, diarrhoea, jaundice. Infectious diseases can be spread through water. This occurs if the stool or other wastes from person suffering from an infectious intestinal disease, gets mixed with the drinking water used by people living nearby. The cholera-causing microbes will enter new hosts through the water they drink and cause disease in them.

3. Sexually-transmitted diseases such as Syphilis and AIDS. Both of these pathogens are transmitted by sexual contact from one partner to the other. Such sexually transmitted
diseases are not spread by casual physical contact. Casual physical contacts include handshakes or hugs or sports as wrestling or by any of the other ways in which we touch each other socially.

4. Spread of disease through vectors. Many animals which live with us may carry diseases. These animals carry the infecting agents from a sick person to another potential host. These animals act as intermediate host and are called vectors. Mosquitoes (Anopheles) are vector of a disease called malaria and dengue fever.

Question 2.
Show the common modes of transmission of diseases.
Answer:
Modes of transmission of diseases

Question 3.
What are general ways of preventing infectious diseases?
Answer:
General ways of preventing infectious diseases:
The general ways of preventing infections mostly relate to preventing exposure. Public hygiene is one basic key to the prevention of infectious diseases.

The following practices are adopted in this method of prevention of diseases:

1. For air-borne microbes, we can prevent exposure by providing living conditions that are not overcrowded.
2. For water-borne microbes, we can prevent exposure by providing safe drinking water.
3. For vector-borne infections, we can provide clean environment. Such a clean environment would not allow mosquito breeding.

Question 4.
Explain acute and chronic diseases.
Answer:
Acute and Chronic diseases: The manifestations of disease will be different depending on a number of factors. One of the most obvious factors that determine how we perceive the disease is its duration. Some diseases last for only very short periods of time, and these are called acute diseases.

The common cold lasts only a few days. Other ailments can last for a long time, even as much as a lifetime and are called chronic diseases. An example is the infection causing elephantiasis, which is very common in some parts of India.

Question 5.
Make a table showing organ specific and tissue-specific manifestation of diseases.
Answer:
Organ-specific and tissue-specific manifestations depend on the target organ which the microbes target after their entry. They are as follows:

Target organ	Specific manifestation
1. Lungs	Cough, breathlessness, chest pain and may be bloody sputum as in TB and lung cancer.
2. Liver	Inflammation of liver cells leading to jaundice characterized by yellowness of skin and eyes as in Hepatitis.
3. Intestine	Inflammation of intestinal mucosa leading to acute diarrhoea and dehydration as in cholera.
4. Nasal chambers	Inflammation of nasal mucosa leading to sneezing, bronchitis, coughing, fever, etc. as in influenza.
5. Brain	Headaches, vomiting, fits or unconsciousness.

Question 6.
Expand AIDS. Explain causes/modes of transmission, effects, incubation period, diagnosis, symptoms and preventive measures of AIDS. What is the significance of 1st December?
Answer:
AIDS (Acquired Immuno-Deficiency Syndrome)
(a) Cause. AIDS is caused by a retro-virus-HIV (Human immuno-deficiency virus).
In India, it was first reported in the prostitutes of Chennai in 1986. It is a pandemic disease.

(b) Epidemiology (Transmission). Human infection occurs through:

1. Unprotected sexual intercourse,
2. Use of contaminated syringes,
3. Blood transfusion,
4. Organ transplantation,
5. Common razor of the barbers, etc.

Effects: It causes damage, decrease in number of platelets, swollen lymph nodes, ritght sweats, loss of memory, etc. It is a 100% fatal disease.

(c) Incubation period is of about 28 months.

(d) Diagnosis by ELISA test and Western Blot test.

(e) Symptoms: HIV kills the Helper T-lytnphocytes. It is characterized by following symptoms :

1. Prolonged fever.
2. Swollen lymph nodes.
3. Weight loss and loss of appetite.
4. Unexplained bleeding.
5. Loss of memory and mental ability.
6. Patient becomes susceptible to other infectious diseases.
7. Night Sweats.

(f) Preventive measures: Involves educating the high risk groups use of disposable syringes; screening tests of blood, organs, semen, etc; monogamous relationship avoiding prostitution, polygamy and heterosexuality, using condoms, avoiding the use of common razor; etc.

(g) Treatment: Drugs like AZT (Azidothymidine), TIAS injection and Protease inhibitors, etc. are known to suppress AIDS virus. Efforts are on for a vaccine against the virus. Ist December is observed as World AIDS Day.

Question 7.
Make a list of common communicable diseases caused by bacteria, viruses, fungi, protozoans and helminthes.
Answer:
Common communicable diseases. The following table shows some important diseases caused by bacteria, fungi and protozoa.

Name of causative agent	Diseases
1. Bacteria	Pneumonia, Tetanus, Tuberculosis, Cholera, Food Poi­soning, Sexually transmitted diseases.
2. Viruses	Chickenpox, poliomyelitis, Influenza, AIDS.
3. Fungi	Skin diseases, Food poisoning.
4. Protozoans	Malaria, Kala-Azar, Amoebic dysentery, and African sleeping sickness.
5. Helminthes	Taeniasis, Cysticercosis, Ascariasis, Elephantiasis.

Short Answer Type Questions:

Question 1.
Define disease and health?
Answer:

• Disease: It is defined as a condition of the body or a part of it in which functions are disturbed. The word disease means lack of ease and comfort.
• Health: It is a state of complete physical, mental and social well-being. Health is also linked with social environment and cultural background.

Question 2.
What precautions could you take in your school to reduce the incidence of infectious diseases?
Answer:

1. Providing safe and clean drinking water.
2. Providing clean environment to prevent vector-borne infections.
3. Adopting public hygienic measures.

Question 3.
Discuss the principle of prevention of communicable disease.
Answer:
Principles of Prevention: Following three limitations are normally confronted while treating an infectious disease:

• If a person has a disease, his/her body functions are damaged and may never recover completely.
• Treatment will take time, which means that person suffering from a disease is likely to be bedridden for sometime even if given proper treatment.
• The person suffering from an infectious disease can serve as the source from where the infection may spread to other people. It is because of such reasons that prevention of diseases is better than their cure.

Question 4.
Differentiate between infectious and non-infectious diseases.
Answer:
Differences between infectious and non-infectious diseases

Characters	Infectious diseases	Non-infectious diseases
1. Transmission	Can be transmitted from an infected person to a healthy person.	Cannot be transmitted from an infected person to a healthy person.
2. Causative agents	Microorganisms called pathogens.	Deficiency of nutrient or hormone: or degeneration of tissue or hypersen­sitivity of body or tumour formation.
3. Nature	They are brought about by extrinsic factor.	They are brought about by intrinsic factors.
4. Examples……..	Typhoid, Cholera, T.B., AIDS, Malaria etc.	Diabetes, Kwashiorkor, Marasmus, Goitre, Cancer, Allergy etc.

Question 5.
What are congenital diseases? Name the three types. Give examples.
Answer:
Congenital diseases: These are inborn diseases present since birth caused by gene or chromosomal mutation.
Types of congenital diseases:

1. Diseases caused by gene mutation, e.g. Haemophilia, colorblindness, alcaptonuria and sickle cell anaemia.
2. Diseases caused by chromosomal mutation, e.g. Down’s syndrome, Klinefelter’s syndrome, Turner’s syndrome.
3. Diseases caused by environmental factors such as radiations or pollutants. They are non-inheritable,

Question 6.
List various modes of direct transmission of diseases, giving one example of each type.
Answer:

1. By direct contact with an infected person e.g. Leprosy, chickenpox etc.
2. By droplet infection e.g. Diphtheria, tuberculosis, etc.
3. By contact with soil e.g. bacterial cysts of tetanus.
4. By animal bite e.g. rabies viruse.
5. Transplacental transmission e.g. viruses of German measles and AIDS.

Question 7.
What are sources of diseases? Name various sources of diseases, giving one example of each.
Answer:
1. Sources of diseases are those sites which are occupied by the pathogens before entering inside the human body. These are also called reservoirs of infection.
2. Types of sources of diseases:
(a) Carriers or vectors e.g. Plasmodium (female Anopheles).
(b) Soil e.g. bacterial cysts of Clostridium tetani.
(c) Air e.g. bacterial cysts of TB.
(d) Food and water e.g. bacterial cysts of Cholera.

