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Punjab State Board PSEB 11th Class Physics Book Solutions Chapter 9 Mechanical Properties of Solids Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids

PSEB 11th Class Physics Guide Mechanical Properties of Solids Textbook Questions and Answers

Question 1.
A steel wire of length 4.7 m and cross-sectional area 3.0 x 10^-5 m² stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 x 10^-5 m² under a given load. What is the ratio of the Young’s modulus of steel to that of copper?
Solution:
Given, length of the steel wire, L₁ = 4.7 m
Area of cross-section of the steel wire,A₁ = 3.0 x 10^-5 m²
Length of the copper wire, L₂ = 3.5 m
Area of cross-section of the copper wire, A₂ = 4.0 x 10^-5 m²

Change in length = ΔL₁ = ΔL₂ = ΔL
Force applied in both the cases = F
Young’s modulus of the steel wire,
Y₁ = \(\frac{F_{1}}{A_{1}} \times \frac{L_{1}}{\Delta L_{1}} \)
= \(\frac{F \times 4.7}{3.0 \times 10^{-5} \times \Delta L} \) ………………………………. (i)
Young’s modulus of the copper wire,
Y₂ = \(\frac{F_{2}}{A_{2}} \times \frac{L_{2}}{\Delta L_{2}}=\frac{F \times 3.5}{4.0 \times 10^{-5} \times \Delta L}\) ………………………………. (ii)
Dividing eq. (i) by eq. (ii), we get:
\(\frac{Y_{1}}{Y_{2}}=\frac{4.7 \times 4.0 \times 10^{-5}}{3.0 \times 10^{-5} \times 3.5}\) = 1.79:1
The ratio of Young’s modulus of steel to that of copper is 1.79: 1.

Question 2.
Figure given below shows the strain-stress curve for a given material. What are (a) Young’s modulus and (b) approximate yield strengths for this material?

Solution:
(a) It is clear from the given graph that for stress 150 x 10⁶ N/m², strain is 0.002.
∴ Young s modulus, Y = \(\frac{\text { Stress }}{\text { Strain }}\)
= \(\frac{150 \times 10^{6}}{0.002}\) = 7.5 x 10¹⁰ N/m²
Hence, Young’s modulus for the given material is 7.5 x10¹⁰ N/m²

(b) The yield strength of a material is the maximum stress that the material can sustain without crossing the elastic limit. It is clear from the given graph that the approximate yield strength of this material is 300 x 10⁶ N/m² or 3 x 10⁸ N/m².

Question 3.
The stress-strain graphs for materials A and B are shown in figure given below.

The graphs are drawn to the same scale.
(a) Which of the materials has the greater Young’s modulus?
(b) Which of the two is the stronger material?
Solution:
(a) A, for a given strain, the stress for material A is more than it is for > material B, as shown in the two graphs.
Young’s modulus = \(\frac{\text { Stress }}{\text { Strain }}\)
For a given strain, if the stress for a material is more, then Young’s modulus is also greater for that material. Therefore, Young’s modulus for material A is greater than it is for material B.

(b) A, the amount of stress required for fracturing a material, corresponding to its fracture point, gives the strength of that material. Fracture point is the extreme point in a stress-strain curve. It can be observed that material A can withstand more strain than material B. Hence, material A is stronger than material B.

Question 4.
Read the following two statements below carefully and state, with reasons, if it is true or false.
(a) The Young’s modulus of rubber is greater than that of steel;
(b) The stretching of a coil is determined by its shear modulus.
Solution:
(a) False.
Reason: For a given stress, the strain in rubber is more than it is in steel.
Young’s modulus, Y = \(\frac{\text { Stress }}{\text { Strain }} \)
For a constant stress Y ∝ \(\frac{1}{\text { Strain }}\)
Hence, Young’s modulus for rubber is less than it is for steel.

(b) True.
Reason: Shear modulus is the ratio of the applied stress to the change in the shape of a body. The stretching of a coil changes its shape. Hence, shear modulus of elasticity is involved in this process.

Question 5.
Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in figure given below. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m.
Compute the elongations of the steel and the brass wires.

Solution:
Given, diameter of the wires, d = 0.25 cm
Hence, the radius of the wires, r = \(\frac{d}{2} \) = 0.125cm = 0.125 x 10^-2m
Length of the steel wire, L₁ = 1.5 m
Length of the brass wire, L₂ = 1.0 m
Total force exerted on the steel wire,
F₁ = (4 +6)g= 10×9.8 = 98N .

Young’s modulus for steel
Y₁ = \(\frac{\left(\frac{F_{1}}{A_{1}}\right)}{\left(\frac{\Delta L_{1}}{L_{1}}\right)} \)
where, ΔL₁ = Change in the length of the steel wire
A₁ = Area of cross-section of the steel wire = πr²₁

Young’s modulus of steel, Y₁ = 2.0 x 10¹¹ Pa
∴ ΔL₁ = \(\frac{F_{1} \times L_{1}}{A_{1} \times Y_{1}}=\frac{F_{1} \times L_{1}}{\pi r_{1}^{2} \times Y_{1}}\)
= \(\frac{98 \times 1.5}{3.14\left(0.125 \times 10^{-2}\right)^{2} \times 2 \times 10^{11}}\)
= 1.5 x 10^-4 m

Total force on the brass wire
F₂ =6 x 9.8=58.8N
Young’s modulus for brass
Y₂ = \(\frac{\left(\frac{F_{2}}{A_{2}}\right)}{\left(\frac{\Delta L_{2}}{L_{2}}\right)}\)

where, ΔL₂ = Change in length of the steel wire
A₂ = Area of cross-section of the brass wire
∴ ΔL₂ = \(\frac{F_{2} \times L_{2}}{A_{2} \times Y_{2}}=\frac{F_{2} \times L_{2}}{\pi r_{2}^{2} \times Y_{2}}\)
= \(\frac{58.8 \times 1.0}{3.14 \times\left(0.125 \times 10^{-2}\right)^{2} \times\left(0.91 \times 10^{11}\right)}\)
= 1.3 x 10^-4 m
Hence, elongation of the steel wire =1.49 x 10^-4 m
and elongation of the brass wire = 1.3 x 10^-4 m

Question 6.
The edge of an aluminium cube is 10 cm long. One face of the cube is firmly fixed to a vertical wall. A mass of 100 kg Is the attached to the opposite face of the cube. The shear modulus of aluminium is 25 GPa. What is the vertical deflection of this face?
Solution:
Given, edge of the aluminium cube, L = 10cm = 0.1 m
The mass attached to the cube, m =100 kg
Shear modulus (ri) of aluminium = 25GPa =25 x 10⁹ Pa
Shear modulus, η = \(\frac{\text { Shear stress }}{\text { Shear strain }}=\frac{\frac{F}{A}}{\frac{\Delta L}{L}}\)
where, F = Applied force = mg = 100 x 9.8 = 980 N
A = Area of one of the faces of the cube = 0.1 x 0.1 = 0.01 m²

ΔL = Vertical deflection of the cube
ΔL = \(\frac{F L}{A \eta}=\frac{980 \times 0.1}{10^{-2} \times\left(25 \times 10^{9}\right)}\)
= 3.92 x 10^-7 m
The vertical deflection of this face of the cube is 3.92 x 10^-7 m.

Question 7.
Four identical tblloW cylindrical columns of mild steel support a big structure of mass’50,000 kg. The inner and outer radii of each column are 30 cm and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column.
Solution:
Given, mass of the big structure, M = 50000 kg
Inner radius of the column, r = 30 cm = 0.3 m
Outer radius of the column, R = 60 cm = 0.6 m
Young’s modulus of steel, Y = 2 x 10¹¹ Pa
Total force exerted, F =Mg = 50000 x 9.8N
Stress = Force exerted on a single column = \(\frac{50000 \times 9.8}{4}\) = 122500 N

Young’s modulus, Y = \(\frac{\text { Stress }}{\text { Strain }} \)
Strain = \(\frac{F}{\frac{A}{Y}} \)
where, Area, A = π(R² – r²) = π[(0.6)² – (0.3)²]
Strain = \(\frac{122500}{3.14\left[(0.6)^{2}-(0.3)^{2}\right] \times 2 \times 10^{11}}\) = 7.22 x 10^-7
Hence, the compressional strain of each column is 7.22 x 10^-7.
∴Compressional strain of all columns is given by
= 7.22 x 10 ^-7 x 4 = 2.88 x 10^-6.

Question 8.
A piece of copper having a rectangular cross-section of 15.2 minxes 19.2 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain?
Solution:
Given, cross-section area.of copper piece (A) = 15.2 mm x 19.1 mm
= (15.2 x 19.1) x 10 ^-6m²
Force applied (F) = 44500 N
Young’s modulus (Y) =1.1 x 10¹¹ Nm^-2

Young s modulus (Y) = \(=\frac{\text { Longitudinal stress }}{\text { Longitudinal strain }}\)
or Longitudinal strain = \(\frac{\text { Longitudinal stress }}{\text { Young’s modulus }}\)
Young’s modulus
= \(\frac{(F / A)}{Y}=\frac{F}{A Y}\)
= \(\frac{44500}{15.2 \times 19.1 \times 10^{-6} \times 1.1 \times 10^{11}}\)
= 0.0013934.

Question 9.
A steel cable with a radius of 1.5 cm supports a chairlift at a ski area. If the maximum stress is not to exceed 10⁸ Nm ^-2, what is the maximum load the cable can support?
Solution:
Radius of the steel cable, r = 1.5cm = 0.015m
Maximum allowable stress = 10⁸ N m^-2
Maximum stress = \(\frac{\text { Maximum force }}{\text { Area of cross – section }} \)
∴ Maximum force = Maximum stress x Area of cross – section
= 10⁸ x π (0.015)²
= 7.065 x 10⁴ N
Hence, the cable can support the maximum load of 7.065 x 10⁴ N.

Question 10.
A rigid bar of mass 15 kg is supported symmetrically by three wires each 2.0 m long. Those at each mid are of copper and the middle one is of iron. Determine the ratio of their diameters if each is to have the same tension.
Solution:
The tension force acting on each wire is the same. Thus, the extension in each case is the same. Since the wires are of the same length, the strain will also be the same.
The relation for Young’s modulus is given as:
Y = \(\frac{\text { Stress }}{\text { Strain }}=\frac{\frac{F}{A}}{\text { Strain }}=\frac{\frac{4 F}{\pi d^{2}}}{\text { Strain }}\) ……………………………. (i)
where, F = Tension force
A = Area of cross-section
d = Diameter of the wire
It can be inferred from equation (i) that Y ∝ \(\frac{1}{d^{2}}\)
Young’s modulus for iron, Y₁ = 190 x 10⁹ Pa
Diameter of the iron wire = d₁
Young’s modulus for copper, Y₂ = 110 x 10⁹ Pa
Diameter of the copper wire = d₂
Therefore, the ratio of their diameters is given as:
\(\frac{d_{2}}{d_{1}}=\sqrt{\frac{Y_{1}}{Y_{2}}}=\sqrt{\frac{190 \times 10^{9}}{110 \times 10^{9}}}=\sqrt{\frac{19}{11}}\)
= 1.31:11.

