Important Questions

Punjab State Board Syllabus PSEB 11th Class Biology Important Questions Pdf in English Medium and Punjabi Medium are part of PSEB Solutions for Class 11.

PSEB 11th Class Biology Important Questions in Punjabi English Medium

• Chapter 1 The Living World Important Questions
• Chapter 2 Biological Classification Important Questions
• Chapter 3 Plant Kingdom Important Questions
• Chapter 4 Animal Kingdom Important Questions
• Chapter 5 Morphology of Flowering Plants Important Questions
• Chapter 6 Anatomy of Flowering Plants Important Questions
• Chapter 7 Structural Organisation in Animals Important Questions
• Chapter 8 Cell: The Unit of Life Important Questions
• Chapter 9 Biomolecules Important Questions
• Chapter 10 Cell Cycle and Cell Division Important Questions
• Chapter 11 Transport in Plants Important Questions
• Chapter 12 Mineral Nutrition Important Questions
• Chapter 13 Photosynthesis in Higher Plants Important Questions
• Chapter 14 Respiration in Plants Important Questions
• Chapter 15 Plant Growth and Development Important Questions
• Chapter 16 Digestion and Absorption Important Questions
• Chapter 17 Breathing and Exchange of Gases Important Questions
• Chapter 18 Body Fluids and Circulation Important Questions
• Chapter 19 Excretory Products and their Elimination Important Questions
• Chapter 20 Locomotion and Movement Important Questions
• Chapter 21 Neural Control and Coordination Important Questions
• Chapter 22 Chemical Coordination and Integration Important Questions

Punjab State Board PSEB 11th Class Physics Important Questions Chapter 7 System of Particles and Rotational Motion Important Questions and Answers.

PSEB 11th Class Physics Important Questions Chapter 7 System of Particles and Rotational Motion

very short answer type questions

Question 1.
(n – 1) equal point masses each of mass m are placed at the vertices of a regular n-polygon. The vacant vertex has a position vector \(\vec{a}\) with respect to the centre of the polygon. Find the position vector of centre of mass. (NCERT Exemplar)
Solution
Suppose, \(\vec{b}\) be the position vector of centre of mass of regular n-polygon. As (n – 1) equal point masses each of mass m are placed at (n – 1) vertices of regular polygon, therefore
\(\frac{(n-1) m b+m a}{(n-1+1) m}\) = 0
⇒ (n – 1)mb + ma = 0
⇒ b = \(\frac{-a}{(n-1)}\)

Question 2.
If net torque on a rigid body is zero, does it linear momentum necessary remain conserved?
Answer:
The linear momentum remain conserved if the net force on the system is zero.

Question 3.
When is a body lying in a gravitation field in stable equilibrium?
Answer:
A body in a gravitation field will be in stable equilibrium, if the vertical line through its centre of gravity passes through the base of the body.

Question 4.
Is centre of mass and centre of gravity body always coincide?
Ans.
No, if the body is large such that g varies from one point to another, then centre of gravity is offset from centre of mass.
But for small bodies, centre of mass and centre of gravity lies at their geometrical centres.

Question 5.
Why is moment of inertia also called rotational inertia?
Answer:
The moment of inertia gives a measure of inertia in rotational motion. So, it is also called rotational inertia.

Question 6.
In a flywheel, most of the mass is concentrated at the rim. Explain why?
Answer:
Concentration of mass at the rim increases the moment of inertia and thereby brings uniform motion.

Question 7.
Does the radius of gyration depend upon the speed of rotation of the body?
Answer:
No, it depends only on the distribution of mass of the body.

Question 8.
Can the mass of body be taken to be concentrated at its centre of mass for the purpose of calculating its rotational inertia?
Answer:
No, the moment of inertia greatly depends on the distribution of mass about the axis of rotation.

Short answer type questions

Question 1.
Does angular momentum of a body in translatory motion is zero?
Solution:
Angular momentum of a body is measured with respect to certain origin.

So, a body in translatory motion can have angular momentum.
It will be zero, if origin lies on the line of motion of particle.

Question 2.
Figure shows momentum versus time graph for a particle moving along x – axis. In which region, force on the particle is large. Why?

Solution:
Net force is given by F = \(\frac{d p}{d t}\)
Also, rate of change of momentum = slope of graph.
As from graph, slope AB = slope CD
And slope (BC) = slope (DE) = 0
So, force acting on the particle is equal in regions AB and CD and in regions BC and DE (which is zero).

Question 3.
Two cylindrical hollow drums of radii R and 2J2, and of a common height h, are rotating with angular velocities ω (anti-clockwise) and ω (clockwise), respectively.
Their axes, fixed are parallel and in a horizontal plane separated by (3R + δ). They are now brought in contact (δ → 0).
(i) Show the frictional forces just after contact.
(ii) Identify forces and torques external to the system just after contact.
(iii) What would be the ratio of final angular velocities when friction ceases? (NCERT Exemplar)
Solution:

(ii) F’ = F = F” where F and F” are external forces through support.
F_net = 0
External torque = F x 3 R, anti-clockwise.

(iii) Let ω₁ and ω₂ be final angular velocities (anti-clockwise and clockwise respectively).
Finally, there will be no friction.
Hence, Rω₁ = 2Rω₂ ⇒ \(\frac{\omega_{1}}{\omega_{2}}\) = 2

Question 4.
Angular momentum of a system is conserved if its M.I. is changed. Is its rotational K. E. also conserved?
Solution:
Kinetic energy of rotation = \(\frac{1}{2}\)Iω² = \(\frac{1}{2}\) (Iω)ω = \(\frac{1}{2}\)Lω

L = Iω is constant, if moment of inertia (I) of the system changes. It means as I changes, then ω also changes.
Hence K.E. of rotation also changes with the change in I. In other words, rotational K.E. is not conserved.