Question 8.
Differentiate symptoms and signs.
Answer:
Differences between symptoms and signs:

Symptoms	Signs
1. They indicate presence of disease. 2. They are collective manifestations of a number of diseases of a particular part or organ.	1. They provide information about the presence of particular diseases. 2. They are distinct for different diseases.

Question 9.
Differentiate acute diseases and chronic diseases.
Answer:
Differences between acute diseases and chronic diseases:

Acute diseases	Chronic diseases
1. They occur very rapidly but for only short period. 2. Do not cause major effects on general health. 3. Examples: Common cold. Cough.	1. These diseases last for a long time and could be dangerous. 2. They have prolonged and major effects on general health. 3. Examples: T.B., Cancer, Diabetes, Arthritis.

Question 10.
How do antibiotics function?
Answer:
Action of antibiotics: Antibiotics commonly block biochemical pathways important for bacteria. Many bacteria, for example, make a cell-wall to protect themselves. The best illustration is action of penicillin, ft blocks the bacterial processes that build the cell-wall. As a result, the growing bacteria become unable to make cell-walls, and die easily.

Human cells don’t make a cell-wall anyway, so penicillin cannot have such an effect on us. Penicillin will have this effect on any bacteria that use such processes for making cell-walls. Similarly, many antibiotics work against many species of bacteria rather than simply working against one.

Question 11.
Why do antibiotics not work against viral infection?
Answer:
Viruses do not have their own metabolic pathways at all, and that is the reason why antibiotics do not work against viral infections. In case of common cold, taking antibiotics does not reduce the severity or the duration of the disease.

Question 12.
What is hydrophobia? How does it occur in man? Why is it called a neurotrophic disease?
Answer:

1. Hydrophobia is another name of a viral disease called Rabies.
2. It is caused by a RNA-virus, Rabies vires, which is injected in the human being along with saliva of rabid animals like dogs, cats, monkeys etc.
3. Because the virus damages the motor neurons of brain and spinal cord, therefore causes paralysis and death.

Question 13.
What are vaccines?
Answer:
Vaccines are produced by deliberate infection of animals, recombinant DNA techniques coupled with hybridomas have opened up the way for custom-made monoclonal antibodies for preventive and therapeutic use. The vaccination prepares the body to fight against the attack.

Question 14.
Give a few examples of vector-borne diseases.
Answer:
Vector-borne diseases:

Vector	Disease
Tse Tse fly (Glossina) Sandfly (Phlebotomus) Female mosquito (Anopheles) Rat flea (Xenopsilla) Aedes mosquito	African sleeping sickness Kala-azar and Oriental sore Malaria Bubonic plague Yellow fever, Dengue.

Question 15.
Write the name of causal organism of the following diseases:

1. Malaria
2. Rabies
3. Influenza
4. Tuberculosis
5. Typhoid

Answer:

Disease	Casual Organism
1. Malaria 2. Rabies 3. Influenza 4. Tuberculosis 5. Typhoid	Plasmodium vivax Rabies-virus Myxovirus influenzae Mycobacterium tuberculosis Salmonella typhosa

Question 16.
Draw simple diagrams to show the structure of Staphylococcus, Heliobacter, SARS, Leishmania and Trypanosoma.
Answer:
Structure of disease-causing agents

Question 17.
Name the infectious disease that leads to immunodeficiency and wasting of body parts. Give the scientific name of the pathogen causing the disease and mention the body organs it primarily affects.
Answer:

1. AIDS is characterized by immunodeficiency and wasting of body parts.
2. It is caused by Human Immunodeficiency Virus (HIV).
3. HIV attacks helper T-lymphocytes, so causing cell-mediated immunodeficiency, so making the body more prone to various infections.

Question 18.
Which disease is called epidemic jaundice? List its main symptoms? How can it be prevented?
Answer:

1. Epidemic jaundice is commonly called Hepatitis-A.
2. It is characterized by yellowing of skin, urine and stool due to damage of liver cells and overproduction of bilirubin.
3. It can be prevented by proper sanitation, use of boiled or ozonised water and intramuscular injection of human immunoglobulins.

Question 19.
Explain the differences between active and passive immunization.
Answer:
Differences between active immunization and passive immunization:

Active Immunization	Passive Immunization
1. Antigens are introduced from outside which trigger off the formation of antibodies in the body.	1. Ready-made antibodies are introduced into the body.
2. It does not provide immediate relief.	2. It provides immediate relief.
3. Immunity thus achieved is long-lasting.	3. It is not long-lasting.

Question 20.
How is health at risk in a cyclone?
Answer:
Health is at risk in case of cyclone because:

1. Social environment is disturbed as it is an important factor in case of individual health.
2. Garbage collected in places is source of multiplication of microbes and breeding place for various vectors.
3. Stagnant water will provide breeding surface for mosquitoes and other such disease spreading agents.

Question 21.
Show by simple diagram how airborne diseases are easier to catch the person who is near the infected person.
Answer:

Fig. Air-transmitted diseases are easier to catch the closer we are to the infected person. However, in closed areas, the droplet nuclei recirculate and pose a risk to everybody. Overcrowded and poorly ventilated housing is therefore a major factor in the spread of airborne diseases.

Question 22.
Draw a diagram showing structure of HIV Virus.
Answer:
Structure of HIV Virus

Question 23.
What is meant by hydrophobia (rabies)? Write its four symptoms. Suggest four preventive measures to check this disease.
Answer:
Rabies: It is caused by bite of rabid or mad dog and other rabid animals.
Causative agent: Rabies virus present in saliva of dog.
Symptoms:

1. High fever.
2. Severe headache.
3. Painful contraction of muscles of throat and chest.
4. Fear of water.

Preventive measures:

1. Wound should be cleaned.
2. Immunise dogs and cats.
3. Kill highly rabid dogs.

Question 24.
A person is suffering from chest pain, breathlessness, loss of body weight, persistent cough and produces blood stained sputum.

1. Name the disease and its causative agent.
2. Mention two means of its transmission.
3. Name the vaccine used to prevent this disease.
4. Who discovered this disease?

Answer:

1. Person is suffering from pulmonary tuberculosis: Causative agent Bacterium namely Mycobacterium tuberculosis.
2. Modes of transmission: It is a communicable disease. Droplet infection during sneezing or otherwise.
3. BCG vaccine can prevent TB.
4. Robert Koch (1882).

Very Short Answer Type Questions:

Question 1.
Health and diseases are complex problems in which group of organisms?
Answer:
Human.

Question 2.
List four special activities which occur in human body?
Answer:

1. Heartbeat
2. Breathing with lungs
3. Working of brain
4. Excretion in kidneys.

Question 3.
What will happen due to malfunctioning of kidney?
Answer:
Toxic substances will accumulate in the body.

Question 4.
Why is food necessary?
Answer:
Food is necessary for cell, tissue functions and maintenance.

Question 5.
What is health?
Answer:
Health is a state of being well enough to function well physically, mentally and socially.

Question 6.
List two other factors which affect health.
Answer:
Personal and community issues both matters for health.

Question 7.
Define disease.
Answer:
Any condition which interferes with the normal functions of the body and impairs the health. It literally means being uncomfortable.

Question 8.
What do you mean by symptoms of a disease? Give example.
Answer:
Symptoms of a disease are things we feel as being wrong, e.g. headache, cough, loose motions etc.

Question 9.
What are the signs of disease?
Answer:
Signs indicate a little more definite indications of presence of disease.

Question 10.
What are two types of diseases on the basis of duration?
Answer:

• Acute diseases
• Chronic diseases

Question 11.
What do you understand by acute diseases?
Answer:
The diseases which last only for a short period of time.

Question 12.
Define chronic diseases.
Answer:
Diseases which last for long time, even life time are called chronic diseases.

Question 13.
Give two examples of acute diseases.
Answer:
1. Cough and cold
2. Flu

Question 14.
Write example of chronic disease.
Answer:
Elephantiasis.

Question 15.
Which kind of diseases are more harmful to the body?
Answer:
Chronic diseases.

Question 16.
What is the cause of dysentary?
Answer:
Contaminated food and water.

Question 17.
Name two types of diseases on the basis of their occurrence.
Answer:

1. Congenital diseases
2. Acquired diseases

Question 18.
Why are communicable diseases called infectious diseases?
Answer:
Because these are caused by the infection and multiplication of some kind of micro-organisms like bacteria, viruses etc.