Question 11.
A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle.
The cross-sectional area of the wire is 0.065 cm². Calculate the elongation of the wire when the mass is at the lowest point of its path.
Solution:
Given, mass, m = 14.5kg
Length of the steel wire, l = 1.0 m
Angular velocity, ω = 2 rev / s
Cross-sectional area of the wire, a = 0.065cm² = 0.065 x 10^-4 m²
Let δl be died elongation of the wire when the mass is at the lowest point of its path.
When the mass is placed at the position of the vertical circle, the total force on the mass is :
F = mg+mlω² ,
= 14.5 x 9.8 +14.5x 1 x (2)² = 200.1 N

Young’s modulus = \(\frac{\text { Stress }}{\text { Strain }} \)
Y = \(\frac{\frac{F}{A}}{\frac{\Delta l}{l}}=\frac{F}{A} \cdot \frac{l}{\Delta l}\)
∴ Δl = \(\frac{F l}{A Y}\)

Young’s modulus for steel = 2 x 10¹¹ Pa
∴ Δl = \(\frac{200.1 \times 1}{0.065 \times 10^{-4} \times 2 \times 10^{11}} \) = 1539.23 x 10^-7
= 1.539 x 10^-4
Hence, the elongation of the wire is 1.539 x 10^-4 m.

Question 12.
Compute the bulk modulus of water from the following data: Initial volume =100.0 litre, Pressure increase =100.0atm (1 atm = 1.013 x 10⁵ Pa), Final volume = 100.5 litre. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large.
Solution:
Initial volume, V₁ = 100.0 l x 10^-3 m³
Final volume, V₂ = 100.5l = 100.5 x 10^-3 m^-3
Increase in volume, V = V₂ — V₁ = 0.5 x 10^-3 m³
Increase in pressure, Δ_p =100.0 atm = 100 x 1.013 x 10⁵ Pa
Bulk modulus = \( \frac{\Delta p}{\frac{\Delta V}{V_{1}}}=\frac{\Delta p \times V_{1}}{\Delta V}\)
= \(\frac{100 \times 1.013 \times 10^{5} \times 100 \times 10^{-3}}{0.5 \times 10^{-3}}\) = 2.206 x 10⁹ Pa
Bulk modulus of air = 1.0 x 10⁵ Pa
∴ \(\frac{\text { Bulk modulus of water }}{\text { Bulk modulus of air }}=\frac{2.026 \times 10^{9}}{1.0 \times 10^{5}}\)
= 2.026 x 10⁴
This ratio is very high because air is more compressible than water.

Question 13.
What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is 103 x 10³ kgm^-3?
Solution:
Let the given depth be h.
Pressure at the given depth, p = 80.0 atm = 80 x 1.01 x 10⁵ Pa
Density of water at the surface, ρ₁ = 1.03 x 10³ kg m^-3
Let ρ₂ be the density of water at the depth h.
Let V₁ be the volume of water of mass m at the surface.
Let V₂ be the volumé of water of mass m at the depth h.
Let ΔV be the change in volume.

ΔV = V₁ – V₂ = \(m\left(\frac{1}{\rho_{1}}-\frac{1}{\rho_{2}}\right)\)
∴ Volumetric strain= \(\frac{\Delta V}{V_{1}}=m\left(\frac{1}{\rho_{1}}-\frac{1}{\rho_{2}}\right) \times \frac{\rho_{1}}{m}\)
∴ \(\frac{\Delta V}{V_{1}}=1-\frac{\rho_{1}}{\rho_{2}}\) ………………………………. (i)

Bulk modulus, B = \(\frac{p V_{1}}{\Delta V}\)
\(\frac{\Delta V}{V_{1}}=\frac{p}{B}\)
Compressibility of water = \(\frac{1}{B}=45.8 \times 10^{-11} \mathrm{~Pa}^{-1}\)
∴ \(\frac{\Delta V}{V_{1}}=80 \times 1.013 \times 10^{5} \times 45.8 \times 10^{-11}\) = 3.71 x 10^-3 ………….(ii)
From equations (i) and (ii), we get

Therefore, the density of water at the given depth (h) is 1.034 x 10³ kg m^-3.

Question 14.
Compute the fractional change in volume of a glass slab, when subjected to a hydraulic pressure of 10 atm.
Solution:
Hydraulic pressure exerted on the glass slab, p = 10 atm = 10 x 1.013 x 10⁵ Pa
Bulk modulus of glass, B = 37 x 10⁹ Nm^-2
Bulk modulus, B = \(\frac{p}{\frac{\Delta V}{V}}\)
where, \(\frac{\Delta V}{V}\) = Fractional change in volume
∴ \(\frac{\Delta V}{V}=\frac{p}{B}=\frac{10 \times 1.013 \times 10^{5}}{37 \times 10^{9}}\)
= 2.73 x 10^-5
Hence, the fractional change in the volume of the glass slab is
2.73 x 10^-5 = 2.73 x 10^-3% = 0.0027%

Question 15.
Determine the volume contraction of a solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of 7.0x 10⁶ Pa.
Solution:
Length of an edge of the solid copper cube, l =10 cm = 0.1 m
Hydraulic pressure, p = 7.0 x 10⁶ Pa
Bulk modulus of copper, B = 140 x 10⁹ Pa
Bulk modulus, B = \(\frac{P}{\frac{P}{\Delta V}}\)
where, \(\frac{\Delta V}{V}\) = Volumetric strain
ΔV = Change in volume
V =,Original volume
ΔV = \(\frac{p V}{B}\)
Original volume of the cube, V = l³

Therefore, the volume contraction of the solid copper cube is 0.05 cm³.

Question 16.
How much should the pressure on a litre of water be changed to compress by 0.10%?
Solution:
Volume of water, V =1 L
It is given that water is to be compressed by 0.10%.
∴ Fractional change, \(\frac{\Delta V}{V}=\frac{0.1}{100 \times 1}=10^{-3}\)
Bulk modulus, B = \(\frac{p}{\frac{\Delta V}{V}}\)
p = B x \( \frac{\Delta V}{V}\)

Bulk modulus of water, B = 2.2 x 10⁹ Nm^-2
p = 22 x 10⁹ x 10^-3
=2.2 x 10⁶ Nm^-2
Therefore, the pressure on water should be 2.2 x 10⁶ Nm^-2.

Additional Exercises

Question 17.
Anvils made of single crystals of diamond, with the shape as shown in figure given below, are used to investigate behaviour of materials under very high pressures. Flat faces at the narrow end of the anvil have a diameter of 0.50 mm, and the wide ends are subjected to a compressional force of 50,000 N. What is the pressure at the tip of the anvil?

Solution:
Diameter of the cones at the narrow ends, d = 0.50 mm = 0.5 x 10^-3m
radius, r = \( \frac{d}{2}\) = 0.25 x 10^-3 m
Compressional force, F = 50000 N
Pressure at the tip of the anvil,
p = \(\frac{\text { Force }}{\text { Area }}=\frac{F}{\pi r^{2}}\)
= \(\frac{50000}{3.14 \times\left(0.25 \times 10^{-3}\right)^{2}}\)
= 2.55 x 10¹¹ Pa
Therefore, the pressure at the tip of the anvil is 2.55 x 10¹¹ Pa.

Question 18.
A rod of length 1.05 m having negligible mass is supported at its ends by two wires of steel (wire A) and aluminium (wire B) of equal lengths as shown in figure given below. The cross-sectional areas of wires A and B are 1.0 mm² and 2.0mm², respectively. At what point along the rod should a mass m he suspended in order to produce (a) equal stresses and (b) equal strains in both steel and aluminium wires.

Solution:
Given, cross-sectional area of wire A, a₁ = 1.0 mm₂ = 1.0 x 10^-6 m²
Cross-sectional area of wire B, a₂ = 2.0 mm² = 2.0 x 10^-6 m²
Young’s modulus for steel, Y₁ = 2 x 10¹¹ Nm^-2
Young’s modulus for aluminium, Y₂ = 7.0 x 10¹⁰ Nm^-2

Let a small mass m be suspended to the rod at a distance y from the end where wire A is attached.
Stress in the wire = \(\frac{\text { Force }}{\text { Area }}=\frac{F}{a}\)
If the two wires have equal stresses, then,
\( \frac{F_{1}}{a_{1}}=\frac{F_{2}}{a_{2}}\)
where, F₁ = Force exerted on the steel wire

F₂ = Force exerted on the aluminium wire
\(\frac{F_{1}}{F_{2}}=\frac{a_{1}}{a_{2}}=\frac{1}{2}\) …………………………. (i)
The situation is shown in the following figure.

Taking torque about the point of suspension, we have
F₁y = F₂(1.05— y)
\(\frac{F_{1}}{F_{2}}=\frac{(1.05-y)}{y}\) ……………………….(ii)
Using equations (i) and (ii), we can write

In order to produce an equal stress in the two wires, the mass should be suspended at a distance of 0.7 m from the end where wire A is attached.

(b) Young s modulus = \(\frac{\text { Stress }}{\text { Strain }} \)
Strain = \(\frac{\text { Stress }}{\text { Young’s modulus }}=\frac{\frac{F}{a}}{Y}\)
If the strain in the two wires is equal, then,

Taking torque about the point where mass m, is suspended at a distance y₁ from the side where wire A attached, we get
F₁y₁ =F₂(1.05-y₁)
\(\frac{F_{1}}{F_{2}}=\frac{\left(1.05-y_{1}\right)}{y_{1}}\) ……………………………. (iv)
Using equations (iii) and (iv), we get
\(\frac{\left(1.05-y_{1}\right)}{y_{1}}=\frac{10}{7}\)
7(1.05 – y₁) = 10 y₁
⇒ 17 y₁ = 7.35
y₁ = 0.432 m
In order to produce an equal strain in the two wires, the mass should be suspended at a distance of 0.432 m from the end where wire A is attached.