Question 5.
How much fraction of the kinetic energy of rolling is purely
(i) translational, (ii) rotational.
Solution:

Question 6.
Listening to the discussion on causes of pollution and due to which temperature on earth is rising, increase in temperature leads to melting of polar ice, Meenu realised that if each one of us contributed to create pollution free environment, then even small efforts can lead to big results. So, she decided to lead the step and instead of going to school by her car, she joined school bus and also asked her father to go to office using car pool.
(i) What do you think is mainly responsible for global warming?
(ii) If the ice on polar caps of the earth melts due to pollution, how will it affect the duration of the day?
Explain.
(iii) What values does Meenu show?
Answer:
(i) Pollution created by the people of world is the main cause of global warming.
(ii) Earth rotates about its polar axis. When ice of polar caps of earth melts, mass concentrated near the axis of rotation spreads out, therefore moment of inertia, I increases.
As no external torque acts,
∴ L = I = Iω = (\(\frac{2 \pi}{T}\)) = Constant
With increase of I, T will increase i.e., length of day will also increase,
(iii) Meenu is considerate towards environment and her thought of leading the steps to reduce pollution is commendable.

Question 7.
Explain how a cat is able to land on its feet after a fall taking the advantage of principle of conservation of angular momentum?
Answer:
When a cat falls to ground from a height, it stretches its body alongwith the tail so that its moment of inertia becomes high. Since, la is to remain constant, the value of angular speed a decreases and therefore the cat is able to’ land on the ground gently.

Question 8.
A uniform disc of radius R is resting on a table on its rim. The coefficient of friction between disc and table is μ (figure). Now, the disc is pulled with a force \(\overrightarrow{\boldsymbol{F}}\) as shown in the figure. What is the maximum value of \(\overrightarrow{\boldsymbol{F}}\) for which the disc rolls without slipping? (NCERT Exemplar)

Solution:
Let the acceleration of the centre of mass of disc be a, then
Ma = F – f
The angular acceleration of the disc is a = a/R (if there is no sliding).
Then, (\(\frac{1}{2}\)MR²)α = Rf
⇒ Ma = 2f
Thus, f =F/3. Since, there is no sliding.
⇒ f ≤ μ mg ⇒ F ≤ 3μ Mg

Question 9.
Two equal and opposite forces act on a rigid body. Under what condition will the body (i) rotate (ii) not rotate?
Answer:
(i) Two equal and opposite forces acting on a rigid body such that their lines of action do not coincide, constitute a couple. This couple produces the turning effect on the body. Hence, the rigid body will rotate.

(ii) If two equal and opposite forces act in such a way that their lines of action coincide, then these forces cancel out the effect of each other. Hence, the body will not rotate.

Long answer type questions

Question 1.
Find position of centre of mass of a semicircular disc of radius r. (NCERT Exemplar)
Solution:
As semicircular disc is symmetrical about its one of diameter, we take axes as shown. So, now we only have to calculate Y_CM (As X_CM is zero by symmetry and choice of origin).

Now, for a small element OAB, as element is small and it can be treated as a triangle so,
Area of sector OAB = \(\frac{1}{2}\) x r x rdθ
Height of triangle = r
Base of triangle = AB = rdθ
So, its mass dm = \(\frac{1}{2}\)r² dθ.ρ [∵ ρ = \(\frac{\text { mass }}{\text { area }}\)]
As centre of mass of a triangle is at a distance of \(\frac{2}{3}\) from its vertex (at centroid, intersection of medians). So, y = \(\frac{2}{3}\)rsinθ (location of CM of small sector AOB).


So, CM of disc is at a distance of \(\frac{4 r}{3 \pi}\)from its centre on its axis of symmetry.

Question 2.
Obtain an expression for linear acceleration of a cylinder rolling down an inclined plane and hence find the condition for the cylinder to roll down the inclined plane without slipping.
Solution:
When a cylinder rolls down on an inclind plane, then forces involved are (i) Weight mg (ii) Normal reaction N (iii) Friction f
From free body diagam,

From free body diagram,
N – mg cos θ = 0
or N = mg cosθ
Also, if a = acceleration of centre of mass down the plane, then
F_net = ma = mgsin θ – f …………… (i)
As friction produces torque necessary for rotation,
τ = Iα = f R

Punjab State Board PSEB 11th Class Physics Important Questions Chapter 8 Gravitation Important Questions and Answers.

PSEB 11th Class Physics Important Questions Chapter 8 Gravitation

Very Short Answer Type Questions

Question 1.
By which law is the Kepler’s law of areas identical?
Answer:
The law of conservation of angular momentum.

Question 2.
Draw areal velocity versus time graph for mars. (NCERT Exemplar)
Answer:
Areal velocity of planet revolving around the Sun is constant with time (Kepler’s second law).

Question 3.
At what factor between the two particles gravitational force does not depend?
Answer:
Gravitational force does not depend upon the medium between the two particles.

Question 4.
Two particles of masses m₁ and m₂ attract each other gravitationally and are set in motion under the influence of the gravitational force? Will the centre of mass move?
Answer:
Since the gravitational force is an internal force, therefore the centre of mass would not move.