Question 19.
What are congenital diseases?
Answer:
Diseases present in the body from the birth. They are mostly hereditary disorders.

Question 20.
Name a disease which is no longer chronic disease.
Answer:
Peptic ulcer.

Question 21.
Name the bacterium responsible for peptic ulcer.
Answer:
Helicobacter pylori.

Question 22.
Who discovered that Helicobacter pylori causes peptic ulcer and were awarded Nobel prize?
Answer:
Robin Warren and Barry Marshall.

Question 23.
Name a few disease-causing microbes.
Answer:
Viruses, Bacteria, Fungi and some Protozoans.

Question 24.
Name any four diseases caused by bacteria.
Answer:
Typhoid, Cholera, Tuberculosis (TB), Anthrax.

Question 25.
List a few common diseases caused by viruses.
Answer:
Common cold, Influenza, Dengue fever and AIDS.

Question 26.
Write examples of protozoanal diseases.
Answer:
Malaria, Kala-azar, Amoebic dysentary.

Question 27.
Name a cutaneous disease caused by fungi.
Answer:
Ringworm.

Question 28.
Name two 100% fatal diseases.
Answer:
Rabies and AIDS.

Question 29.
What is the common name of influenza? Give its causative agent.
Answer:
Influenza is commonly called flu. It is caused by Myxovirus influenza virus.

Question 30.
How does Penicillin acts as a useful antibiotic?
Answer:
Penicillin blocks the pathway that build the cell wall as a result growing bacteria are unable to form a cell wall.

Question 31.
Why antibiotic do not affect cough and cold?
Answer:
Cough and cold are mostly caused by viruses and antibiotics fail to act on viruses.

Question 32.
Name three sexually transmitted diseases.
Answer:
AIDS, Syphilis, Gonorrhea.

Question 33.
Sexually transmitted diseases are not spread by which factors?
Answer:
Handshake, embracing, wrestling.

Question 34.
How do bacteria causing T.B. reach lungs?
Answer:
Through nose during breathing.

Question 35.
How do bacteria causing typhoid enter body.
Answer:
During intake of contaminated food and water.

Question 36.
What is the cause of jaundice?
Answer:
Hepatitis virus.

Question 37.
Who proposed the name malaria from bad air?
Answer:
Macculoch (1827).

Question 38.
Why is rabies called a neurotrophic disease?
Answer:
Because the toxins of Rabies-virus damage the motor neurons of the brain.

Question 39.
Expand the term AIDS.
Answer:
Acquired Immuno Deficiency Syndrome.

Question 40.
Give the full form of HIV.
Answer:
Human-Immuno Deficiency Virus.

Question 41.
What is the significance of December 1?
Answer:
World AIDS Day falls on December 1.

Question 42.
Which disease is characterized by yellowing of skin?
Answer:
Hepatitis (epidemic jaundice).

Question 43.
Name three modes of transmission of AIDS.
Answer:

1. Sexual intercourse with infected partner
2. Use of contaminated syringes and
3. Contaminated blood transfusion.

Question 44.
Who prepared the first vaccine?
Answer:
Edward Jenner.

Question 45.
Why is it difficult to make antiviral substances?
Answer:
Viruses do not have their own biochemical pathways, instead they utilize the machinery of cells they attack, therefore, it is difficult to make antiviral substances.

Question 46.
Define vectors.
Answer:
Vectors: They are organisms which spread the disease-causing agents from infected person to a healthy person.

Question 47.
Encephalitis attack which organ of body?
Answer:
Brain.

Question 48.
List a few symptoms of encephalitis.
Answer:
Headache, Vomitting, Unconsciousness.

Question 49.
Name a disease which does not occur after one attack?
Answer:
Small-pox.

Question 50.
Which disese have been eliminated from the world.
Answer:
Small pox.

Question 51.
List some common modes of spread of disease.
Answer:

1. Direct contact
2. Air
3. Indirect contact
4. Insect bites
5. Contaminated food and water
6. Rabid animal.

Question 52.
Name the two ways of preventing diseases.
Answer:

• General way is preventing exposure.
• Strong immune system.

Question 53.
Name a few diseases for which vaccines are available.
Answer:
Whooping cough, Diphtheria, Measeles, Polio and Tuberculosis.

Question 54.
What is immune system?
Answer:
The system of animal body, which protects it from various infectious agents and cancer is termed immune system.

Question 55.
List two features for individual health.
Answer:
Good economic conditions of jobs and stress-free life are needed for individual health.

Punjab State Board PSEB 9th Class Science Important Questions Chapter 12 Sound Important Questions and Answers.

PSEB 9th Class Science Important Questions Chapter 12 Sound

Long Answer Type Questions:

Question 1.
(A) Name two different types of waves.
(B) Give an experiment to explain the formation of transverse waves.
(C) Define transverse waves.
(D) What should be the conditions for the production of transverse waves?
(E) Give examples of transverse waves.
(F) Define crest and trough.
Answer:
(A) Types of Waves: Waves are classified according to the direction of vibration of particles of the medium. It can be either in the direction parallel to the direction of propagation of wave or in a direction perpendicular to the direction of propagation. In this way waves are of two types:

• Transverse waves
• Longitudinal waves.

(B) Formation of Transverse waves. To understand the formation of transverse waves attach one end of a long string to the hook fixed in the wall as shown in the figure, Hold the other end of the string in your hand. The coloured threads of length 10 cm each with the string at equal distance as shown in the figure.
image
Now give jerk to the slinky. An upward maund or hump is formed in the string which travels along the string towards the fixed end. Such sudden disturbance that lasts for short duration is called pulse i.e. the particles of string move along with the disturbance in the perpendicular direction.

As shown in fig. (c) if we continuously give up and down jerks to the free end of the string a number of waves begin to travel along the string forming a wave train. Each part of the string vibrates up and down while the waves travel along the string. So the waves in the string are transverse in nature.

The points (c, c, c …………… ) of maximum displacement in the upward direction are called crests and the points (T, T, T…………) of maximum displacement in the downward direction are called troughs.

(C) Transverse Wave: These are the waves in which particles of the medium vibrate (up and down) in a direction perpendicular to the direction of propagation of wave.

(D) Conditions necessary for formation of transverse waves:

1. The medium should have property of inertia.
2. The medium should have property of elasticity so that the particles can come back to their original positions after being disturbed.
3. The medium should have minimum frictional force between its particles so that the particles may keep vibrating for a long.
4. Vibrations of plucked stretched string of a violin.

(E) Examples of Transverse Waves.

1. In loose string or spring: If a string is held in hand and the other end is tied to a fixed support and it is continuously moved up and down then transverse waves are produced.
2. Wave on the surface of still water: If a pebble is dropped gently on the surface of still water of a pond then ripples are produced on the water surface.
3. A cork floating on the surface of water would then begin to vibrate up and down then transverse waves are produced. The cork is not displaced with the waves but keeps tossing up and down,

(F) Crests and Troughs:

• Crests: In transverse waves, the particles of the medium which have maximum displacement in the upward positive direction along Y-axis are called crests.
• Troughs: In transverse waves, the particles of the medium which have maximum displacement in the downward negative direction along Y-axis are called troughs.

Question 2.
(A) Define longitudinal wave.
(B) Arrange an experiment to demonstrate the formation of longitudinal wave.
(C) In reference to longitudinal wave, define compressions and rarefactions.
Answer:
(A) Longitudinal wave. Those waves in which the particles of the medium vibrate about their mean position in the direction of propagation of disturbance are called longitudinal waves and the wave motion is called the longitudinal wave motion. Sound in air gets propagated in the form of longitudinal wave motion consisting of regions of compressions and rarefactions.

(B) Experiment: Consider, a tuning fork struck gently with a rubber pad so that its prong begins to vibrate [Fig. (a)]. As prong moves towards right, it compresses the layer of air in contact with it. As air has elasticity, the compressed air tends to relieve itself of its strain and moves towards the right to compress the next layer and so on.

Thus, a wave of compression moves towards the right. At the point of compression, there is an increase of pressure and is shown in form of crest C. At the point of rarefaction of concentration of particles is least and has been shown as trough R. When the prong moves towards left, a region of reduced pressure or rarefaction is produced towards right [Fig. (b)].
Examples:

• Hearing in man,
• Vibrating tuning fork,
• Beating diaphragm of drum.