Question 19.
A mild steel wire of length 1.0 m and cross-sectional area 0.50 x 10^-2 cm² is stretched, well within its elastic limit, horizontally between two pillars. A mass of 100 g is suspended from the mid-point of the wire. Calculate the depression at the mid-point.
Solution:

Length of the steel wire = 1.0 m
Area of cross-section, A = 0.50 x 10^-2 cm² = 0.50 x 10^-6 m²
A mass 100 g is suspended from its mid-point.
m = 100 g = 0.1kg
Hence, the wire dips, as shown in the given figure.

Original length = XZ
Depression = l
The length after mass m, is attached to the wire = XO +OZ
Increase in the length of the wire:
Δl = (XO + OZ)-XZ
Where

Expanding and neglecting higher terms, we get:
Δl = \(\frac{l^{2}}{0.5}\)
Strain = \(\frac{\text { Increase in length }}{\text { Original length }}\)
Let T be the tension in the wire.

∴ mg = 2Tcos θ
Using the figure, it can be written as
Cos θ = \(\frac{l}{\left[(0.5)^{2}+l^{2}\right]^{\frac{1}{2}}}=\frac{l}{(0.5)\left[1+\left(\frac{l}{0.5}\right)^{2}\right]^{\frac{1}{2}}}\)
Expanding the expression and eliminating the higher terms, we get
Cos θ = \(\frac{l}{(0.5)\left(1+\frac{l^{2}}{2(0.5)^{2}}\right)}\)
\(\left(1+\frac{l^{2}}{2(0.5)^{2}}\right)\) ≈ 1 for small l

Hence, the depression at the mid-point is 0.0107 m.

Question 20.
Two strips of metal are riveted together at their ends by four rivets, each of diameter 6.0 mm. What is the maximum tension’ that can be exerted by the riveted strip if the shearing stress on the rivet is not to exceed 6.9 x 10⁷ Pa?
Assume that each rivet is to carry one-quarter of the load.
Solution:
Diameter of the metal strip, d = 6.0 mm = 6.0 x 10^-3 m
Radius, r = \(\frac{d}{2}\) = 3.0 x 10^-3 m
Maximum shearing stress = 6.9 x 10⁷ Pa
Maximum stress = \(\frac{\text { Maximum load or force }}{\text { Area }}\)
Maximum force = Maximum stress x Area
= 6.9 x 10⁷ x π x (r)²
= 6.9 x 10⁷ x 3.14 x (3 x10^-3)²
= 1949.94 N
Each rivet carries one quarter of the load.
∴ Maximum tension on each rivet = 4 x 1949.94 = 7799.76 N.

Question 21.
The Marina trench is located in the Pacific Ocean, and at one place it is nearly eleven km beneath the surface of the water. The water pressure at the bottom of the trench is about 1.1 x 10⁸ Pa. A steel ball of initial volume 0.32m is dropped into the ocean and falls to the bottom of the trench. What is the change in the volume of the ball when it reaches to the bottom?
Solution:
Water pressure at the bottom, p = 1.1 x 10⁸ Pa
Initial volume of the steel ball, V = 0.32 m³
Bulk modulus of steel, B = 1.6 x 10¹¹ Nm^-2
The ball falls at the bottom of the Pacific Ocean, which is 11 km beneath the surface.
Let the change in the volume of the ball on reaching the bottom of the trench be ΔV.
Bulk modulus, B = \( \frac{p}{\frac{\Delta V}{V}}\)
ΔV = \(\frac{p V}{B}=\frac{1.1 \times 10^{8} \times 0.32}{1.6 \times 10^{11}}\)
= 2.2 x 10^-4 m³
Therefore, the change in volume of the ball on reaching the bottom of the trench is 2.2 x 10^-4 m³.

Punjab State Board PSEB 11th Class Physics Book Solutions Chapter 6 Work, Energy and Power Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Physics Chapter 6 Work, Energy and Power

PSEB 11th Class Physics Guide Work, Energy and Power Textbook Questions and Answers

Question 1.
The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative:
(a) work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket.
(b) work done by gravitational force in the above case,
(c) work done by friction on a body sliding down an inclined plane,
(d) work done by an applied force on a body moving on a rough horizontal plane with uniform velocity,
(e) work done by the resistive force of air on a vibrating pendulum in bringing it to rest.
Solution:
(a) Positive
In the given case, force and displacement are in the same direction. Hence, the sign of work done is positive. In this case, the work is done on the bucket.

(b) Negative
In the given case, the direction of force (vertically downward) and displacement (vertically upward) are opposite to each other. Hence, the sign of work done is negative.

(c) Negative
Since the direction of frictional force is opposite to the direction of motion, the work done by frictional force is negative in this case.

(d) Positive
Here the body is moving on a rough horizontal plane. Frictional force opposes the motion of the body. Therefore, in order to maintain a uniform velocity, a uniform force must be applied to the body. Since the applied force acts in the direction of motion of the body, the work done is positive.

(e) Negative
The resistive force of air acts in the direction opposite to the direction of motion of the pendulum. Hence, the work done is negative in this case.

Question 2.
A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1.
Compute the
(a) work done by the applied force in 10 s,
(b) work done by friction in 10 s,
(c) work done by the net force on the body in 10 s,
(d) change in kinetic energy of the body in 10 s, and interpret your results.
Solution:
Mass of the body, m = 2 kg
Applied force, F = 7 N
Coefficient of kinetic friction, μ = 0.1
Initial velocity, u = 0
Time, t = 10 s
The acceleration produced in the body by the applied force is given by Newton’s second law of motion as:
a’ = \(\frac{F}{m}\) = \(\frac{7}{2}\) = 3.5 m/s²
Frictional force is given as:
f = μ mg = 0.1 × 2 × 9.8 = -1.96 N
The acceleration produced by the frictional force:
a” = –\(\frac{1.96}{2}\) = -0.98 m/s²
Total acceleration of the body:
a = a’ + a”
= 3.5 + (-0.98) = 2.52 m/s²
The distance travelled by the body is given by the equation of motion:
s = ut + \(\frac{1}{2}\) at²
= 0 + \(\frac{1}{2}\) × 2.52 × (10)² = 126 m
(a) Work done by the applied force, W_a = F × s = 7x 126 = 882 J
(b) Work done by the frictional force, W_f = f × s = -1.96 × 126 = -247 J
(c) Net force = 7 + (-1.96) = 5.04 N
Work done by the net force, W_net = 5.04 × 126 = 635 J
(d) From the first equation of motion, final velocity can be calculated as:
υ = u + at
= 0 + 2.52 × 10 = 25.2 m/s
Change in kinetic energy = \(\frac{1}{2}\) mυ ² – \(\frac{1}{2}\) mu²
= \(\frac{1}{2}\) × 2(υ² – u²) = (25.2)² – 0² = 635 J

Question 3.
Given in figure below are examples of some potential energy functions in one dimension. The total energy of the particle is indicated by a cross on the ordinate axis. In each case, specify the regions, if any, in which the particle cannot be found for the given energy. Also, indicate the minimum total energy the particle muqt have in each case. Think of simple physical contexts for which these potential energy shapes are relevant.


Solution:
Total energy of a system is given by the relation:
E = P.E. + K.E.
> K.E. = E – P.E.
Kinetic energy of a body is a positive quantity. It cannot be negative. Therefore, the particle will not exist in a region where K.E. becomes negative.
(i) In the given case, the potential energy (V₀) of the particle becomes greater than total energy (E) for x > a. Hence, kinetic energy becomes negative in this region. Therefore, the particle will not exist in this region. The minimum total energy of the particle is zero.

(ii) In the given case, the potential energy (V₀) is greater than total energy (E) in all regions. Hence, the particle will not exist in this region.

(iii) In the given case, the condition regarding the positivity of K.E. is satisfied only in the region between x > a and x<b.
The minimum potential energy in this case is -V₁. Therefore, K.E. = E-(-V₁) = E + V₁.
Therefore, for the positivity of the kinetic energy, the total energy of the particle must be greater than -Vj. So, the minimum total energy the particle must have is -V₁.

(iv) In the given case, the potential energy (V₀) of the particle becomes greater than the total energy (E) for –\(\frac{b}{2}\)< x <\(\frac{b}{2}\) and –\(\frac{a}{2}\)< x <\(\frac{a}{2}\).
Therefore, the particle will not exist in these regions.
The minimum potential energy in this case is -V₁. Therefore, K.E. = E – (-V₁ ) = E + V₁. Therefore, for the positivity of the kinetic energy, the total energy of the particle must be greater than -V₁. So, the minimum total energy the particle must have is -V₁.

Question 4.
The potential energy function for a particle executing linear simple harmonic motion is given by V (x) = kx² / 2, where k is the force constant of the oscillator. For k = 0.5 Nm^-1, the graph of V (x) versus x is shown in figure below. Show that a particle of total energy 1 J moving under this potential must ‘turn back’ when it reaches x = ±2m.

Solution:
Total energy of the particle, E = 1 J
Force constant, k = 0.5 Nm^-1
Kinetic energy of the particle, K = \(\frac{1}{2}\)mυ²
According to the conservation law:
E = V + K
1 = \(\frac{1}{2}\)kx² + \(\frac{1}{2}\)mυ ²
At the moment of ‘turn back’, velocity (and hence K) becomes zero.
> 1 = \(\frac{1}{2}\)kx²
\(\frac{1}{2}\) × 0.5 ײ = 1
x² = 4
x = ±2
Hence, the particle turns back when it reaches x = ±2 m.

Question 5.
Answer the following:
(a) The casing of a rocket in flight burns up due to friction. At whose expense is the heat energy required for burning obtained? The rocket or the atmosphere?
(b) Comets move around the sun in highly elliptical orbits. The gravitational force on the comet due to the sun is not normal to the comet’s velocity in general. Yet the work done by the gravitational force over every complete orbit of the comet is zero. Why?
(c) An artificial satellite orbiting the earth in very thin atmosphere loses its energy gradually due to dissipation against atmospheric resistance, however small. Why then does its speed increase progressively as it comes closer and closer to the earth?
(d) In fig. (i) the man walks 2 m carrying a mass of 15 kg on his hands. In fig. (ii), he walks the same distance pulling the rope behind him. The rope goes over a pulley, and a mass of 15 kg hangs at its other end. In which case is the work done greater?

Solution:
(a) The burning of the casing of a rocket in flight (due to friction) results in the reduction of the mass of the rocket.
According to the conservation of energy:
Total Energy (T.E.) = Potential energy (P.E.) + Kinetic energy (K.E.)
= mgh + \(\frac{1}{2}\) mυ²
The reduction in the rocket’s mass causes a drop in the total energy. Therefore, the heat energy required for the burning is obtained from the rocket.

(b) Gravitational force is a conservative force. Since the work done by a conservative force over a closed path is zero, the work done by the gravitational force over every complete orbit of a comet is zero.