Question 5.
Work done in moving a particle round a closed path under the action of gravitation force is zero. Why?
Answer:
Gravitational force is a conservative force which means that work done by it, is independent of path followed.

Question 6.
What would happen if the force of gravity were to disappear suddenly?
Answer:
The universe would collapse. We would be thrown away because of the centrifugal force. Eating, drinking and in fact all activities would become impossible.

Question 7.
Why a body weighs more at poles and less at equator?
Answer:
The value of g is more at poles than at the equator. Therefore, a body weighs more at poles than at equator.

Question 8.
Give a method for the determination of the mass of the moon.
Solution:
Soli By making use of the relation, gm = \(\frac{G M_{m}}{R_{m}^{2}} \)

Short Answer Type Questions

Question 1.
A planet moving along an elliptical orbit is closest to the Sim at a distance r₁ and farthest away at a distance of r₂.
If v₁ and v₂ are the linear velocities at these points respectively, then find the ratio \(\frac{v_{1}}{v_{2}}\).
Solution:
From the law of conservation of angular momentum
mr₁v₁ = mr₂v₂
⇒ r₁v₁ = r₂v₂ or
\(\frac{v_{1}}{v_{2}}=\frac{r_{2}}{r_{1}}\)

Question 2.
A mass M is broken into two parts, m and (M – m). How is m related to M so that the gravitational force between two parts is maximum?
Solution:
Let =m,m₂ =M – m
F = G\(\frac{m(M-m)}{r^{2}}=\frac{G}{r^{2}}\left(M m-m^{2}\right)\)
Differentiating w.r:t. m, \(\frac{d F}{d m}=\frac{G}{r^{2}}(M-2 m)\)
For F to be maximum, \(\frac{d F}{d m}\) = 0

m₁ = m₂ = M/2

Question 3.
Two stationary particles of masses M₁ and M₂ are a distance d apart. A third particle lying on the line joining the particles, experiences no resultant gravitational force. What is the distance of this particle from M₁?
Solution:
The force on m towards Mi is F =G \(\frac{M_{1} m}{r^{2}}\)
The force on m towards Mi is F = G \(\frac{M_{2} m}{(d-r)^{2}} \)

Equating two forces, we have

So, distance of an particle from m is . r = d
r = d \(\left(\frac{\sqrt{M_{1}}}{\sqrt{M_{1}}+\sqrt{M_{2}}}\right)\).

Question 4.
Aspherical planet has mass M_p and clinometer D_p. A particle of mass m falling freely near the surface of this planet will experience an acceleration due to gravity, equal to whom?
Solution:
Force is given by
F = \(-\frac{G M m}{R^{2}}=\frac{G M_{p} m}{\left(D_{P} / 2\right)^{2}}=\frac{4 G M_{P} m}{D_{P}^{2}}\)
\(\frac{F}{m}=\frac{4 G M_{P}}{D_{P}^{2}}\)

Question 5.
What is the gravitational potential energy of a body at height h from the Earth surface?
Solution:
Gravitational potential energy, i. e.,
U_h = \(-\frac{G M m}{R+h}=-\frac{g R^{2} m}{R+h}\)
[where g = \(\frac{G M}{R^{2}}\) ]
= – \(\frac{g R^{2} m}{R\left(1+\frac{h}{R}\right)}=-\frac{m g R}{1+\frac{h}{R}}\)
.
Question 6.
An artificial satellite is moving in a circular orbit around the Earth with a speed equal to half the magnitude of escape velocity from Earth.
Determine
(i) the height of satellite above Earth’s surface.
(ii) if the satellite is suddenly stopped, find the speed with
which the satellite will hit the Earth’s surface after falling down.
Solution:
Escape velocity = \(\sqrt{2 g R}\), where g is acceleration due to gravity on surface of Earth and R the radius of Earth.
Orbital velocity = \(\frac{1}{2} v_{e}=\frac{1}{2} \sqrt{2 g R}=\sqrt{\frac{g R}{2}} \) …………………. (i)

(i) If h is the height of satellite above Earth

h=R
(ii) If the satellite is stopped in orbit, the kinetic energy is zero and its
potential energy is – \(\frac{G M m}{2 R}\)
Total energy =-\(\frac{G M m}{2 R}\)

Let v be its velocity when it reaches the Earth.
Hence the kinetic energy = \(\frac{1}{2} m v^{2}\)
Potential energy = – \(\frac{G M m}{2 R}\)

Question 7.
Why do different planets have different escape velocities?
Solution:
Escape velocity, v = \(\sqrt{2 g R}=\sqrt{\frac{2 G M}{R}}\)
Thus escape velocity of a planet depends upon (i) its mass (M) and
(ii) its size (R).
As different planets have different masses and sizes, so they have different escape velocities.

Question 8.
Under what circumstances would your weight become zero?
Answer:
The weight will become zero under the following circumstances
(i) during free fall
(ii) at the centre of the Earth
(iii) in an artificial satellite
(iv) at a point where gravitational pull of Earth is equal to the gravitational pull of the Moon.