(C) Compressions and Rarefactions:
Compression: The region of high pressure in the longitudinal wave so that the particles of the medium are closer to each other than the normal distance between them. The higher the pressure, the higher is the number of particles per unit volume i.e. higher is the density, is called compression. In fig it is denoted by ‘c’.

Rarefaction: The region of low pressure in the longitudinal wave so that the particles of the medium are far away from each other than the normal distance between them, is called Rarefaction. In fig it is denoted by ‘R’.

Question 3.
Establish the relation between wave velocity, wavelength and frequency of a wave.
Solution:
Suppose
υ = Wave velocity
ν = Frequency of the wave (i.e. frequency of vibrating particles of the medium)
λ = Wavelength of wave.
T = Time period of a vibrating particle (i.e. time taken by particle of the medium to complete 1 vibration)
Distance travelled by the wave during T seconds = λ
∴ Distance covered by the wave in unit time (1 s) = \(\frac{\lambda}{\mathrm{T}}\)
But distance covered in unit time is wave velocity
∴ υ = \(\frac{\lambda}{\mathrm{T}}\)
or υ = \(\frac {1}{T}\) × λ
∴ υ = ν × λ [∵ \(\frac {1}{T}\) = ν]
i. e. Wave velocity = Frequency × Wavelength
This relation is true both for longitudinal and transverse waves.

Question 4.
Distinguish between sound waves and light waves.
Answer:
Difference between sound waves and light waves

Sound Waves	Light Waves
1. Sound waves are mechanical waves.	Light waves are electromagnetic waves.
2. Sound waves are longitudinal waves in which the direction of vibration of particles of the medium is same as that of propagation of wave.	Light waves are transverse waves in which the vibration of the particles of the medium is in a direction perpendicular to the direction of propagation of wave.
3. Sound waves can not travel through vacuum. These require some material medium.	Light waves can travel through vacuum.
4. Speed of sound waves in air is 340 m s-1.	Speed of light waves in air is very large. It is 3 × 108 m s-1.
5. Sound waves are produced due to vibrations of particles of related medium.	Light waves depend upon the change in electromagnetic fields.
6. Sound waves have low frequency and large wavelength.	Light waves have high frequency and small wavelength.
7. Sound waves cannot be polarised.	Light waves can be polarised.
8. Sound waves produce effect on our ears.	Light waves produce effect on our eyes.
9. The velocity of sound waves is independent of the wavelength of wave.	The velocity of lightwave depends upon the wavelength of the wave.

Question 5.
Explain the classification of sound waves on the basis of frequency range.
Answer:
1. Audible waves: Those sound waves to which human ears can respond (i.e. those sounds waves which can be heard by human beings). In human beings the audible range is from 20 Hz to 20,000 Hz. As the human being advances in age his ears become less sensitive to sounds of high frequencies. These waves are produced by vibrations in air column, tuning fork and violin.

2. Ultrasonic waves: The sound waves having frequency above 20 KHz (i.e. 20,000 Hz) are called ultrasonic waves or ultrasonics. Insects of some species can hear such sound waves. Bats, dolphins, etc. produce such sound waves.

3. Infrasonic waves: Those sound waves which have frequency less than 20 Hz are called infrasonic waves or infrasonics. Whale and elephants produce infrasonic waves. These waves are produced before the main high frequency waves of earthquake occur on hearing these waves the animals become terrorised and become impatient.

Question 6.
What are the laws of reflection of sound? How will you prove these laws experimentally?
Answer:
Like light, sound also obeys the laws of reflection these are:
1. The angle made by incident sound and reflected sound with the normal to the reflecting surface at the point of incidence are always equal.
i.e. \(\angle i=\angle r\)
2. The incident sound, the normal at the point of incidence and the reflected sound all lie in the same plane.

Experimental Verification:
Take two cardboard tubes A and B about 1 m long and 5 cm in diameter. Mount the tubes as shown in figure facing metal plate as shown in fig. Place a watch at the mouth of the tube A and try to hear the sound by applying ear close to the end of the tube B.

Place a screen S made of cardboard or of some other absorbing material in between the two tubes to prevent sound from reaching our ear directly. It will be observed that the sound is maximum when angles made by tubes A and B with normal are equal i.e., \(\angle i=\angle r\)

Question 7.
List the three characteristics of sound waves. State the factors on which each of these characteristics depends.
Answer:
Characteristics of sound: The three characteristics of sound are

1. Loudness
2. Pitch and
3. Quality or timbre.

1. Loudness: It is the response differently i.e., one sound louder than the other of ear to the intensity of sound. It distinguishes between a loud sound and soft (low) sound. Even two sounds of equal intensity, may hear. Loudness depends on two main factors.
Factors on which sound depends:
(a) Intensity of sound
(b) Sensitivity of ear.
Graphs given below show the wave shape of a loud sound and a soft sound having the same frequency.

2. Pitch: Pitch is the sensation which helps a listener to distinguish between a high and a low note. Pitch depends on frequency. The faster the vibration of the source of sound, the higher is the frequency and higher is the pitch,

The voice of a child or a lady is shriller than that of a man i.e., the pitch of a lady’s sound is higher than that of a man. The mosquito’s sound is of high frequency and hence high pitch.

3. Quality or timbre. The quality or timber of sound is that characteristic which helps us, to distinguish one sound from another having the same pitch and loudness. It is due to the quality of sound that one can recognise the voice of friend without seeing him.

Short Answer Type Questions:

Question 1.
What is periodic motion? Give three examples.
Answer:
Periodic motion: The motion of a body that repeats itself regularly after a fixed interval of time is called periodic motion. Such type of motion is vibratory motion.
Examples of periodic motion.

1. Motion of the earth around the sun.
2. The motion of a swing which moves to and fro (left and right) about its mean position.
3. The motion of a simple pendulum.
4. The motion of the hands of a clock.

Question 2.
Define oscillatory motion. Give examples.
Answer:
Oscillatory or vibratory motion. If a body moves to and fro repeatedly about a fixed position (called mean position), its motion is said to be oscillatory or vibratory motion.
Examples of oscillatory motion:

1. Motion of the pendulum of a wall clock.
2. Motion of a swing.

Question 3.
Differentiate between transverse waves and longitudinal waves.
Answer:
Differences between transverse and longitudinal waves:

Transverse waves	Longitudinal waves.
1. In transverse waves, the particles of the medium vibrate perpendicular to the direction of wave motion.	In longitudinal waves, the particles of the medium vibrate along the direction of wave motion.
2. These waves travel in the form of alternate crest and troughs.	These waves travel in the form of alternate compressions and rarefactions.
3. These waves can be transmitted through solids or liquid surfaces.	These waves can be transmitted through all the three media, viz (i) solids, (ii) liquids and (iii) gases.
4. They do not cause pressure changes in the medium through which they pass. Example. Waves formed over water surface.	They cause changes in the pressure of the different parts of the medium through which they pass. Example. Sound waves in air,

Question 4.
How is sound propagated? Can it be propagated through vacuum? out of solid, liquid and gas in which medium speed of sound is maximum and in which it is least?
Answer:
The propagation of sound is in the form of transverse waves therefore sound waves consist of compressions and rarefactions. The source of sound is always in vibrating state. The sound emitted by a vibrating source can always propagates through a medium.

Sound can not travel through vacuum. Transverse waves can travel through all the three mediums solid, liquid and gas. This happens due to elasticity of the medium. Solids are more elastic than liquids and gases. It has been proved experimentally that the speed of sound is maximum in solids the least in gases.

Question 5.
What are applications of ultrasound?
Answer:
Applications of ultrasound:

1. Glaton whistle. It is used by hunters. When hunter and hound (hunting dog) get separated an hunter wants to call back dog to help him catch the prey, he blows the Glaton whistle which produces only ultrasonic waves.
2. These waves can be heard by dog but not by other animals and birds of the forest.
3. Bats judge the distance of prey or the coming obstacle by sending these waves. By observing the time taken by waves to travel back, they can find the distance of the obstacle/ prey.
4. Ultrasound waves are used to find the depth of the sea.
5. Ultrasonic waves are used by doctors for scanning different parts of the body.
6. These waves are used by dentists to compress the silver filled in the cavity of teeth.
7. These are used to clean parts located in hard-to-reach places. Objects are cleaned in cleaning solution and ultrasonics are passed in the solution.
8. Particles of dust, grease get detached due to high-frequency vibrations and get into cleaning solution.