(c) When an artificial satellite, orbiting around earth, moves closer to earth, its potential energy decreases because of the reduction in the height. Since the total energy of the system remains constant, the reduction in P.E. results in an increase in K.E. Hence, the velocity of the satellite increases. However, due to atmospheric friction, the total energy of the satellite decreases by a small amount.

(d) Case (i)
Mass, m = 15 kg
Displacement, s = 2 m
Work done, W = Fs cosθ
where, θ = Angle between force and displacement
= mgs cosθ = 15 × 2 × 9.8 cos 90°
= 0 ( cos90° = 0)

Case (ii)
Mass, m = 15 kg Displacement, s = 2 m
Here, the direction of the force applied on the rope and the direction of the displacement of the rope are same.
Therefore, the angle between them, θ =0°
Since, cos0° = 1
Work done, W = Fs cosθ = mgs
= 15 × 9.8 × 2 = 294J
Hence, more work is done in the second case.

Question 6.
Underline the correct alternative:
(a) When a conservative force does positive work on a body, the potential energy of the body increases/decreases/ remains unaltered.
(b) Work done by a body against friction always results in a loss of its kinetic/potential energy.
(c) The rate of change of total momentum of a many-particle system is proportional to the external force/sum of the internal forces on the system.
(d) In an inelastic collision of two bodies, the quantities which do not change after the collision are the total kinetic energy/total linear momentum/total energy of the system of two bodies.
Solution:
(a) Decreases, A conservative force does a positive work on a body when it displaces the body in the direction of force. As a result, the body advances toward the centre of force. It decreases the separation between the two, thereby decreasing the potential energy of the body.

(b) Kinetic energy, The work done against the direction of friction reduces the velocity of a body. Hence, there is a loss of kinetic energy of the body.

(c) External force, Internal forces, irrespective of their direction, cannot produce any change in the total momentum of a body. Hence, the total momentum of a many-particle system is proportional to the external forces acting on the system.

(d) Total linear momentum, The total linear momentum always remains conserved whether it is an elastic collision or an inelastic collision.

Question 7.
State if each of the following statements is true or false. Give reasons for your answer.
(a) In an elastic collision of two bodies, the momentum and energy of each body is conserved.
(b) Total energy of a system is always conserved, no matter what internal and external forces on the body are present.
(c) Work done in the motion of a body over a closed loop is zero for every force in nature.
(d) In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system.
Solution:
(a) False, In an elastic collision, the total energy and momentum of both the bodies, and not of each individual body, is conserved.

(b) False, Although internal forces are balanced, they cause no work to be done on a body. It is the external forces that have the ability to do work. Hence, external forces are able to change the energy of a system.

(c) False, The work done in the motion of a body over a closed loop is zero for a conservation force only.

(d) True, In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system. This is because in such collisions, there is always a loss of energy in the form of heat, sound, etc.

Question 8.
Answer carefully, with reasons:
(a) In an elastic collision of two billiard balls, is the total kinetic energy conserved during the short time of collision of the balls (i. e., when they are in contact)?
(b) Is the total linear momentum conserved during the short time of an elastic collision of two balls?
(c) What are the answers to (a) and (b) for an inelastic collision?
(d) If the potential energy of two billiard balls depends only on the separation distance between their centres, is the collision elastic or inelastic? (Note, we are talking here of potential energy corresponding to the force during collision, not gravitational potential energy).
Solution:
(a) No, In an elastic collision, the total initial kinetic eneigy of the bails will be equal to the total final kinetic energy of the balls. This kinetic energy is not conserved at the instant the two balls are in contact with each other. In fact, at the time of collision, the kinetic energy of the balls will get converted into potential energy.

(b) Yes, In an elastic collision, the total linear momentum of the system always remains conserved.

(c) No, In an inelastic collision, there is always a loss of kinetic energy, i. e., the total kinetic energy of the billiard balls before collision will always be greater than that after collision.
Yes, The total linear momentum of the system of billiards balls will remain conserved even in the case of an inelastic collision.

(d) Elastic, In the given case, the forces involved are conservation. This is because they depend on the separation between the centres of the billiard balls. Hence, the collision is elastic.

Question 9.
A body is initially at rest. It undergoes one-dimensional motion with constant acceleration. The power delivered to it at time t is proportional to
(i) \(t^{\frac{1}{2}}\)
(ii) t
(iii) \(t^{\frac{3}{2}}\)
(iv) t²
Solution:
(ii) t
Let a body of mass m which is initially at rest undergoes one-dimensional motion under a constant force F with a constant acceleration a.
Acceleration (a) = \(\frac{F}{m}\) …………. (i)
Using equation of motion, υ = u + at
⇒ υ = 0 + \(\frac{F}{m}\).t …………. (∵ u = 0)
⇒ υ = \(\frac{F}{m}\)t …………… (ii)
Power delivered (P) = Fυ
Substituting the value from eq. (ii), we get
⇒ P = F × \(\frac{F}{m}\) × t
⇒ P = \(\frac{F^{2}}{m}\)t
Dividing and multiplying by m in R.H.S.
P = \(\frac{F^{2}}{m^{2}}\) × mt= a²mt [Using eq.(i)]
As mass m and acceleration a are constant.
∴ P ∝ t

Question 10.
A body is moving unidirectionally under the influence of a source of constant power. Its displacement in time t is proportional to
(i) \(t^{\frac{1}{2}}\)
(ii) t
(iii) \(t^{\frac{3}{2}}\)
(iv) t²
Solution:
(iii) \(t^{\frac{3}{2}}\)
Power is given by the relation:
P = Fυ
= maυ = mυ \(\frac{d v}{d t}\) = Constant (say,k )
υ dv = \(\frac{k}{m}\) dt
Integrating both sides:
\(\frac{v^{2}}{2}=\frac{k}{m}\)t
υ = \(\sqrt{\frac{2 k t}{m}}\)
For displacement x of the body, we have:
υ = \(\frac{d x}{d t}=\sqrt{\frac{2 k}{m}} t^{\frac{1}{2}}\)
dx = k’\(t^{\frac{1}{2}}\)dt
where, k’ = \(\sqrt{\frac{2 k}{3}}\) = New constant
On integrating both sides, we get:
x = \(\frac{2}{3} k^{\prime} t^{\frac{3}{2}}\)
x ∝ \(t^{\frac{3}{2}}\)

Question 11.
A body constrained to move along the z-axis of a coordinate system is subject to a constant force F given by F = -î + 2ĵ + 3k̂N
where î, ĵ, k̂ are unit vectors along the x-, y- and z-axis of the system respectively. What is the work done by this force in moving the body a distance of 4 m along the z-axis?
Solution:
Force exerted on the body, F = -î + 2ĵ + 3k̂N
Displacement, s = 4 k̂ m
Work done, W = F.s
= (-î + 2ĵ + 3k̂).(4k̂)
= 0 + 0 + 3 × 4 = 12 J
Hence, 12 J of work is done by the force on the body.

Question 12.
An electron and a proton are detected in a cosmic ray experiment, the first with kinetic energy 10 keV, and the second with 100 keV. Which is faster, the electron or the proton? Obtain the ratio of their speeds, (electron mass = 9.11 × 10^-31 kg, proton mass = 1.67 × 10^-27 kg, 1 eV = 1.60 × 10^-19 J).
Solution:
Mass of the electron, m_e = 9.11 × 10^-31 kg
Mass of the proton, m_p =1.67 × 10^-27 kg
Kinetic energy of the electron, E_Ke =10 keV = 10⁴ eV
= 10⁴ × 1.60 × 10^-19
1.60 × 10^-15 J
Kinetic energy of the proton, E_Kp = 100 keV = 10⁵ eV = 1.60 × 10^-14 J
For the velocity of an electron v_e, its kinetic energy is given by the relation:

Question 13.
A rain drop of radius 2 mm falls from a height of500 m above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) until at half its original height, it attains its maximum (terminal) speed, and moves with uniform speed thereafter. What is the work done hy the gravitational force on the drop in the first and second half of its journey? What is the work done by the resistive force in the entire journey if its speed on reaching the ground is 10 ms^-1?
Solution:
Radius of the rain drop, r = 2 mm = 2 × 10 ^-3 m
Volume of the rain drop, V = \(\frac{4}{3}\) πr³
= \(\frac{4}{3}\) × 3.14 × (2 × 10^-3)³ m^-3
Density of water, ρ = 10³ kgm^-3
Mass of the rain drop, m = ρV
= \(\frac{4}{3}\) × 3.14 × (2 × 10^-3)³ × 10³
Gravitational, F = mg
= \(\frac{4}{3}\) × 3.14 × (2 × 10^-3)³ × 10³ 9.8N
The work done by the gravitational force on the drop in the first half of its journey.
W₁ = Fs
= \(\frac{4}{3}\) × 3.14 × (2 × 10^-3 )³ × 10³ × 9.8 × 250
= 0.082 J

This amount of work is equal to the work done by the gravitational force on the drop in the second half of its journey, i. e., W_II, = 0.082 J.
As per the law of conservation of energy, if no resistive force is present, then the total energy of the rain drop will remain the same.
∴ Total energy at the top:
E_T = mgh + 0
= \(\frac{4}{3}\) × 3.14 × (2 × 10^-3)³ × 10³ × 9.8 × 500
= 0.164 J
Due to the presence of a resistive force, the drop hits the ground with a velocity of 10 m/s.
∴ Total energy at the ground:
E_g = \(\frac{1}{2}\) mυ² + 0
= \(\frac{1}{2}\) × \(\frac{4}{3}\) × 3.14 × (2 × 10^-3)³ × 10³ × 9.8 × (10)²
= 1.675 × 10^-3 J = 0.001675
∴ Resistive force = E_g – E_t = 0.001675 – 0.164 = -0.162 J

Question 14.
A molecule in a gas container hits a horizontal wall with speed 200 m s^-1 and angle 30° with the normal, and rebounds with the same speed. Is momentum conserved in the collision? Is the collision elastic or inelastic?
Solution:
Yes; Collision is elastic
The momentum of the gas molecule remains conserved Whether the collision is elastic or inelastic.
The gas molecule moves with a velocity of 200 m/s and strikes the stationary wall of the container, rebounding with the same speed.
It shows that the rebound velocity of the wall remains zero. Hence, the total kinetic energy of the molecule remains conserved during the collision. The given collision is an example of an elastic collision.