Long Answer Type Questions

Question 1.
A mass m is placed at P, a distance h along the normal through the centre O of a thin circular ring of mass M and radius r. If the mass is removed further away such that OP becomes 2h, by what factor the force of gravitation will decrease, if h = r? (NCERT Exemplar)

Solution:
Consider a small element of the ring of mass dM, gravitational force between dM and m, distance x apart in figure i.e.,
dF = \(\frac{G(d m) m}{x^{2}}\)

dF can be resolved into two rectangular components.
(i) dF cos θ along PO and
(ii) dF sinθ perpendicular to PO (given figure)
The total force (F) between the ring and mass (m) can be obtained by integrating the effects of all the elements forming the ring, whereas all the components perpendicular to PO cancel out i.e., ∫dFsinθ=O, the component along PO add together to give F i.e.,

Question 2.
A satellite is to be placed in equatorial geostationary orbit around the Earth for communication.
(i) Calculate height of such a satellite.
(ii) Find out the minimum number of satellites that are needed to cover entire Earth so that at least one of satellite is visible from any point on the equator.
[M = 6 x 10 ²⁴ kg, R = 6400 km, T = 24 h, G = 6.67 x 10^-11SI (NCERT Exemplar)
Solution:
(i) As, according co Kepler’s third law, we get
T² = \(\frac{4 \pi^{2} r^{3}}{G M}\)
⇒ r = \( \left(\frac{G M T^{2}}{4 \pi^{2}}\right)^{1 / 3}\)

As we known =R +h
h=r-R
h=4.23 x 10⁷ m – 6.4 x 10⁶ m
h = 3.59 x 10⁷ m

(ii) In ΔOES,cos θ = \(\frac{O A}{O S}=\frac{R}{R+h}\)
= \(\frac{1}{\left(1+\frac{h}{R}\right)}\)
= \(\frac{1}{(1+5.609)}\)
=0.1513
(as,\(\frac{h}{R}=\frac{3.59 \times 10^{7} \mathrm{~m}}{6.4 \times 10^{6} \mathrm{~m}}\) = 5.609)
where, θ ≈ 81° or 2θ = 162°
Number of satellites required to cover entire the Earth.
= \(\frac{360^{\circ}}{162^{\circ}}=2.2\) ≈ 3.

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 Physics Important Questions Chapter 9 Mechanical Properties of Solids Important Questions and Answers.

PSEB 11th Class Physics Important Questions Chapter 9 Mechanical Properties of Solids

Very Short Answer Type Questions

Question 1.
Two identical solid balls, one of ivory and the other of wet clay, are dropped from the same height on the floor. Which will rise to a greater height after striking the floor and why?
Answer:
The ball of ivory will rise to a greater height because ivory is more elastic than wet-clay.

Question 2.
Is it possible to double the length of a metallic wire by applying a force over it?
Answer:
No, it is not possible because, within elastic limit, strain is only order of 10^-3, wires actually break much before it is stretched to double the length.

Question 3.
Is stress a vector quantity? (NCERT Exemplar)
Stress = \(\frac{\text { Magnitude of internal reaction force }}{\text { Area of cross – section }}\)
Therefore, stress is a scalar quantity, not a vector quantity.

Question 4.
What does the slope of stress versus strain graph indicate?
Answer:
The slope of stress (on y-axis) and strain (on x-axis) gives modulus of elasticity.
The slope of stress (on x-axis) and strain (on y-axis) gives the reciprocal of modulus of elasticity.

Question 5.
Stress and pressure are both forces per unit area. Then in what respect does stress differ from pressure?
Answer:
Pressure is an external force per unit area, while stress is the internal restoring force which comes into play in a deformed body acting transversely per unit area of a body.

Question 6.
What is the Young’s modulus for a perfect rigid body?
Solution:
Young’s modulus (Y) = \(\frac{F}{A} \times \frac{l}{\Delta l}\)
For a perfectly rigid body, change in length Δl = 0
∴ Y = \(\frac{F}{A} \times \frac{l}{0}\) = ∞
Therefore, Young’s modulus for a perfectly rigid body is ∞.

Question 7.
What is Bulk modulus for a perfectly rigid body?
Solution:
Bulk modulus (B) = \(\frac{p}{\Delta V / V}=\frac{p V}{\Delta V}\)
For perfectly rigid body, change in volume ΔV = 0
∴ B = \(\frac{p V}{0}\) = ∞
Therefore, Bulk modulus for a perfectly rigid body is ∞.

Short Answer Type Questions

Question 1.
Explain why steel is more elastic than rubber?
Solution:
Consider two pieces of wires, one of steel and the other of rubber. Suppose both are of equal length (L) and of equal area of cross-section (a). Let each be stretched by equal forces, each being equal to F. We find that the change in length of the rubber wire (l_r) is more than that of the steel (l_s)i.e.,l_r>l_s.
If Y_s and Y_r are the Young’s moduli of steel and rubber respectively, then from the definition of Young’s modulus,

i. e,, the Young’s modulus of steel is more than that of rubber. Hence steel is more elastic than rubber.
Or
Any material which offers more opposition to the deforming force to change its configuration is more elastic.

Question 2.
Elasticity is said to be internal property of matter. Explain.
Answer:
When a deforming force acts on a body, the atoms of the substance get displaced from their original positions. Due to this, the configuration of the body (substance) changes. The moment, the deforming force is removed, the atoms return to their original positions and hence, the substance or body regains its original configuration. That is why, elasticity is said to be internal property of matter.