Question 6.
Define the terms wave and wave motion.
Answer:
Wave and wave motion. A wave is a pattern of disturbance which travels through a medium due to repeated vibrations of the particles of the medium, the disturbance being handed over from one particle to the next. The motion of the disturbance is called wave motion.

Question 7.
Distinguish between a wave pulse and a periodic wave.
Answer:

Pulse	Periodic wave
1. A pulse is a wave produced by a sudden disturbance of short duration.	A periodic wave produced by continuous and regular vibrations of the particles of the medium.
2. Due to pulse, the medium oscillates for a short time and then returns to its undisturbed position.	Due to periodic wave, medium vibrates for a long time after being disturbed.
3. It is not repetitive.	It repeats itself after a fixed interval.
4. It is formed in a small portion of the medium.	It spreads over the entire length of the medium.

Question 8.
What are mechanical or elastic waves? Give examples.
Answer:
Mechanical waves: The waves which require a material medium for their propagation are called mechanical waves. They are also called elastic waves because their propagation depends on the elastic properties of the medium.

Examples of mechanical waves:

1. Sound waves in air.
2. Waves over water surface
3. Waves produced during an earthquake. These are known as seismic waves.

Question 9.
State two factors on which the speed of sound depends.
Answer:
The speed of sound through a medium depends on following two factors:

• Nature of the medium.
• Temperature of the medium.

Question 10.
Explain in brief the dependence of speed of sound on nature of material medium and temperature.
Answer:
Speed of sound in different media. Sound travels fastest through solids and slowest through gases. This is because elasticity of solids is much greater than that of liquids which in turn, is greater than that of gases.

Effect of Temperature. The speed of sound increases with the increase in temperature of the medium through which sound travels.

Question 11.
Define the terms time period and frequency of an oscillating body. Give their units and write the relation between them.
Answer:
Time period: The time taken by an oscillating body to complete one oscillation is called its time period. It is denoted by T. Its SI unit is second(s).
Frequency: The number of oscillations or vibrations completed by an oscillating body in one second is called its frequency. It is denoted by ν (Greek letter nu).
SI unit of frequency = per second (s^-1) = cycles per second (cps)
= hertz (Hz).
Relation between time period and frequency.
Let T = time period of an oscillating body. Then,
number of oscillations completed in “T second = 1
number of oscillations completed in 1 second = \(\frac {1}{T}\)
But number of oscillations completed in 1 second = frequency (ν)
∴ ν = \(\frac {1}{T}\)
Hence frequency is equal to the reciprocal of time period.

Question 12.
Give two practical applications of reflection of sound waves.
Answer:
Two practical applications of reflection of sound waves:
1. Reflecting boards. In large halls or auditorium large wooden boards are fixed behind the speaker in the form of a concave cylinder. The sound waves on striking the reflecting boards get reflected parallel to the principal axis reaches everyone in the auditorium so that everyone can hear clearly. The working of reflective board is based on reflection.

2. Ear trumpet. The sound energy received by the wide end of the trumpet is reflected into much smaller area thereby increasing the intensity of sound. Hence, a person who is hard of heamg, can hear the sound distinctly.

Question 13.
Distinguish between the terms music and noise.
Answer:

• Music: The sound which has a pleasing sensation to the ears is called music. It is produced by regular and periodic vibrations, without any sudden change in loudness.
• Example: The sound produced by plucking the string of a sitar, sound from a tabla, etc.
• Noise: The sound which has a displeasing on the ears is called noise. It is produced by vibrations at irregular intervals and with sudden change in loudness.
• Example: The sound produced by an explosion.

Question 14.
How can bats ascertain distances, directions, nature and size of the obstacles without eyes?
Answer:
Bats can produce and receive ultrasonic waves. During flight, a bat emits ultrasonic waves. The bat receives back these waves after being reflected by the obstacle in its path. From the time interval between transmission and reception of ultrasonic waves, the bat gets information about the distance, nature of obstacle and its direction of location. Hence bats can move about freely even in total darkness.

Question 15.
It is observed that some animals get disturbed before earthquake. How?
Answer:
Earthquakes produce low-frequency infrasound before the mainshock waves begin. These infrasonic waves are out of our audible range. But animals are able to detect these waves and hence some animals get disturbed before earthquakes.

Important Formulae:

1. Wave Velocity (υ) = ν × λ
2. Frequency (ν) = \(\frac {1}{T}\)
3. Wave length (λ) = v × T
4. Total Distance = Velocity × Time

Numerical Problems (Solved):

Question 1.
What will be the frequency of Mohan’s heart when it beats 75 times in 1 minute?
Solution:
Time taken for 75 heart beats = 1 min
= 60 s
and time taken for 1 heartbeat = \(\frac {60}{75}\)
∴ Time period (T) = 0.80 s
We know, frequency (ν) = \(\frac{\text { 1 }}{\text { Time Period(T) }}\)
= \(\frac {1}{0.80}\)
= 1.25 Hertz (Hz)

Question 2.
A boat strikes waves of ocean having crest 100 m away. The wave velocity of crest is 20 m s^-1. What is the frequency of waves striking the boat?
Solution:
Wave length (λ) = 100 m
Wave velocity (υ) = 20 m/s
Frequency (ν) = ?
We know, υ = ν × λ
20 = v × 100
or ν = \(\frac {20}{100}\)
= 0.2 Hertz (Hz)

Question 3.
A source of wave produces 40 crests in 0.4 s. Find the frequency of wave.
Solution:
A wave cycle consists of one crest and one trough.
Since the given source produces 40 crests, it produces 40 wave cycles in 0.4 s.
No. of wave cycles produced in 0.4 s = 40
No. of wave cycles produced in 1 s = \(\frac {40}{4}\)
= 100
∴ Frequency of wave = 100 Hz

Question 4.
A source produces a sound of wavelength. 1.7 x 10^-2 m. If its velocity is 343.4 m s -1, then find frequency of sound,
Solution:
Velocity, υ = 343.4 ms^-1
Wave length, λ = 1.7 × 10^-2 m;
Frequency, ν = ?
We know, υ = ν × λ
ν = \(\frac {υ}{λ}\)
= \(\frac{343.4}{1.7 \times 10^{-2}}\)
∴ Frequency, ν = 2.02 × 10⁴ Hz

Question 5.
What will be the frequency of the wave, if its time period is 0.05 s?
Solution:
Here Time Period T = 0.05 s
Frequency, ν = ?
But Frequency, ν = \(\frac{\text { 1 }}{\text { Time Period(T) }}\)
= \(\frac {1}{0.05}\)
= \(\frac {100}{5}\)
= 20Hz

Question 6.
A distance displacement of a periodic wave is shown in a graph, if velocity of the wave is 320 m s^-1 , then find (a) wavelength (b) frequency.
Solution:
(a) Wavelength = Distance between two consecutive crests
= 50 – 10
= 40 cm
= \(\frac {40}{100}\)m/s
= 0.4 m

(b) Velocity of wave, υ = 320m/s
Wavelength, λ = 0.4 cm
Frequency, ν = ?
We know, υ = ν × λ
or ν = \(\frac {υ}{λ}\)
= \(\frac {320}{0.4}\)
Frequency, ν = 800 Hz

Question 7.
Longitudinal waves is produced on a spring. This wave travels with a velocity of 30 cm/s and its frequency is 20 Hz. What is the minimum distance between two consecutive compressions?
Solution:
Here, wave velocity, υ = 30 cm/s
= 0.30 m/s
Frequency, ν = 20 Hz
Wave length, λ = Distance between two consecutive compressions = ?
υ = ν × λ
or λ = \(\frac {υ}{ν}\)
= \(\frac {0.30}{20}\) = 0.15 m

Question 8.
A message was transmitted from boat which returned to the sender after reflection from the bottom of the sea in 0.8 s. If the velocity of sound in water is 1500 ms^-1 then find the depth of sea.
Solution:
Velocity of sound, υ = 1500 ms^-1
Time taken, t = 0.8 s
Distance travelled by sound = Velocity of wave × Time
Total distance travelled by sound (2d) = 1500 × 0.8 = 1200 m
∴ Depth of sea = \(\frac {1200}{2}\)
= 600 m

Question 9.
The frequency of a tuning fork is 600 Hertz. What will be its time period?
Solution:
Time period of tuning fork = \(\frac{\text { 1 }}{\text { Frequency of Tuning fork }}\)
= \(\frac {1}{400}\)
= 0.0025 s