Question 15.
A pump on the ground floor of a building can pump up water to fill a tank of volume 30 m3 in 15 min. If the tank is 40 m above the ground, and the efficiency of the pump is 30%, how much electric power is consumed by the pump?
Solution:
Volume of the tank, V = 30 m³
Time of operation, t = 15 min = 15 × 60 = 900 s
Height of the tank, h = 40 m
Efficiency of the pump, η = 30%
Density of water, ρ = 10³ kg/m³
Mass of water, m = ρ V = 30 × 10³ kg
Output power can be obtained as:
P₀ = \(\frac{\text { Work done }}{\text { Time }}=\frac{m g h}{t}\)
= \(\frac{30 \times 10^{3} \times 9.8 \times 40}{900}\) = 13.067 × 10³ W
For input power P_i efficiency η, is given by the relation :
η = \(\frac{P_{o}}{P_{i}}\) = 30%
P_i = \(\frac{13.067}{30}\) × 100 10³
= 0.436 × 10² W = 43.6 kW

Question 16.
Two identical ball bearings in contact with each other and resting on a frictionless table are hit head-on by another ball bearing of the same mass moving initially with a speed V. If the collision is elastic, which of the following figure is a possible result after collision?

Solution:
It can be observed that the total momentum before and after collision in each case is constant.
For an elastic collision, the total kinetic energy of a system remains conserved before and after collision.
For mass of each ball bearing m, we can write:
Total kinetic energy of the system before collision:
= \(\frac{1}{2}\)mV² + \(\frac{1}{2}\)(2m)0
= \(\frac{1}{2}\)mV²

Case (i)
Total kinetic energy of the system after collision:
= \(\frac{1}{2}\)m × 0 + \(\frac{1}{2}\)(2m) (\(\frac{V}{2}\))²
= \(\frac{1}{4}\)mV²
Hence, the kinetic energy of the system is not conserved in case (i).

Case (ii)
Total kinetic energy of the system after collision:
= \(\frac{1}{2}\) (2m) × 0 + \(\frac{1}{2}\)mV²
= \(\frac{1}{2}\) mV²
Hence, the kinetic energy of the system is conserved in case (ii).

Case (iii)
Total kinetic energy of the system after collision:
= \(\frac{1}{2}\)(3m) (\(\frac{V}{3}\))²
= \(\frac{1}{6}\) mV²
Hence, the kinetic energy of the system is not conserved in case (iii).

Question 17.
The bob A of a pendulum released from 30° to the vertical hits another hob B of the same mass at rest on a table as shown in’ figure given below. How high does the bob A rise after the collision? Neglect the size of the bobs and assume the collision to be elastic.

Solution:
Bob A will not rise at all
In an elastic collision between two equal masses in which one is stationary, while the other is moving with some velocity, the stationary mass acquires the same velocity, while the moving mass immediately comes to rest after collision. In this case, a complete transfer of momentum takes place from the moving mass to the stationary mass. Hence, bob A of mass m, after colliding with bob B of equal mass, will come to rest, while bob B will move with the velocity of bob A at the instant of collision.

Question 18.
The bob of a pendulum is released from a horizontal position. If the length of the pendulum is 1.5 m, what is the speed with which the bob arrives at the lowermost-point, given that it dissipated 5% of its initial energy against air resistance?
Solution:
Length of the pendulum, l = 1.5 m
Mass of the bob = m
Energy dissipated = 5%
According to the law of conservation of energy, the total energy of the system remains constant.

At the horizontal position:
Potential energy of the bob, E_P = mgl
Kinetic energy of the bob, E_K = 0
Total energy = mgl …………. (i)

At the lowermost point (mean position):
Potential energy of the bob, E_P = 0
Kinetic energy of the bob, E_K = \(\frac{1}{2}\)mυ²
Total energy, E_x = \(\frac{1}{2}\)mυ² ………….. (ii)
As the bob moves from the horizontal position to the lowermost point, 5% of its energy gets dissipated.
The total energy at the lowermost point is equal to 95% of the total energy at the horizontal point, i. e.,
\(\frac{1}{2}\)mυ² = \(\frac{95}{100}\) × mgl
υ = \(\sqrt{\frac{2 \times 95 \times 1.5 \times 9.8}{100}}\) = 5.28 m/s

Question 19.
A trolley of mass 300 kg carrying a sandbag of 25 kg is moving uniformly with a speed of 27 km/h on a frictionless track. After a while, sand starts leaking out of ahole on the floor of the trolley at the rate of 0.05 kg s^-1 . What is the speed of the trolley after the entire sand bag is empty?
Solution:
The sand bag is placed on a trolley that is moving with a uniform speed of 27 km/h. The external forces acting on the system of the sand bag and the trolley is zero. When the sand starts leaking from the bag, there will be no change in the velocity of the trolley. This is because the leaking action does not produce any external force on the system. This is in accordance with Newton’s first law of motion. Hence, die speed of the trolley will remain 27 km/h.

Question 20.
A body of mass 0.5 kg travels in a straight line with velocity υ = ax^3/2 where a = 5 m^-1/2s^-1. What is the work done by the net force during its displacement from x = 0 to x = 2 m?
Solution:
Mass of the body, m = 0.5 kg
Velocity of the body is governed by the equation, υ = a \(x^{\frac{3}{2}}\) where,
a = 5m^-1/2s^-1
Initial velocity, u (at x = 0) = 0
Final velocity, υ (at x = 2 m) = 5 × (2)^3/2 m/s = 10√2 m/s
Work done, W = Change in kinetic energy
= \(\frac{1}{2}\)m(υ² – u²)
= \(\frac{1}{2}\) × 0.5[(10√2)² – (0)²]
= \(\frac{1}{2}\) × 0.5 × 10 × 10 × 2
= 50 J

Question 21.
The blades of a windmill sweep out a circle of area A. (a) If the wind flows at a velocity v perpendicular to the circle, what is the mass of the air passing through it in time t? (b) What is the kinetic energy of the air? (c) Assume that the windmill converts 25% of the wind’s energy into electrical energy, and that A = 30 m², υ = 36km/h and the density of air is 1.2 kg m^-3. What is the electrical power produced?
Solution:
(a) Area of the circle swept by the windmill = A
Velocity of the wind = υ
Density of air = ρ
Volume of the wind flowing through the windmill per sec = Aυ
Mass of the wind flowing through the windmill per sec = ρ Aυ
Mass m, of the wind flowing through the windmill in time t = ρ A υ t

(b) Kinetic energy of air = \(\frac{1}{2}\) mυ²
= \(\frac{1}{2}\) (ρAυt)v² = \(\frac{1}{2}\)ρ Aυ³t

(c) Area of the circle swept by the windmill, A = 30 m²
Velocity of the wind, υ =36 km/h = 36 × \(\frac{5}{18}\) m/s
= 10 m/s [1 km/s = \(\frac{5}{18}\) m/s]
Density of air, ρ = 1.2 kg m^-3
Electric energy produced = 25% of the wind energy
= \(\frac{25}{100}\) ×Kinetic energy of air
= \(\frac{1}{8}\)ρAυ³t

\(\) = \(\frac{1}{8}\)ρAυ³
= \(\frac{1}{8}\) × 1.2 × 30 × (10)³
= 4.5 × 10³ W = 4.5 kW

Question 22.
A person trying to lose weight (dieter) lifts a 10 kg mass, one thousand times, to a height of 0.5 m each time. Assume that the potential energy lost each time she lowers the mass is dissipated, (a) How much work does she do against the gravitational force? (b) Fat supplies 3.8 × 10⁷ J of energy per kilogram which is converted to mechanical energy with a 20% efficiency rate. How much fat will the dieter use up?
Solution:
(a) Mass of the weight, m = 10 kg
Height to which the person lifts the weight, h = 0.5
m Number of times the weight is lifted, n = 1000
∴ Work done against gravitational force:
= n(mgh)
=1000 × 10 × 9.8 × 0.5
= 49 × 10³ J = 49 kJ

(b) Energy equivalent of 1 kg of fat = 3.8 × 10⁷ J
Efficiency rate = 20%
Mechanical energy supplied by the person’s body
= \(\frac{20}{100}\) × 3.8 × 10⁷ J
= \(\frac{1}{5}\) × 3.8 × 107 J 5
Equivalent mass of fat lost by the dieter
= \(\frac{1}{\frac{1}{5} \times 3.8 \times 10^{7}}\) × 49 × ³
= \(\frac{245}{3.8}\)× 3.8 × 107
= × 10^-4
= 6.45 × 10^-3 kg

Question 23.
A family uses 8 kW of power, (a) Direct solar energy is incident on the horizontal surface at an average rate of 200 W per square meter. If 20% of this energy can be converted to useful electrical energy, how large an area is needed to supply 8 kW? (b) Compare this area to that of the roof of a typical house.
Solution:
(a) Power used by the family, P = 8 kW = 8 × 10³ W
Solar energy received per square meter = 200 W
Efficiency of conversion from solar to electrical energy = 20%
Area required to generate the desired electricity = A
As per the information given in the question, we have
8 × 10³ = 20% × (A × 200)
= \(\frac{20}{100}\) × A × 200
> A = \(\frac{8 \times 10^{3}}{40}\) = 200m²

(b) In order to compare this area to that of the roof of a typical house, let ‘a’ be the side of the roof
∴ area of roof =a × a = a²
Thus a² = 200 m²
or a = \(\sqrt{200 \mathrm{~m}^{2}}\) = 14.14 m
∴ area of roof = 14.14 × 14.14 m²
Thus 200 m² is comparable to the roof of a typical house of dimensions 14.14 m × 14.14 m = 14 m × 14 m.

Question 24.
A bullet of mass 0.012 kg and horizontal speed 70 ms^-1 strikes a block of wood of mass 0.4 kg and instantly comes to rest with respect to the block. The block is suspended from the ceiling by means of thin wires. Calculate the height to which the block rises. Also, estimate the amount of heat produced in the block.
Solution:
Mass of the bullet, m = 0.012 kg
Initial speed of the bullet, u_b = 70 m/s
Mass of the wooden block, M = 0.4 kg
Initial speed of the wooden block, u_B = 0
Final speed of the system of the bullet and the block = υ
Applying the law of conservation of momentum,
mu_b +MU_B = (m + M)υ
012 × 70 + 0.4 × 0 = (0.012 + 0.4)υ
∴ υ = \(\frac{0.84}{0.412}\) = 2.04 m/s

For the system of the bullet and the wooden block,
Mass of the system, m’ = 0.412 kg
Velocity of the system = 2.04 m/s
Height up to which the system rises = h
Applying the law of conservation of energy to this system,
Potential energy at the highest point
= Kinetic energy at the lowest point
m’ gh = \(\frac{1}{2}\)m’ υ²
h = \(\frac{1}{2}\)(\(\frac{v^{2}}{g}\))
= \(\frac{1}{2}\) (\(\frac{(2.04)^{2}}{9.8}\)) = 0.2123m
The wooden block will rise to a height of 0.2123 m.
Heat produced = Kinetic energy of the bullet
– Kinetic energy of the system
= \(\frac{1}{2}\) mu_b– \(\frac{1}{2}\)m’ υ²
= \(\frac{1}{2}\) × 0.012 × (70)² – \(\frac{1}{2}\) × 0.412 × (2.04)²
= 29.4 – 0.857 = 28.54 J

Question 25.
Two inclined frictionless tracks, one gradual and the other steep meet at A from where two stones are allowed to slide down from rest, one on each track (see figure). Will the stones reach the bottom at the same time? Will they reach there with the same speed? Explain. Given θ₁ = 30°, θ₂ = 60°and h = 10m, what are the speeds and times taken by the two stones?