Question 3.
A wire elongates by l mm when a load W is hanged from it. If the wire goes over a pulley and two weights W each are hung at the two ends, then what will be the elongation of the wire in mm?
Solution:
According to Hooke’s law,
Modulus of elasticity, E = \(\frac{W}{A} \times \frac{L}{l}\)
where, L = original length of the wire
A- cross-sectional area of the wire
Elongation, l = \(\frac{W L}{A E}\) ………………………… (i)
On either side of the wire, tension is W and length is L/2.
Δl = \(\frac{W L / 2}{A E}=\frac{W L}{2 A E}=\frac{l}{2}\) [from eq.(i)]
Total elongation in the wire = \(\frac{l}{2}+\frac{l}{2}\) = l

Question 4.
A bar of cross-section A is subjected to equal and opposite tensile forces at its, ends. Consider a plane section of the bar whose normal makes an angle θ with the axis of the bar.
(i) What is the tensile stress on this plane?
(ii) What is the shearing stress on this plane?
(iii) For what value of θ is the tensile stress maximum?
(iv) For what value of θ is the shearing stress maximum?

Solution:
(i) The resolved part of F along the normal is the tensile force on this plane and the resolved part parallel to the plane is the shearing force on the plane.
∵ Area of MO plane section = A sec θ
Tensile stress = \(\frac{\text { Force }}{\text { Area }}=\frac{F \cos \theta}{A \sec \theta}=\frac{F}{A} \cos ^{2} \theta\)
= [ ∵ sec θ = \(\frac{1}{\cos \theta}\)]

(ii) Shearing stress applied on the top face
So, F = F sinθ
Shearing stress = \(\frac{\text { Force }}{\text { Area }}=\frac{F \sin \theta}{A \sec \theta}\)
= \(\frac{F}{A}\) sinθcosθ
= \(\frac{F}{2 A} \sin 2 \theta\) [∵ sin 2θ = 2sinθcosθ]

(iii) Tensile stress will be maximum when cos²θ is maximum i.e., cosθ = 1 or θ=0°.

(iv) Shearing stress will be maximum when sin20 is maximum i.e., sin2θ = 1 or 2θ = 90° or θ = 45°.

Question 5.
What is an elastomer? What are their special features?
Answer:
Elastomers are those substances which can be stretched to cause large strains.Substances like tissue of aorta, rubber etc., are elastomers.

The stress-strain curve for an elastomer is shown in figure. Although elastic region is very large but the materials does not obey Hooke’s law over most of the region. Moreover, there is no well-defined plastic region.

Question 6.
The Young’s modulus for steel is much more than that for rubber. For the same longitudinal strain, which one will have greater tensile stress? (NCERT Exemplar)
Solution:
Young’s modulus (Y) = \( \frac{\text { Stress }}{\text { Longitudinal strain }}\)
For same longitudinal strain, Y ∝ stress

Question 7.
Why are the springs made of steel and not of copper?
Answer:
A spring will be better one if a large restoring force is set up in it on being deformed, which in turn depends upon the elasticity of the material of the spring. Since the Young’s modulus of elasticity of steel is more than that of copper, hence, steel is preferred in making the springs.

Question 8.
Identical springs of steel and copper are equally stretched. On which, more work will have to be done? (NCERT Exemplar)
Solution:
Work done in stretching a wire is given by
W =- \(\frac{1}{2}\) F x Δl
As springs of steel and copper are equally stretched.
Therefore, for same force (F).
W ∝ Δl …………………………………… (i)

Young’s modulus (Y) = \(\frac{F}{A} \times \frac{l}{\Delta l}\)
or Δl = \(\frac{F}{A} \times \frac{l}{Y}\)
As both spring are identical,
∴ Δl ∝ \(\frac{1}{Y}\) …………………………………. (ii)
From eqs. (i) and (ii), we get W ∝ \(\frac{1}{Y}\) .
∴ \(\frac{W_{\text {steel }}}{W_{\text {copper }}}=\frac{Y_{\text {copper }}}{Y_{\text {steel }}}<1\)
[as Y_steel > Y_copper]
or W_steel < W_Copper
Therefore, more work will be done for stretching copper spring.

Long Answer Type Questions

Question 1.
A steel wire of length 21 and cross-sectional area A is stretched within elastic limit as shown in figure. Calculate the strain and stress in the wire.

Solution:
Total length L =21. Increase in length of the wire, when it is stretched from its mid-point.

From Pythagoras theorem, BC² =l² + x²
BC= \(\sqrt{l^{2}+x^{2}}\)
Similarly, AC = \(\sqrt{l^{2}+x^{2}}\)

Since x<< l, so using Binomial expansion, we have
\(\left(1+\frac{x^{2}}{l^{2}}\right)^{1 / 2}=\left(1+\frac{x^{2}}{2 l^{2}}\right)\)
[Neglecting terms containing higher powers of x]
∴ ΔL = 2l\(\left(1+\frac{x^{2}}{2 l^{2}}\right)-2 l=\frac{x^{2}}{l}\)
Hence Strain = \(\frac{\Delta L}{L}=\frac{x^{2}}{l \times 2 l}=\frac{x^{2}}{2 l^{2}}\)

Stress = \(\frac{F}{A}=\frac{\text { Tension }}{\text { Area }} \)
So, area of cross section of wire having radius r is πr²
Stress = \(\frac{T}{\pi r^{2}}\)

Question 2.
Consider a long steel bar unde a tensile stress due to forces F acting at the edges along the length of the bar (figure). Consider a plane making an angle θ with the length. What are the tensile and shearing stresses on this plane?

(a) For what angle is the tensile stress a maximum?
(b) For what angle is the shearing stress a maximum? (NCERT Exemplar)
Solution:
Consider the adjacent diagram.
Let the cross-sectional area of the bar be A. Consider the equilibrium of the plane aa’.
A force F must be acting on this plane making an angle \(\frac{\pi}{2}\) – θ with the normal ON. Resolving F into components, along the plane (FP) and normal to the plane.