Question 10.
A stone is dropped in a 44.1 m deep well. If the sound produced by striking of stone with the water surface is heard after 3.13 s then find the velocity of wave in air.
Solution:
Given : Depth of well(h) = 44.1 m
Acceleration due to gravity (g) = 9.8 m s^-2
Time taken by stone go to the surface (t)= 313 s of water and sound to return
Suppose stone takes t₁ time to reach the water surface, then from

Question 11.
A man claps near a cliff and echo is heard after 5 s. If the velocity of sound is 346 m s^-1, then what will be the distance between the man and the cliff?
Solution:
Velocity of sound (v) = 346 m s^-1
Time taken for echo to be heard (t) = 5 s
Distance travelled by sound (S) = υ × t
= 346 m s^-1 × 5 s
= 1730 m
Sound took 5 s to travel twice the distance between man and cliff
Distance between man and cliff = \(\frac {1730}{2}\)
= 865 m

Question 12.
A ship produces ultrasonic sound which is collected in 3.42 s after reflection from the surface of sea. If the velocity of ultrasonics is 1531 m s^-1, then what is the distance of sea surface from sea?
Solution:
Time taken from transmission to collection of sound (t) = 3.42 s
Velocity of ultrasonics in sea water (υ) = 1531 m s^-1
Distance travelled by transmitted sound= 2d where d is the depth of sea
2d = velocity of sound × time
2d = 1531 m s^-1 × 3.42 s
2d = 5236 m
d = \(\frac {5236}{2}\)
= 2618 m
∴ Distance of ship from sea surface (d) = 618 m

Very Short Answer Type Questions:

Question 1.
What is sound?
Answer:
Sound: It is a kind of energy which produces in us the sensation of hearing.

Question 2.
In which medium the velocity of sound is more – Solids or Gases?
Answer:
In solids, velocity of sound is more i.e. sound travels faster in solids than in gases.

Question 3.
What frequency of sound is audible to human ear?
Answer:
Human ear can hear sound of frequency from 20 Hz to 20,000 Hz.

Question 4.
What is the nature of sound Longitudinal wave or Transverse wave?
Answer:
Longitudinal wave.

Question 5.
What should be the properties of the medium for producing sound waves?
Answer:
The medium should have the property of (i) inertia and (ii) elasticity.

Question 6.
What is the relation between frequency, wavelength and wave velocity?
Answer:
Wave velocity = Frequency × Wavelength.

Question 7.
What is the unit of frequency?
Answer:
Hertz (Hz).

Question 8.
What is the relation between frequency (ν) and time period (T)?
Answer:
ν = \(\frac {1}{T}\)

Question 9.
On dropping a pebble in still water, what type of waves are produced on the surface of water?
Answer:
Transverse waves.

Question 10.
What kind of sound waves are produced in air?
Answer:
Longitudinal waves.

Question 11.
What is the full form of SONAR?
Answer:
The full form of SONAR is Sound Navigation and Ranging.

Question 12.
What is seismograph for?
Answer:
Seismograph is a device used to measure intensity of earthquake.

Question 13.
Which scale measures the intensity of earthquake measured?
Answer:
Richter scale.

Question 14.
Earthquake of what intensity is considered safe on Richter Scale.
Answer:
Earthquake of intensity upto 5 on Richter Scale is considered safe.

Question 15.
What is the cause for production of sound?
Answer:
Vibrations.

Question 16.
What is the time for persistence of hearing?
Answer:
It is \(\frac {1}{10}\) = 0.1 s.

Question 17.
What is the velocity of sound on moon?
Answer:
Sound cannot travel on moon because moon has no atmosphere.

Question 18.
Which animal can hear infrasonics?
Answer:
Elephant.

Question 19.
What is audible range for human beings?
Answer:
20 Hz to 20,000 Hz.

Question 20.
What is the minimum distance of the obstacle from the source of sound for hearing distinct echo?
Answer:
17.2 m.

Question 21.
Which has a higher pitch, whistle or a drum?
Answer:
Whistle has higher pitch.

Question 22.
A violin and a sitar may have the same frequency, yet we can distinguish between their notes. Why?
Answer:
This is on account of the difference in quality (timbre) of sound produced by them.

Punjab State Board PSEB 9th Class Science Important Questions Chapter 11 Work and Energy Important Questions and Answers.

PSEB 9th Class Science Important Questions Chapter 11 Work and Energy

Long Answer Type Questions:

Question 1.
What is potential energy? Deduce an expression for the gravitational potential energy of a body of mass’m’ placed at height ‘h’ above the ground. Give some practical examples.
Answer:
Potential Energy: It is the energy possessed by an object when it is raised above or below the surface of earth.
Mathematical Expression: Consider a body of mass ‘m’ raised to a height ‘h’ above the surface of earth to a position ‘B’. In order to lift the body to this height we have to apply minimum force equal to mg, the weight of the body in the upward direction from position ‘A’.

∴ Work done on the body against the force of gravity (W) = Force × Displacement
W = F × S
But F = mg and S = h
W = mg × h
= mgh

The work done (W = mgh) on the body is against the force of gravity
∴ E_p = W = mgh
Since this work done is against the force of gravity, so this energy is called gravitational potential energy. This potential energy does not depend on the path along which the body is moved.

Practical Examples:
1. In olden days there were no digital watches and the watches used to be wound up. When its spring is open then the watch does not work. When its spring is wound up then it starts working.

As is shown in Fig. on winding the spring closes and potential energy is stored up. This potential energy due to change in size and shape of the spring helps to move the hands of watch.
2. Try to open a spring with your both hands. Work is, therefore, being done by your hands. By doing this the spring extends in its length and as such potential energy is stored in it.

Question 2.
What is kinetic energy? Derive a mathematical expression for kinetic energy.
Answer:
Kinetic Energy: It is defined as the energy possessed by a body due to velocity, if body is not in motion then it does not possess kinetic energy.
Mathematical Expression: Suppose a football of mass ‘m’ is at rest and force ‘F’ is applied. Due to the force acting on it, the football covers a distance ‘S’ in’t’ seconds and attains a velocity υ. Thus, acceleration produced in the football is ‘a’

Work done on the football W = F × S ………..(1)
According to Newton’s second law of motion,
F = m × a …………. (2)

Equation (8) proves that the work done on the football is stored in it as kinetic energy.
∴ kinetic energy stored = Work done = \(\frac {1}{2}\)mv²

Question 3.
What is law of conservation of energy? Explain the law with the help of an example and prove its reality.
Answer:
Law of conservation of energy: According to this law total energy always remain constant or total energy always remains the same. Though one form of energy can be transformed to some other form of energy but total energy still remains constant.

Explanation with example: Throw a ball vertically upwards from the surface of earth. You do some work on the ball. This work gets stored in the form of potential energy. This is called gravitational potential energy. As the ball moves up and up its velocity continue to decrease and its kinetic energy continue to decrease and potential energy goes on increasing. When ball reaches the maximum height its kinetic energy becomes minimum i.e. zero but potential energy becomes maximum. We can prove mathematically that at any position in the path of motion of the ball the sum of potential energy and kinetic energy remains same.

Mathematical Proof for law of conservation of energy:
Consider a ball of mass 10 kg situated at a point 30 m above the surface of earth i.e. point A. Let it be dropped from point A or shown in fig.
At point A
potential energy of ball (P.E.) = m × g × h
= 10 × 10 × 30
= 3000 J
Since, ball is at rest initially thus, at point A it kinetic energy (K.E.) = 0
∴ Total mechanical energy of ball = P.E. + K.E.
= 3000 + 0 = 3000 J ………….(i)

At point B
Ball is above the surface of earth by 20 m.
potential energy (P.E.) = m × g × h
= 10 × 10 × 20
= 2000 J
Using υ² – u² = 2gh
υ² = u² + 2gh
υ² = 0 + 2 × 10 × 10
∴ υ² = 200
∴ Kinetic energy at point B (K.E.) = \(\frac {1}{2}\)mυ²
= \(\frac {1}{2}\) × 10 × (200)
= 1000 J
∴ Total mechanical energy of ball at point B = potential energy + kinetic energy
= 2000 + 1000
= 3000 J. ……………… (ii)
Similarly, at point C and D total mechanical energy will be 3000 J.
These equation prove that total energy is always conserved.