Solution:
No; the stone moving down the steep plane will reach the bottom first Yes; the stones will reach the bottom with the same speed The given situation can be shown as in the following figure:

Here, the initial height (AD) for both the stones is the same (h).
Hence, both will have the same potential energy at point A.
As per the law of conservation of energy, the kinetic energy of the stones at points B and C will also be the same, i. e.,
\(\frac{1}{2}\) mυ₁² = \(\frac{1}{2}\)mυ₂²
υ₁ = υ₁ = υ, say
where,
m = Mass of each stone
υ = Speed of each stone at points B and C
Hence, both stones will reach the bottom with the same speed, υ.

For stone I: Net force acting on this stone is given by:
F_net = ma₁ = mgsinθ₁
a₁ = gsinθ₁

For stone II: a₂ = gsinθ₂
∵ θ₂ > θ₁
∴ sinθ₂ > sinθ₁
∴ a₂ > a₁
Using the first equation of motion, the time of slide can be obtained as:
υ = u + at
∴ t = \(\frac{v}{a}\) (∵ u = 0)
For stone I: t₁ = \(\frac{v}{a_{1}}\)
For stone II: t₂ = \(\frac{v}{a_{2}}\)
∵ a₂ > a₁
∴ t₂ < t₁
Hence, the stone moving down the steep plane will reach the bottom first. The speed (υ) of each stone at points B and C is given by the relation obtained from the law of conservation of energy.

Question 26.
A1 kg block situated on a rough incline is connected to a spring of spring constant 100 Nm^-1 as shown in figure given below. The block is released from rest with the spring in the unstretched position. The block moves 10 cm down the incline before coming to rest. Find the coefficient of friction between the block and the incline. Assume that the spring has a negligible mass and the pulley is frictionless.

Solution:
Mass of the block, m = 1 kg
Spring constant, k = 100 N m^-1
Displacement in the block, x = 10 cm = 0.1 m
The given situation can be shown as in the following figure.

At equilibrium:
Normal reaction, R = mg cos 37°
Frictional force, f = μR = p mg cos 37°
where, μ is the coefficient of friction
Net force acting on the block = mg sin 3 7° – f
= mgsin37° – μmgcos37°
= mg (sin37° – μcos37°)
At equilibrium, the work done by the block is equal to the potential energy of the spring, i.e.,
mg (sin37° – μcos37°) x = \(\frac{1}{2}\)kx ²
1 × 9.8 (sin 37° – μcos37°) = \(\frac{1}{2}\) × 100 × 0.1
0.6018 – μ × 0.7986 = 0.5102
0.7986 μ = 0.6018 – 0.5102 = 0.0916
∴ μ = \(\frac{0.0916}{0.7986}\) = 0.115

Question 27.
A bolt of mass 0.3 kg falls from the ceiling of an elevator moving down with an uniform speed of 7 m^-1. It hits the floor of the elevator (length of the elevator = 3m) and does not rebound. What is the heat produced by the impact? Would your answer be different if the elevator were stationary?
Solution:
Mass of the bolt, m = 0.3 kg
Speed of the elevator = 7 m/s
Height, h = 3 m
Since the relative velocity of the bolt with respect to the lift is zero, at the time of impact, potential energy gets converted into heat energy.
Heat produced = Loss of potential energy
= mgh = 0.3 × 9.8 × 3 = 8.82 J
The heat produced will remain the same even if the lift is stationary. This is because of the fact that the relative velocity of the bolt with respect to the lift will remain zero.

Question 28.
A trolley of mass 200 kg moves with a uniform speed of 36 km/h on a frictionless track. A child of mass 20 kg runs on the trolley from one end to the other (10 m away) with a speed of 4 m s^-1 relative to the trolley in a direction opposite to the its motion, and jumps out of the trolley. What is the final speed of the trolley? How much has the trolley moved from the time the child begins to run?
Solution:
Mass of the trolley, M = 200 kg
Speed of the trolley, υ = 36 km/h = 36 × \(\frac{5}{18}\)m/s = 10 m/s
Mass of the boy, m = 20 kg
Initial momentum of the system of the boy and the trolley
= (M + m)υ
= (200 + 20) × 10
= 2200 kg-m/s
Let v’ be the final velocity of the trolley with respect to the ground.
Final velocity of the boy with respect to the ground = υ’ – 4
Final momentum = Mυ’ + m(υ’ – 4)
= 200 υ’ + 20υ’ – 80
= 220 υ’ – 80
As per the law of conservation of momentum,
Initial momentum = Final momentum
2200 =220 υ’ – 80
∴ υ’ = \(\frac{2280}{220}\) = 10.36 m/s
Length of the trolley, l = 10 m
Speed of the boy, υ” = 4 m/s
Time taken by the boy to run, t = \(\frac{10}{4}\) = 2.5 s
∴ Distance moved by the trolley = υ’ × t= 10.36 × 2.5 = 25.9 m

Question 29.
Which of the following potential energy curves in the given figure cannot possibly describe the elastic collision of two billiard balls? Here r is the distance between centres of the balls.

Solution:
(i), (ii), (iii), (iv), and (vi)
The potential energy of a system of two masses is inversely proportional to the separation between them. In the given case, the potential energy of the system of the two balls will decrease as they come closer to each other. It will become zero [i.e., V (r) = 0] when the two balls touch each other, i.e., at r = 2R, where R is the radius of each billiard ball. The potential energy curves given in figures (i), (ii), (iii), (iv), and (vi) do not satisfy these two conditions. Hence, they do not describe the elastic collisions between them.

Question 30.
Consider the decay of a free neutron at rest: n → p + e^–. Show that the two-body decay of this type must necessarily give an electron of fixed energy and, therefore, cannot account for the observed continuous energy distribution in the β-decay of a neutron or a nucleus (see figure).

[Note: The simple result of this exercise was one among the several arguments advanced by W. Pauli to predict the existence of a third particle in the decay products of P-decay. This particle is known as
neutrino. We now know that it is a particle of intrinsic spin \(\frac{1}{2}\) (like e^–, p or n), but is neutral, and either massless or having an extremely small mass
(compared to the mass of electron) and which interacts very weakly with matter. The correct decay process of neutron is: n → p + e^– + υ]
Solution:
The decay process of free neutron at rest is given as:
n → p + e^–
From Einstein’s mass-energy relation, we have the energy of electron as
Δ𝜏 mc²
where,
Δ𝜏 m = Mass defect
= Mass of neutron – (Mass of proton + Mass of electron)
c = Speed of light
Δ𝜏 m and c are constants. Hence, the given two-body decay is unable to explain the continuous energy distribution in the p-decay of a neutron or a nucleus. The presence of neutrino v on the LHS of the decay correctly explains the continuous energy distribution.
Here, Δ𝜏 m= mass of neutron – (mass of proton + mass of electron)
= 1.6747 × 10^-24 -(1.6724 × 10^-24 + 9.11 × 10^-28)
= (1.6747 – 1.6733) × 10^-24
= 0.0014 × 10^-24 gms ‘
∴ E = 0.0014 × 10^-24 × (3 × 10¹⁰)² ergs
= 0.0126 × 10^-4 ergs
= [Latex]\frac{0.0126 \times 10^{-4}}{1.6 \times 10^{-12}}[/Latex]
( ∵ 1 eV = 1.6 × 10^-19 J = 1.6 × 10^-19 × 10⁷ ergs)
= 0.007875 × 10 ⁸ eV
= 0.7875 × 10⁶ eV
= 0.79 MeV

A positron has the same mass as an electron but an opposite charge of+e. When an electron and a positron come close to each other, they destroy each other. Their masses, are converted into energy according to Einstein’s relation and the energy so obtained is released in the form of y-rays and is given by
E’ = mc² = 2 × 9.1 × 10^-31 kgx (3 × 10⁸ ms^-1)²
= 1.638 × 10^-13 J
= \(\frac{1.638 \times 10^{-13}}{1.6 \times 10^{-13}}\) MeV
= 1.02MeV (∵1 MeV = 1.6 × 10^-13 eV)
= Minimum energy a photon must possess for pair production.

Alternate method
Decay of a free neutron at rest,
n → p + e^–
Let Δ m be the mass defect during this process.
∴ Mass defect (Δ m) = (Mass of neutron) – (Mass of proton and electron) This mass defect is fixed and hence electron of fixed energy should be produced. Therefore, two-body decay of this type cannot explain the observed continuous energy distribution in the (3-decay of a neutron in a nucleus.

Punjab State Board PSEB 11th Class Biology Book Solutions Chapter 2 Biological Classification Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Biology Chapter 2 Biological Classification

PSEB 11th Class Biology Guide Biological Classification Textbook Questions and Answers

Question 1.
Discuss how classification systems have undergone several changes over a period of time?
Answer:
(i) Linnaeus proposed a two kingdom system of classification with Plantae and Animalia kingdoms was developed that included all plants and animals respectively. But as this system did not distinguish between the eukaryotes and prokaryotes, unicellular and multicellular organisms and photosynthetic (green algae) and non-photosynthetic (fungi) organisms, so scientists found it an inadequate system of classification. Classification systems for the living organisms have hence, undergone several changes over time.

(ii) The two kingdom system of classification was replaced by three kingdom system, then by four and finally by five kingdom system of classification of RH Whittaker (1969).

(iii) The five kingdoms included Monera, Protista, Fungi, Plantae and Animalia. This is the most accepted system of classification of living organisms.

(iv) But, Whittaker has not described viruses and lichens. Then Stanley described viruses, viroids, etc.
Thus, over a period of time, classification systems have undergone several changes.

Question 2.
State two economically important uses of:
(a) Heterotrophic bacteria
(b) Archaebacteria
Answer:
(a) Heterotrophic Bacteria

• Maintain soil fertility by nitrogen fixation, ammonification and nitrification, e.g., Rhizobium bacteria (in the root nodules of legumes).
• The milk products such as butter, cheese, curd, etc., are obtained by the action of bacteria. The milk contains bacterial forms like Streptococcus lacti, Escherichia coli, Lactobacillus lactis and Clostridium sp., etc.