F_P = F cosθ
F_N = Fsinθ
Let the area of the face aa’ be A’, then
\(\frac{A}{A^{\prime}}\) = sinθ’
∴ A’= \(\frac{A}{\sin \theta}\)

(a) For tensile stress to be maximum, sin²θ =1
⇒ sinθ = 1
⇒ θ = \(\frac{\pi}{2}\)
(b) For shearing stress to be maximum,
sin 2θ = 1
⇒ 2θ = \(\frac{\pi}{2}\)
⇒ θ = \(\frac{\pi}{4}\)

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 11 Thermal Properties of Matter Important Questions and Answers.

PSEB 11th Class Physics Important Questions Chapter 11 Thermal Properties of Matter

Very Short Answer Type Questions

Question 1.
Is it correct to call heat as the energy in transit?
Answer:
Yes, it is perfect correct to call heat as the energy in transit because it is continuously flowing on account of temperature differences between bodies or parts of a system.

Question 2.
Why should a thermometer bulb have a small heat capacity?
Answer:
The thermometer bulb having small heat capacity will absorb less heat from the body whose temperature is to be measured. Hence, the temperature of that body will practically remain unchanged.

Question 3.
Why is a gap left between the ends of two railway lines in a railway track?
Answer:
It is done to accommodate the linear expansion of railway line during summer. If the gap is not left in summer, the lines will bend causing a threat of derailment.

Question 4.
Why water is used as an coolant in the radiator of cars?
Answer:
Because specific heat of water is very high due to this it absorbs a large amount of heat. This helps in maintaining the temperature of the engine low.

Question 5.
Black body radiation is white. Comment.
Answer:
The statement is true. A black body absorbs radiations of all wavelengths. When heated to a suitable temperature, it emits radiations of all wavelengths. Hence, a black body radiation is white.

Question 6.
White clothes are more comfortable in summer while colourful clothes are more comfortable in winter. Why?
Answer:
White clothes absorb very little heat radiation and hence they are comfortable in summer. Coloured clothes absorb almost whole of the incident radiation and keep the body warm in winter.

Question 7.
Can we boil water inside in the earth satellite?
Answer:
No, the process of transfer of heat by convection is based on the fact that a liquid becomes lighter on becoming hot and rise up. In condition of weightlessness, this is not possible. So, transfer of heat by convection is not possible in the earth satellite.

Question 8.
What is the difference between the specific heat and the molar specific heat?
Answer:
The specific heat is the heat capacity per unit mass whereas the molar specific heat is the heat capacity per mole.

Question 9.
Calorimeters are made of metals not glass. Why?
Answer:
This is because metals are good conductors of heat and have low specific heat capacity.

Question 10.
Calculate the temperature which has numeral value of Celsius and Fahrenheit scale. (NCERT Exemplar)
Answer:
Let Q be the value of temperature having same value an Celsius and Fahrenheit scale.
\(\frac{{ }^{\circ} F-32}{180}=\frac{{ }^{\circ} C}{100}\)
⇒ Let F = C = Q
⇒ \(\frac{Q-32}{180}=\frac{Q}{100}\) = Q= 40°C or -40°F

Short Answer Type Questions

Question 1.
In what ways are the gas thermometers superior to mercury thermometers?
Answer:
A gas thermometer is more superior to a mercury thermometer, as its working is independent of the nature of gas (working substance) used. As the variation of pressure (or volume) with temperature is uniform, the range, in which temperature can be measured with a gas thermometer is quite large. Further, a gas thermometer is more sensitive than mercury thermometer.

Question 2.
The difference between length of a certain brass rod and that of a steel rod is claimed to be constant at all temperatures. Is this possible?
Solution:
Yes, it is possible to describe the difference of length to remain constant. So, the change in length of each rod must be equal at all temperature. Let α_b and α_s be the length of the brass and the steel rod and a band as be the coefficients of linear expansion of the two metals. Let there is change in temperature be ΔT.
Then, α_bL_bΔT = α_sL_sΔT
or α_bL_b=α_sL_s => L_b/L_s=α_s/α_b
Hence, the lengths of the rods must be in the inverse ratio of the coefficient of linear expansion of their materials.

Question 3.
Two identical rectangular strips-one of copper and the other of steel are riveted to form a bimetallic strip. What will happen on heating?
Solution:
The coefficient of linear expansion of copper is more than steel. On heating, the expansion in copper strip is more than the steel strip. The bimetallic strip will bend with steel strip on inner (concave) side.

Question 4.
What kind of thermal conductivity and specific heat requirements would you specify for cooking utensils?
Solution:
A cooking utensil should have (i) high conductivity, so that it can conduct heat through itself and transfer it to the contents quickly, (ii) low specific heat, so that it immediately attains the temperature of the source.

Question 5.
Woollen clothes are warm in winter. Why?
Solution:
Woollen fibres enclose a large amount of air in them. Both wool and air are bad conductors of heat. The small coefficient of thermal conductivity prevents the loss of heat from our body due to conduction. So, we feel warm in woollen clothes.

Question 6.
Why rooms are provided with the ventilators near the roof?
Solution:
It is done so to remove the harmful impure air and to replace it by the cool fresh air. The air we breathe out is warm and so it is lighter. It rises upwards and can go out through the ventilator provided near the roof. The cold fresh air from outside enters the room through the doors and windows. Thus, the convection current is set up in the air.