Question 4.
If the force acting on the object is not in the direction of motion then how will you consider the work done? Explain giving example and also tell when will the work done be minimum and when it will be maximum?
Answer:
When the force acting on the object is not in the direction of motion. A gardener pushes the lawn mower in the forward direction. He applies force F on the handle HH’ facing the ground.

As is clear from the figure, the gardener is not applying force in the horizontal direction but applies force in a direction making angle θ with the direction of force. In such situation the force acting on the machine drives the roller in the horizontal direction from A to B.
Here the effective force is F cos θ and not F and the vertical component F sin θ balances the weight of the lawn mower.
∴ Work done by the lawnmower (W) = Component of Force × Displacement
= F cos θ × S
1. When the force acts along the direction of displacement then θ = 0° and cos θ = cos 0° = 1
∴ W = F cos θ × S
= F × 1 × S
W = F × S
This time the work done will be maximum.

2. When the force acting on the object is in a direction perpendicular to the direction of motion then θ = 90° and cos θ = cos 90° = 0
∴ W = F cos θ x S
= F cos 90° x S
= F x 0 x S
⇒ W = 0
This time the work done is minimum.

Short Answer Type Questions:

Question 1.
What is work? How can you calculate it? Also give the unit of work.
Answer:
Work: Work done by a force or by an object in the product of applied force and the displacement taking place in the direction of applied force.
If F = force acting on the body
S = displacement of the body in the direction of force
Work done by force, W = F × S ……(i)
we know, when a force acts on a body acceleration is produced in the body.
Thus, if m = mass of body
a = acceleration produced in the body then from Newton’s second law of motion
F = m × a ….(ii)
Using (i) and (ii),
W = m × a × S ………….(iii)

Unit of work: When force is measured in Newton (N) and displacement in metre (m), the
work = Newton × Metre
W = N × m
W = Joule (J)
∴ S.I. unit of work is Joule (J) and C.G.S. unit is erg.
1 Joule = 10⁷ erg.

Question 2.
Show by giving an example that if force acting in the body does not produce any displacement then the work done will be zero?
Answer:
This statement can be understood by the following example.
If a child tries to push a car by applying his, maximum force but the car does not displace even through 1 centimeter, then in the language of physics we can say that the child has done no work.
In this situation suppose the child applies force F and displacement S = 0 then
∴ Work done by the child, W = F × S
= F × 0
⇒ W = 0

Question 3.
A stone tied to one end of string is moved in a circle. How much work is done by the centripetal force in this circular motion?
Answer:
A stone is made to move in a circle as shown the fig. then the finger with which you hold the string will experience some force. The force acting on the moving stone is known as centripetal force.

This force acts radially inwards towards the centre of the circle. If in this situation the stone gets detached then it will fly off in the direction tangential along AT or BT as shown in fig. Now, the centripetal force acts perpendicular to the direction of motion, thus no work is done by the centripetal force.

Question 4.
What is power? Write its SI unit also.
Answer:
Power. Harish and Karan climbed up a tree to pluck 60 mangoes each. They started climbing at the same moment of time. Harish got his 60 mangoes in 30 minutes whereas Karan got his 60 mangoes in 60 minutes. That means Karan took more time to do the same work. Both did the same work, both had same energy but their power was not same i.e. Power of Harish is more than that of Karan.

Power: Power of an object or a machine is its time rate of doing work i.e. Power is defined as rate of doing work with time.
∴ Power = \(\frac{\text { Work done }}{\text { time taken }}\)
or P = \(\frac{\text { w }}{\text { t }}\)
If work is measured in joule and time in second then unit of power is ‘watt’.
∴ 1 watt = \(\frac{\text { 1 Joule }}{\text { 1 second }}\)
commercial unit of power is kilowatt.
1 kilowatt = 1000 watt = 1000 joule/second

Question 5.
Prove by an experiment that mechanical energy can be transformed into heat energy.
Or
When we hammer a nail into a wooden block the nail gets heated up? Why?
Answer:
Place a nail on a wooden block and hammer it, some part of the nail goes into the wooden block, when nail is completely into the block due to hammering, when nail in further hammered then nail, hammer and block all gets heated up. When hammer is lifted up for hammering, its potential energy increases due to its position.

When it falls on the nail then its whole energy is transformed into kinetic energy of the nail and the nail goes into the wooden block. When nail is completely embedded into the block then mechanical energy of the hammer converts into heat energy of the nail, block and hammer meaning thereby that mechanical energy gets converted or transformed into heat energy.

Question 6.
Differentiate between Potential Energy and Kinetic Energy.
Answer:
Difference between Potential energy (P.E.) and Kinetic energy (K.E.)

Potential Energy	Kinetic Energy
1. The potential energy of an object depends upon its position and size.	The kinetic energy of an object is due to its motion or velocity.
2. Potential energy (P.E.) = m × g × h	Kinetic energy (K.E.) = \(\frac {1}{2}\)mυ2
3. Potential energy of an object depends upon its height above the ground or its depth below the ground surface.	The kinetic energy of an object depends upon its velocity.

Question 7.
A horse and a dog are running with the same velocity. If the mass of horse is ten times the mass of the dog then what will be ratio of their kinetic energy?
Solution:
Suppose the mass of dog = m
∴ Mass of the horse = 10m
Velocity of horse = Velocity of dog = υ (say)

Question 8.
Two masses m and 2m are dropped from height h and 2h. On reaching the ground, which will have a greater kinetic energy and why?
Answer:
K.E. of mass m = P.E. lost by mass m = mgh
K.E. of mass 2m = P.E. lost by mass 2m
= 2m × g × 2h = 4mgh
Hence, mass 2 m will have a greater kinetic energy on reaching the ground.

Question 9.
Two objects having same mass’m’ are moving with velocities υ and 2υ. Find ratio of their kinetic energies.
Answer:
Let bodies A and B each have same mass m and their respective velocities are υ and 2υ

Question 10.
In what conditions the physical capacity of a man to do work decreases?
Answer:
After sickness and in old age the physical capacity of a man to do work decreases because the energy of his body muscles becomes less.

Question 11.
Explain clearly the difference between energy and power with the help of an example.
Answer:

• Energy: Energy is the ability to do work and is equal to the magnitude of total work which could be done by the energy. It has no relation with the time.
• Power: It is the rate of doing work. It has no relation with the magnitude of total work.
• Example: A worker taken one hour to complete a piece of work, whereas the other worker takes 2 hours to complete the same piece of work. In this case, both the workers are doing the same work i.e. both are consuming same amount of energy.

But first worker did the same work in half the time as compared to other worker thus. First worker has doubled the power than the other worker.

Important Formulae:

1. Kinetic Energy = \(\frac {1}{2}\) mυ²
2. Potential Energy = mgh
   Here m = Mass of the object; g = Acceleration due to gravity, h = Height
3. Work (W) = Force (F) × Displacement (S)
4. Power (P) = \(\frac{\text { Work Done (W) }}{\text { Time taken (t) }}\)
   Or Power (P) = \(\frac{\text { F × S }}{\text { t }}\)
   = F × \(\frac{\text { S }}{\text { t }}\) (∵ υ = \(\frac{\text { S }}{\text { t }}\))
   = F × υ
5. 1 Joule = 1 Newton × 1 metre
6. 1 watt = \(\frac{\text { 1 Joule }}{\text { 1 Second }}\)
7. Horse Power = 746 watt
8. 1 Kilowatt hour = 36,00,000 Joule = 3.6 × 10⁶ Joule
9. 1 watt hour = 3600 Joule

Numerical Problems (Solved):

Question 1.
J of energy is applied to lift a box of mass 0.5 kg. How much high it would be raised?
Solution:
Potential energy (P.E.) = 1 J
Mass of the box (m) = 0.5 kg
Acceleration due to gravity, (g) = 10 m/s²
Height to which box is raised, (h) = ?
We know, potential energy (P.E.) = m × g × h
1 = 0.5 × 10 × h
1 = 5 × h
or h = \(\frac {1}{5}\) = 0.2
∴ Height to which box would be raised, (h) = 0.2 m

Question 2.
A woman pulls water-filled bucket weighing 5 kg from a well 10 m deep in 10 s. What is her power?
Solution:
Mass of water filled bucket, (m) = 5 kg
Depth of well(h) = 10 m
Acceleration due to gravity, (g) = 10 m s^-2
Work done by woman(W) = P.E
= m × g × h
= 5 × 10 × 10
= 500 J
Time (t) = 10 s
Now, power = \(\frac{\text { Total work done(W) }}{\text { Total time taken(t) }}\)
= \(\frac {500J}{10s}\)
= 50 J/s
= 50 watt