(b) Archaebacteria

• Methanogens are responsible for the production of methane (biogas) from the dung of animals.
• Archaebacteria help in the degradation of waste materials.

Question 3.
What is the nature of cell walls in diatoms?
Answer:
In case of diatoms, the cell wall forms two thin overlapping cells, which fit together as in a soap box. The cell wall is made up of silica. Due to siliceous nature of cell wall, it is known as diatomite or diatomaceous Earth. Diatomaceous Earth is a whitish, highly porous, chemically inert, highly absorbant and fire proof substance.

Question 4.
Find out what do the terms ‘algal bloom’ and ‘red tides’ signify?
Answer:
Sometimes, greqn algae such as Chlorella, Scenedesmus and Spirogyra, etc,, grow in excess in water bodies and impart green colour to the water. These are called algal blooms. Red dianoflagellates (Gonyaulax) grow in abundance in sea and impart red colour to the ocean. This looks like red tides. Both due to algal blooms and ‘red tide’ the animal life declines due to toxins and deficiency of oxygen inside water.

Question 5.
How are viroids different from viruses?
Answer:
Viroids different from viruses as follows:

Virus	Viroids
1. These are smaller than bacteria.	Smaller than viruses.
2. Both RNA and DNA present.	Only RNA is present.
3. Protein coat present.	Protein coat absent.
4. Causes diseases like mumps and AIDS.	Causes plant diseases like spindle tuber diseases – potato.

Question 6.
Describe briefly the four major groups of Protozoa.
Answer:
Protozoans are divided into four phyla on the basis of locomotory organelles – Zooflagellata, Sarcodina, Sporozoa and Ciliates.
(i) Zooflagellates: These protozoans possess one to several flagella for locomotion. Zooflagellates are generally uninucleate, occasionally multinucleate.
The body is covered by a firm pellicle. There is also present cyst formation.
Examples: Giardia, Trypanosoma, Leishmania and Trichonympha, etc.

(ii) Sarcodines: These protozoans possess pseudopodia for locomotion. Pseudopodia are of four types, i.e., lobopodia, filopodia, axopodia and reticulopodia. Pseudopodia are also used for engulfing food particles. Sarcodines are mostly free living, found in freshwater, sea water and on damp soil only a few are parasitic. Nutrition is commonly holozoic. Sarcodines are generally uninucleates. Sarcodines are of four types – Amoeboids (i.e.,Amoeba, etc.), radiolarians (i.e., Acanthometra, etc.), foraminiferans (i.e., Elphidium, etc.) and heliozoans (i.e., Actinophrys, etc.).

(iii) Sporozoans: All of them are endoparasites. Locomotoryorganelles (cilia, flagella, pseudopodia, etc.) are absent. Nutrition is parasitic (absorptive), Phogotrophy is rare. The body is covered with an elastic pellicle or cuticle. Nucleus is single. Contractile vacuoles are absent. Life cyle consists of two distinct asexual and sexual phases. They may be passed in one (monogenetic) or two different hosts (digenetic),e.g., Plasmodium, Monocystis, etc.

(iv) Ciliates: These are aquatic, actively moving organisms because of the presence of thousands of cilia. They have a cavity (gullet) that opens to the outside of the cell surface. The coordinated movements of rows of cilia causes the water laden with food to enter into the gullet, e.g., Paramecium.

Question 7.
Plants are autotrophic. Can you think of some plants that are partially heterotrophic?
Answer:
Plants are autotrophs, i.e., they prepare their own food through the process of photosynthesis. But, in nature there are also some other plants which are partjally heterotrophic, i.e., they partially depend upon another organisms for food requirements, e.g.,
(i) Loranthus and Viscum are partial stem parasites which have leathery leaves. They attack several fruit and forest trees and with the help of their haustoria draw sap from the xylem tissue of the host.

(ii) Insectivorous plants have special leaves to trap insects. The trapped insects are killed and digested by proteolytic enzymes secreted by the epidermis of the leaves, e.g., pitcher plant.

(iii) Parasitic plant, e.g., Cuscutta develops haustoria, which penetrate, the vascular bundles of the host plant to absorb water and solutes.

Question 8.
What do the terms phycobiont and mycobiont signify?
Answer:
In case of lichens (t. e., an association of algae and fungi), the algal partner which is capable of carrying out photosynthesis is known as phycobiont, whereas the fungal partner which is heterotrophic in nature is known as mycobiont.

Question 9.
Give a comparative account of the classes of Kingdom Fungi under the following:
(i) Mode of nutrition
(ii) Mode of reproduction

Fungal Class	Mode of Nutrition	Mode of Reproduction
Myxomycetes	Heterotrophic and mostly saprophytic	Asexual and sexual reproduction
Phycomycetes	Mostly parasites	Asexual and sexual methods
Zygomycetes	Mostly saprophytic	Asexual and sexual reproduction
Ascomycetes	Saprophytes or parasites	Asexual and sexual reproduction
Basidiomycetes	Saprophytes or parasites	Asexual and sexual method
Deuteromycetes	Saprophytes or parasites	Only asexual reproduction

Question 10.
What are the characteristic features of Euglenoids?
Answer:
The characteristic features of euglenoids are as follows :

• They occur in freshwater habitats and damp soils.
• A single long flagella present at the anterior end.
• Creeping movements occur by expansion and expansion of their body known as euglenoid movements.
• Mode of nutrition is holophytic, saprobic or holozoic.
• Reserve food material is paramylum.
• Euglenoids are known as plant and animal.
  Plant characters of them are as follows:
  (a) Presence of chloroplasts with chlorophyll.
  (b) Holophytic nutrition
  Animal characters of them are as follows:
  (a) Presence of pellicle, which is made up of proteins and not a cellulose.
  (b) Presence of stigma.
  (c) Presence of contractile vacuole.
  (d) Presence of longitudinal binary fission.
• Under favourable conditions euglenoids multiply by longitudinal binary fission, e.g., Euglena, Phacus, Paranema, etc.

Question 11.
Give a brief account of viruses with respect to their structure ’ and nature of genetic material. Also name four common viral diseases.
Answer:
Viruses are non-cellular, ultramicroscopic, infectious particles. They are made up of envelope, capsid, nucleoid and occasionally one or two enzymes. Viruses possess an outer thin loose covering called envelope. The central portion of nucleoid is surrounded by capsid that is made up of ( smaller sub-units known as capsomeres.

The nucleic acid present in the viruses is known as nucleoid. It is the r infective part of the virus which utilises the host cell machinery. The
genetic material of viruses is of four types –

(i) Double stranded DNA (dsDNA) as found in pox virus, hepatitis-B virus and herpes virus, etc.
(ii) Single stranded DNA (ssDNA) occur in coliphage Φ, coliphage Φ x 174.
(iii) Double stranded RNA (dsRNA) occur in Reo virus,
(iv) Single stranded RNA (ssRNA) occur in TMV virus, polio virus, etc.
Four common viral diseases are – (i) Polio, (ii) AIDS, (iii) Hepatitis-B (iv) Rabies.

Question 12.
Organise a discussion in your class on the topic—Are viruses living or non-living?
Answer:
Viruses are intermediate between living and non-living objects. They resemble non-living objects in:

• Lacking protoplast. ‘
• Ability to get crystallised.
• High specific gravity which is found only in non-living objects.
• Absence of respiration and energy storing system.
• Absence of growth and division.
• Cannot live independent of a living cell.

They resemble living objects in:

• Presence of genetic material (DNA or RNA).
• Property of mutation.
• Irritability.
• Can grow and multiply inside the host cell.

Punjab State Board PSEB 11th Class Physics Book Solutions Chapter 8 Gravitation Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Physics Chapter 8 Gravitation

PSEB 11th Class Physics Guide Gravitation Textbook Questions and Answers

Question 1.
Answer the following:
(a) You can shield a charge from electrical forces by putting it inside a hollow conductor. Can you shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means?
(b) An astronaut inside a small spaceship orbiting around the earth cannot detect gravity. If the space station orbiting around the earth has a large size, can he hope to detect gravity?
(c) If you compare the gravitational force on the earth due to the sun to that due to the moon, you would find that the Sun’s pull is greater than the moon’s pull. (You can check this yourself using the data available in the succeeding exercises). However, the tidal effect of the moon’s pull is greater than the tidal effect of sun. Why?
Answer:
(a) No, Gravitational influence of matter on nearby objects cannot be screened by any means. This is because gravitational force unlike electrical forces is independent of the nature of the material medium. Also, it is independent of the status of other objects.

(b) Yes, If the size of the space station is large enough, then the astronaut will detect the change in Earth’s gravity (g).

(c) Tindal effect depends inversely upon the cube of the distance while gravitational force depends inversely on the square of the distance. Since the distance between the Moon and the Earth is smaller than the distance between the Sun and the Earth, the tidal effect of the Moon’s pull is greater than the tidal effect of the Sun’s pull.

Question 2.
Choose the correct alternative:
(a) Acceleration due to gravity increases/decreases with increasing altitude.
(b) Acceleration due to gravity increases/decreases with increasing depth (assume the earth to be a sphere of uniform density).
(c) Acceleration due to gravity is independent of mass of the earth/mass of the body.
(d) The formula -G Mm(1 /r₂ -1/r₁) is more/less accurate than the formula mg (r₂ – r₁) for the difference of potential energy between two points r₂ and r₁ distance away from the centre of the earth.
Solution:
(a) Decreases,
Explanation : Acceleration due to gravity at altitude h is given by the relation:
g_h = \(\left(1-\frac{2 h}{R_{e}}\right) g\)
where, R_e = Radius of the Earth
g = Acceleration due to gravity on the surface of the Earth
It is clear from the given relation that acceleration due to gravity decreases with an increase in height.

(b) Decreases,
Explanation : Acceleration due to gravity at depth d is given by the relation:
g_d = \(\left(1-\frac{d}{R_{e}}\right) g\)
It is clear from the given relation that acceleration due to gravity decreases with an increase in depth.

(c) Mass of the body,
Explanation : Acceleration due to gravity of body of mass m is given by the relation:
g = \(\frac{G M_{e}}{R^{2}}\)
where, G = Universal gravitational constant
M_e = Mass of the Earth
R_e = Radius of the Earth
Hence, it can be inferred that acceleration due to gravity is independent of the mass of the body.

(d) More,
Explanation : Gravitational potential energy of two points r₂ and r₁ distance away from the centre of the Earth is respectively given by:
V(r₁) = -GmM
V(r₂) = – \(\frac{G m M}{r_{2}}\)
V = V (r₂) – V(r₁) = -GmM\(\left(\frac{1}{r_{2}}-\frac{1}{r_{1}}\right)\).
Hence, this formula is more accurate than the formula mg (r₂ – r₁).