Question 7.
Why it is much hotter above a fire than by its side?
Solution:
Heat carried away from a fire sideways mainly by radiation. Above the fire, heat is carried by both radiation and convection of air but convection carries much more heat than radiation. So, it is much hotter above a fire than by its sides.

Question 8.
How does tea in a Thermo flask remain hot for a long time?
Solution:
The air between the two walls of the Thermo flask is evacuated. This prevents heat loss due to conduction and convection. The loss of heat due to radiation is minimised by silvering the inside surface of the double wall. As the loss of heat due to the three prócesses is minimised and the tea remains hot for a long time.

Question 9.
100 g of water is supercooled to -10°C. At this point, due to some disturbance mechanised or otherwise, some of it suddenly freezes to ice. What will be the temperature of the resultant mixture and how much mass would freeze? [S_w = 1 cal/g/°C and L^w_fusion =80 cal/g/°C] (NCERT Exemplar)
Answer:
Gwen, mass of water (m) = 100
Change in temperature, ΔT =0 – (-10) = 10°C
Specific heat of water (S_w) =1 cal/g/°C
Latent heat of fusion of water L^w_fusion = 80 cal/g
Heat required to bring water in supercooling from —10° C to 0°C.
Q = ms_wΔT
=100 x 1 x 10 = 1000cal
Let m gram of ice be melted.
∴ Q = mL
or m= \(\frac{Q}{L}\) = \(\frac{1000}{80}\) =12.5g
As small mass of ice is melted, therefore the temperature of the mixture will remain 0°C.

Long Answer Type Questions

Question 1.
Show that the coefficient of volume expansion for a solid substance is three times its coefficient of linear expansion.
Solution:
Consider a solid in the form of a rectangular parallelopiped of sides a, b and c respectively so that its volume V = abc.
If the solid is heated so that its temperature rises by ΔT, then increase in its sides will be
Δa=a.αΔT, Δb=b.α.ΔT and Δc=c. α . ΔT
or a’ =a+Δa =a(1 +α ΔT)
b’=b+Δb = b(l +α ΔT)
and c’ =c + Δc=c (1 +a.ΔT)
∵ New volume, V’ = V + ΔV = a’ b’ c’ = abc (1+ α . Δ T)³
∴ Increase in volume,
ΔV=V’ -V=[abc(1+α ΔT)³ -abc]
∴ Coefficient of volume expansion,

However, as a has an extremely small value for solids, hence terms containing higher powers of a may be neglected. Therefore, we obtain the relation γ =3 α i. e., coefficient of volume expansion of a solid is three times of its coefficient of linear expansion.

Question 2.
Distinguish between conduction, convection and radiation.
Solution:

Conduction	Convection	Radiation
1. It is the transfer of heat by direct physical contact.	1. It is the transfer of heat by the motion of a fluid.	1. It is the transfer of heat by electromagnetic waves.
2. It is due to temperature differences. Heat flows from high-temperature region to low temperature region.	2. It is due to difference in density. Heat flows from low-density region to high-density region.	2. It occurs from all bodies at temperatures above 0 K.
3. It occurs in solids through molecular collisions, without actual flow of matter.	3. It occurs in fluids by actual flow of matter.	3. It can take place at large distances and does not heat the intervening medium.
4. It is a slow process.	4. It is also a slow process.	4. It propagates at the speed of light.
5. It does not obey the laws of reflection and refraction.	5. It does not obey the laws of reflection and refraction.	5. It obeys the laws of reflection and refraction.

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 12 Thermodynamics Important Questions and Answers.

PSEB 11th Class Physics Important Questions Chapter 12 Thermodynamics

Very Short Answer Type Questions

Question 1.
Can a system be heated and its temperature remain constant? (NCERT Exemplar)
Answer:
If the system does work against the surroundings so that it compensates for the heat supplied, the temperature can remain constant.

Question 2.
Air pressure in a car tyre increases during driving. Explain. (NCERT Exemplar)
Answer:

• During driving, temperature of the gas increases while its volume remains constant.
• So, according to Charles’ law, at constant V, p ∝ T.
• Therefore, pressure of gas increases.

Question 3.
Write conditions for an isothermal process.
Answer:
The conditions for an isothermal process are :

• The walls should be diathermic.
• The process should be quasi-static.

Question 4.
Why air quickly leaking out of a balloon becomes cooler?
Answer:
Leaking of air is adiabatic expansion and adiabatic expansion produces cooling.

Question 5.
If a refrigerator’s door is kept open, will the room become cool or hot? Explain.
Answer:
Here, heat removed is less than the heat supplied and hence the room become hotter.

Question 6.
Is reversible process is possible in nature?
Answer:
A reversible process is never possible in nature because of dissipative forces and condition for a quasi-static process is not practically possible.

Question 7.
On what factors, the efficiency of a Carnot engine depends?
Answer:
On the temperature of source of heat and the sink.

Question 8.
Which thermodynamic law put restrictions on the complete conversion of heat into work?
Answer:
According to second law of thermodynamics, heat energy cannot converted into work completely.

Short Answer Type Questions

Question 1.
What are the limitations of the first law of thermodynamics?
Answer:
Following are the limitations of the first law of thermodynamics :

• It does not tell us about the direction of flow of heat.
• It fails to explain why heat cannot be spontaneously converted into work.