Question 3.
A boy weighing 50 kg climbs up a vertical height of 100 m. Calculate the amount of work done by him. How much potential energy does he gain? Take r = 9.8 m/s²?
Solution:
Here, m = 50kg, g = 9.8 m/s², h = 100m
Work done by the body = mgh
= 50 × 9.8 × 100
= 49000 J
= 4.9 × 10⁴ J
Gain in P.E. = Work done = 4.9 × 10⁴ J

Question 4.
A man drops a 10 kg rock from the top of a 20 m ladder. What will be its kinetic energy when it reaches the ground? What will be its speed just before it hits the ground? Does the speed depend on the mass of the rock? (Take g = 10 m s^-2)
Solution:
Here u = 0, m = 10 kg, h = 20 m, g = 10 m s^-2
We know, υ² – u² = 2gh
∴ υ^-2 – 0^-2 = 2 × 10 × 20
or υ = \( \sqrt{{400}} \)
= 20 m s^-1
K.E. = \(\frac {1}{2}\)mυ²
= \(\frac {1}{2}\) × 10 × (20)²
= 2000 J
Speed does not depends on the mass of the rock because the acceleration due to gravity under which the rock falls does not depend on mass.

Question 5.
A rocket of 3 × 10⁶ kg mass takes off from a launching pad and acquires a vertical velocity of 1 km/s at an altitude of 25 km. Calculate the potential energy, and the kinetic energy. (Take g = 10 m s^-2).
Solution:

Question 6.
An electric heater of 1000 W is used for 2 hours a day. What is the cost of using it for a month of 28 days, if 1 unit costs ₹ 3.00?
Solution:
Here, P = 1000 W
= 1 kW
Total time, t = 2 × 28 hours
= 56 hours
Total energy consumed = P × t
= 1kW × 56h
= 56kWh
∴ Cost of 1 kWh = ₹ 3.00
∴ Cost of using electricity for Feb = 3 × 56
= ₹ 168.

Question 7.
The power of a motor pump is 5 kW. How much water per minute the pump an raise to height of 20 m? Take g = 10 ms^-2.
Solution:
Here, P = 5 kW = 5000 W, t = 1 min = 60s, h = 20 m, g = 10 ms^-2

Question 8.
Calculate the electricity bill amount for the month of November of a family if 4 tube lights of 40 W each for seven hours, a TV of 150 W for three hours and two bulbs of 60 W each for four hours are used per day. The cost per unit is ₹ 3.50.
Solution:
Power of each tube light = 40 W
∴ Total power of 4 tube lights = 4 × 40 W
= 160 W
Energy consumed by 4 tube lights each day
= 160 W × 7 h
= 1120 Wh
= \(\frac {1120}{1000}\) = 1.112kWh
Energy consumed by T.V. per day = 150W × 3h
= 450 Wh
= \(\frac {450}{1000}\) = 0.45 kWh
Energy consumed by 2 bulbs per day = 60W × 4h × 2
= 480 Wh
= \(\frac {480}{1000}\) = 0.48 kWh
Total energy consumed per day by all appliances = 1.12 + 0.45 + 0.48 = 2.05 kWh
Total energy consumed in 30 days = 2.05 kWh × 30
= 61.50 kWh
Cost of 1 kWh = ₹ 3.50
Cost of 61.50 kWh = 3.50 × 61.50
= ₹ 215.25

Question 9.
A person carrying 10 bricks each of man 2.5 kg. on his head moves to a height 20 metres in 50 seconds. Calculate the power spent in carrying bricks of the person. (g = 10 ms^-2)
Solution:
Total mass (m) = 10 × 2.5 kg
= 25 kg.
Time (t) = 50 s
Displacement (S) = 20m
g = 10 ms^-1
Force exerted by person (F) = mg
= 25kg × 10 ms^-2
= 250 N
Work done = Force × Displacement
= F × S
= 250 N × 20 m
= 5000 J
Power = \(\frac{\text { Work done }}{\text { Time }}\)
= \(\frac {5000J}{50s}\)
= 100 w.

Question 10.
A car of 1000 kg moving with a velocity of 30 m/s stops with uniform acceleration after covering a distance of 50 m on application of brakes. Find the force applied by the brakes on the car and also work done.
Solution:
u = 30 ms^-1
υ = 0
S = 50m
Now υ² – u² = 2aS
(0)² – (30)² = 2 × a × 50
– (30 × 30) = 100 × a
a = –\(\frac {900}{100}\)
∴ a = – 9m/s²
Force, F = m × a
= 1000 × 9
= 9000 N
work done by the brakes, W = F × S
= 9000 × 50
= 450000 N-m
= 4.5 x 10⁵ J

Question 11.
A freely falling hammer of mass 1 kg 7 falls on a nail fixed in a block of wood. If the hammer falls from a height of 1 m then what will be the kinetic energy just before striking the nail? (Take g = 10 m s^-2)
Solution:
Mass of the hammer, (m) = 1kg
Height of the hammer (h) = 1 m
Acceleration due to gravity (g) = 10ms^-2
using υ² – u² = 2gS
υ² – (0)² = 2 × 10 × 1
υ² = 20 ………..(i)
Now kinetic energy of hammer (K.E.) = \(\frac {1}{2}\) × m × υ²
= \(\frac {1}{2}\) × 1 × 20 [From (i)]
= 10 J

Question 12.
A car is moving with a speed of 54 km/h. What will be the kinetic energy of the boy of mass 40 kg sitting in the car?
Solution:
Speed of the boy = speed of the car
= 54 km/h
= 54 × \(\frac {5}{18}\)m s^-1
= 3 × 5 m s^-1
= 15 m s^-1
Mass of the boy (m) = 40kg
∴ kinetic energy (K.E.) of the boy = \(\frac {1}{2}\)mυ²
= \(\frac {1}{2}\) × 40 × (15)²
= 20 × 15 × 15
= 4500 J

Question 13.
1 Joule of energy is used for one heart beat. Calculate the power of the heart if it throbs 72 times in one minute.
Solution:
Work done in 1 heart beat = 1 J
∴ Total work done by heart in 72 beats = 72 × 1 J
= 72 J
Time (t) = 1 min
= 1 × 60 s
= 60 s
Power = \(\frac{\text { Work done }}{\text { Time taken }}\)
= \(\frac {72 J}{60s}\)
= 1.2 J/s
= 1.2 Watt

Very Short Answer Type Questions:

Question 1.
State the relation between commercial unit of energy and joules.
Answer:
1 commercial unit of energy (or 1 kWh) = 3.6 × 10⁶ Joule.

Question 2.
How much work is done on a body of mass 1 kg whirling on a circular path of radius 5m?
Answer:
Work done is zero.

Question 3.
What is the SI unit of power?
Answer:
The SI unit of power is Watt.

Question 4.
A ball is thrown vertically upwards. Its velocity keeps on decreasing. What happens to its kinetic energy when it reaches the maximum height?
Answer:
The kinetic energy of the body changes into its potential energy.

Question 5.
If the heart works 60 joules in one minute, what is its power?
Answer:
Power = \(\frac{\text { Energy }}{\text { Time }}\)
= \(\frac {60J}{60s}\)
= 1 W

Question 6.
Name the term used for the sum of kinetic energy and potential energy of a body.
Answer:
Mechanical energy.

Question 7.
How many joules make one-kilowatt hour?
Answer:
1 kilowatt-hour = 3.6 × 10⁶ J.

Question 8.
What should be the change in velocity of a body required to increase its kinetic? energy to four times of its initial value?
Answer:
The velocity of the body should doubled at constant mass.

Question 9.
Under what conditions the work done by a force is zero inspite of displacement being taking place?
Answer:
When displacement is in a direction perpendicular to the applied force.

Question 10.
What is the power of a machine which does 2000 joules of work in 10 seconds?
Solution:
Power of a machine = \(\frac{\text { Work done }}{\text { time }}\)
= \(\frac {2000J}{10s}\)
= 200 W

Question 11.
What is the SI unit of kinetic energy?
Answer:
Joule.

Question 12.
Water flows down the mountains to the plains. What happens to the potential energy of water?
Answer:
Potential energy of water will decrease. It will change to kinetic energy of water.