Question 2.
Two bodies at different temperatures T₁ and T₂ are brought in contact.
Under what condition, they settle to mean temperature? (after they attain equilibrium)
Answer:
Let m₁ and m₂ are masses of bodies with specific heats s₁ and s₂, then if their temperature after they are in thermal equilibrium is T.

Then, if > T₁> T₂ and assuming no heat loss.
Heat lost by hot body = heat gained by cold body
m₁s₁(T₁-T)=m₂s₂(T-T₂)
⇒ \(\frac{m_{1} s_{1} T_{1}+m_{2} s_{2} T_{2}}{m_{1} s_{1}+m_{2} s_{2}}\) = T[equilibrium temperature]
So for, bodies to settle down to mean temperature,
m₁ = m₂ and s₁ = s₂
means bodies have same specific heat and have equal masses.
Then, T = \(\frac{T_{1}+T_{2}}{2}\) [mean temperature]

Question 3.
When ice melts, then change in internal energy, is greater than the heat supplied, why?
Solution:
When ice melts, volume of water formed is less than that of ice. So, surroundings (environment) does work on the system (ice). And by first law,
ΔQ = ΔW+ΔU
⇒ ΔU = ΔQ-ΔW
(ΔW = negative as work is done on the system)
⇒ ΔU>ΔQ

Question 4.
Calculate the work done for adiabatic expansion of a gas.
Solution:
Consider (say µ mole) an ideal gas, which is undergoing an adiabatic expansion.
Let the gas expands by an infinitesimally small volume dV, at pressure p, then the infinitesimally small work done given by
dW = pdV
The net work done from an initial volume V₁ to final volume V₂ is given by
W= ∫v₁v² pdV
For an adiabatic process, pV^γ = constant = K

For an adiabatic process, K = p₁V^γ = p₂V^γ
For an ideal gas, p₁V₁ = μRT₁ and p₂V₂ = μRT₂.
So, we have
W = \(\frac{1}{(1-\gamma)}\left[\mu R T_{2}-\mu R T_{1}\right]=\frac{\mu R}{(\gamma-1)}\left[T_{1}-T_{2}\right]\)

Question 5.
What is a heat engine? What is the best way to increase efficiency of a heat engine? Is it possible to design a thermal engine that has 100% efficiency?
Solution:
A heat engine is a device (or a combination) which converts heat into work.
Its efficiency, η = \(\frac{\text { Work output }}{\text { Heat input }}=1-\frac{T_{2}}{T_{1}}\)
where, T₂ = temperature of sink
T₁ = temperature of source.
From above expression, we can see that for 100% efficiency, T₂ =0
It is impossible to design a thermal engine that has 100% efficiency because it is not possible to have a sink with kelvin temperature.

Question 6.
An ideal engine works between temperatures T₁ and T₂. It derives an ideal refrigerator that works between temperatures T₃ and T₄. Find the ratio Q₃/Q₁ in terms of T₁, T₂, T_3, and T₄.

Solution:
W = work done by engine = Q₁ – Q₂
and W = work done supplied to refrigerator = Q₃ -Q₄
Q₁ – Q₂ =Q₃ – Q₄
Dividing by Q₁, on both sides of the above equation, we get

Question 7.
Under what condition, an ideal Carnot engine has 100% efficiency?
Solution:
Efficiency of a Carnot engine is given by η = \(\left(1-\frac{T_{2}}{T_{1}}\right)\)
where, T₂ = temperature of sink
and T₁ = temperature of sink source
So for η = 1 or 100%, T₂ = 0 K or heat is rejected into a sink at 0 K temperature.

Question 8.
Draw p-V diagram of a Carnot cycle.
Solution:
p-V diagram for Carnot cycle

Long Answer Type Questions

Question 1.
A cycle followed by a machine (made of one mole of a perfect gas in a cylinder with a piston) is shown in figure

A to B: volume constant B to C: adiabatic
C to D: volume constant D to A : adiabatic
V_C = V_D = 2 V_A = 2 V_B
(i) In which part of the cycle, heat is supplied to the machine from outside?
(ii) In which part of the cycle, heat is being given to the surrounding by the machine?
(iii) What is the work done by the machine in one cycle? Write your answer in terms of P_A’ P_B’ V_A.
(iv) What is the efficiency of the machine?
Take γ = \(\frac{5}{3}\) for the gas and C_V =R for one mole.
Solution:
(i) A to B because T_B > T_A, as p ∝ T [ ∴ V = constant]
(ii) C to D because T_C>T_D, as P ∝ T [∴ V=constant]
(iii) W_AB = \(\int_{B}^{C} \) pdV=O and W_CD =0 [∴ V= constant]
Similarly,

(iv) Heat supplied during process A to B
dQ_AB = dU_AB
Q_AB = \(\frac{3}{2} n R\left(T_{B}-T_{A}\right)=\frac{3}{2}\left(p_{B}-p_{A}\right) V_{A}\)
∴ Efficiency = \(=\frac{\text { Net work done }}{\text { Heat supplied }}=\left[1-\left(\frac{1}{2}\right)^{2 / 3}\right]\)

Question 2.
Explain with the suitable example that a reversible process must be carried slowly and a fast process is necessarily irreversible.
Answer:
A reversible process must pass through equilibrium states which are very close to each other so that when process is reversed, it passes back through these equilibrium states. Then, it is again decompressed or it passes through same equilibrium states, system can be restored to its initial state without any change in surroundings. e.g., If a gas is compressed as shown But a reversible process can proceed without reaching equilibrium in intermediate states.