Question 3.
Suppose there existed a planet that went around the Sun twice as fast as the Earth. What would be its orbital size as compared to that of the Earth?
Solution:
Time taken by the Earth to complete one revolution around the Sun,
T_e = 1 year
Orbital radius of the Earth in its orbit, R_e = 1 AU
Time taken by the planet to complete one revolution around the Sun,
T_p = \(\frac{1}{2} T_{e}=\frac{1}{2}\) year
Orbital radius of the planet = R_p
From Kepler’s third law of planetary motion, we can write

Hence, the orbital radius of the planet will be 0.63 times smaller than that of the Earth.

Question 4.
I₀, one of the satellites of Jupiter, has an orbital period of 1.769 days and the radius of the orbit is 4.22 x 10⁸ m. Show that the mass of Jupiter is about one-thousandth that of the sun.
Solution:
Orbital period of I₀, TI₀ = 1.769 days = 1.769 x 24 x 60 x 60 s
Orbital radius of I_I0, R_I0 = 4.22 x 108 m
Satellite I₀ is revolving around the Jupiter Mass of the latter is given by the relation:
M_j = \(\frac{4 \pi^{2} R_{I_{O}}^{3}}{G T_{I_{O}}^{2}}\) ……………………….. (i)

where M_j = Mass of Jupiter
G = Universal gravitational constant Orbital period of the Earth,
T_e = 365.25 days = 365.25 x 24 x 60 x 60 s Orbital radius of the Earth,
Re =1 AU = 1.496 x 10¹¹ m
Mass of Sun is given as:
M_s = \(\frac{4 \pi^{2} R_{e}^{3}}{G T_{e}^{2}}\) ………………………. (ii)

= 1045.04
∴ \(\frac{M_{s}}{M_{j}} \sim 1000\)
\(M_{s} \sim 1000 \times M_{j}\)
Hence, it can be inferred that the mass of Jupiter is about one thousandth that of the Sun.

Question 5.
Let us assume that our only consists of 2.5 x 10¹¹ stars each of one solar mass. How long will a star at a distance of 50,000 ly from the galactic center take to complete one revolution? Take the diameter of the Milky Way to be 10⁵ ly.
Solution:
Number of stars in our galaxy, = 2.5 x 10¹¹
Mass of each star = one solar mass =2 x 10³⁰ kg
Mass of our galaxy,M =2.5 x 10¹¹ x 2 x 10³⁰ = 5 x 10⁴¹ kg
Diameter of Milky Way, d = 10⁵ ly
Radius of Milky Way,r= 5 x 10⁴ ly
∵ 1 ly=946 x 10¹⁵m
∴ r=5 x 10⁴ x 9.46 x 10¹⁵
=4.73 x 10²⁰m

Since a star revolves around the galactic centre of the Milky Way, its time period is given by the relation:
T = \(\left(\frac{4 \pi^{2} r^{3}}{G M}\right)^{1 / 2}\)

1 year = 365 x 324 x 60 x 60 s
∵ 1 s = \(\frac{1}{365 \times 24 \times 60 \times 60} \) years
∴ 1.12 x 10¹⁶ s = \(\frac{1.12 \times 10^{16}}{365 \times 24 \times 60 \times 60}\) = 3.55 x 10 ⁸ years

Question 6.
Choose the correct alternative:
(a) If the zero of potential energy Is at infinity, the total energy of an orbiting satellite is negative of its kinetic/potential energy.
(b) The energy required to launch an orbiting satellite out of Earth’s gravitational influence Is more/less than the energy required to project a stationary object at the same height (as the satellite) out of Earth’s influence.
Solution:
(a) Kinetic energy
Explanation: Total mechanical energy of a satellite is the sum of its kinetic energy (always positive) and potential energy (maybe negative). At infinity, the gravitational potential energy of the satellite is zero. As the Earth-satellite system is a bound system, the total energy of the satellite is negative.

Thus, the total energy of an orbiting satellite at infinity is equal to the negative of its kinetic energy.

(b) Less
Explanation: An orbiting satellite acquires certain amount of energy that enables it to revolve around the Earth. This energy is provided by its orbit. It requires relatively lesser energy to move out of the influence of the Earth’s gravitational field than a stationary object on the Earth’s surface that initially contains no energy.

Question 7.
Does the escape speed of a body from the Earth depend on
(a) the mass of the body,
(b) the location from where it is projected,
(c) the direction of projection,
(d) the height of the location from where the body is launched?
Answer:
(a) No, we know that the escape velocity of the body is given by
V_e = \(\sqrt{\frac{2 G M}{R}} \) , where M and R are mass and radius of Earth. Thus clearly, it does not depend on the grass of the body as V_e is independent of it.

(b) Yes, we know that V_e depends upon the gravitational potential at the point from where the body is launched. Since the gravitational potential depends on the latitude and height of the point, therefore the escape velocity depends on the location of the point from where it is projected. It can also be experienced as:
V_e = \(\sqrt{2 g r}\) . As g has different values at different heights. Therefore, V_e depends upon the height of location.

(c) No, it does not depend on the direction of projection as V_e is independent of the direction of projection.

(d) Yes, it depends on the height of location from where the body is launched as explained in (b).

Question 8.
A comet orbits the Sun in a highly elliptical orbit. Does the comet have a constant
(a) linear speed,
(b) angular speed,
(c) angular momentum,
(d) kinetic energy,
(e) potential energy,
(f) total energy throughout its orbit? Neglect any mass loss of the comet when it comes very close to the Sun.
Solution:
(a) No, according to law of conservation of linear momentum L = mvr constant, therefore the comet moves faster when it is close to the sun and moves slower when it is farther away from the sun. Therefore, the speed of the comet does not remain constant.

(b) No, as the linear speed varies, the angular speed also varies. Therefore, angular speed of the comet does not remain constant.

(c) Yes, as no external torque is acting on the comet, therefore, according to law of conservation of angular momentum, the angular momentum of the comet remain constant.

(d) No, kinetic energy of the comet = \(\frac{1}{2}\) mν² As the linear speed of the comet changes as its kinetic energy also changes. Therefore, its KE does not remains constant.

(e) No, potential energy of the comet changes as its kinetic energy changes.

(f) Yes, total energy of a comet remain constant throughout its orbit.

Question 9.
Which of the following symptoms is likely to afflict an astronaut in space
(a) swollen feet,
(b) swollen face,
(c) headache,
(d) orientational problem?
Answer:
(b), (c), and (d)
Explanations:
(a) Legs hold the entire mass of a body in standing position due to gravitational pull. In space, an astronaut feels weightlessness because of the absence of gravity. Therefore, swollen feet of an astronaut do not affect him/her in space.

(b) A swollen face is caused generally because of apparent weightlessness in space sense organs such as eyes, ears, nose, and mouth constitute a person’s face. This symptom can affect an astronaut in space.

(c) Headaches are caused because of mental strain. It can affect the working of an astronaut in space.

(d) Space has different orientations. Therefore, orientational problems can affect an astronaut in space.

Question 10.
In the following two exercises, choose the correct answer from among the given ones:
The gravitational intensity at the center of a hemispherical shell of uniform mass density has the direction indicated by the arrow (see figure) (i) a, (ii) b, (iii) c, (iv) 0.

Solution:
(iii) Gravitational potential (V) is constant at all points in a spherical shell.
Hence, the gravitational potential gradient \(\left(\frac{d V}{d r}\right)\) is zero everywhere inside the spherical shell. The gravitational potential gradient is equal to the negative of gravitational intensity. Hence, intensity is also zero at all points inside the spherical shell. This indicates that gravitational forces acting at a point in a spherical shell are symmetric.

If the upper half of a spherical shell is cut out (as shown in the given figure), then the net gravitational force acting on a particle located at center O will be in the downward direction.

Since gravitational intensity at a point is defined as the gravitational force per unit mass at that point, it will also act in the downward direction. Thus, the gravitational intensity at center O of the given hemispherical shell has the direction as indicated by arrow c.

Question 11.
For the above problem, the direction of the gravitational intensity at an arbitrary point P is indicated by the arrow
(i) d,
(ii) e,
(iii) f,
(iv) g.
Solution:
(ii) Gravitational potential (V) is constant at all points in a spherical shell.
Hence, the gravitational potential gradient \(\left(\frac{d V}{d r}\right)\) inside the spherical shell. The gravitational potential gradient is equal to the negative of gravitational intensity. Hence, intensity is also zero at all points inside the spherical shell. Ibis indicates that gravitational forces acting at a point in a spherical shell are symmetric. If the upper half of a spherical shell is cut out (as shown in the given figure), then the net gravitational force acting on a particle at an arbitrary point P will be in the downward direction.

Since gravitational intensity at a point is defined as the gravitational force per unit mass at that point, it will also act in the downward direction. Thus, the gravitational intensity at an arbitrary point P of the hemispherical shell has the direction as indicated by arrow e.

Question 12.
A rocket Is fired from the Earth towards the Sun. At what distance from the Earth’s centre is the gravitational force on the rocket zero? Mass of the Sun = 2 x 10³⁰kg, mass of the Earth = 6 x 10²⁴ kg. Neglect the effect of other planets etc. (orbital radius = 1.5 x 10¹¹ m)
Solution:
Mass of the Sun, M_s = 2 x 10³⁰ kg
Mass of the Earth, M_e = 6 x 10 ²⁴ kg
Orbital radius, r = 1.5 x 10¹¹ m
Mass of the rocket = m

Let x be the distance from the center of the Earth where the gravitational force acting on satellite P becomes zero.
From Newton’s law of gravitation, we can equate gravitational forces acting on satellite P under the influence of the Sun and the Earth as:

= 2.59 x 10⁸ m

Question 13.
How will you ‘weigh the Sun’, that is estimate its mass? The mean orbital radius of the Earth around the Sun is 1.5 x 10⁸ km.
Solution:
Orbital radius of the Earth around the Sun, r = 1.5 x 10⁸ km = 1.5 x 10¹¹ m
Time taken by the Earth to complete one revolution around the Sun,
T = 1 year = 365.25 days = 365.25 x 24 x 60 x 60 s
Universal gravitational constant, G = 6.67 x 10^-11N-m² kg^-2
Thus, mass of the Sun can be calculated using the relation,
M = \(\frac{4 \pi^{2} r^{3}}{G T^{2}}\)
= \(\frac{4 \times(3.14)^{2} \times\left(1.5 \times 10^{11}\right)^{3}}{6.67 \times 10^{-11} \times(365.25 \times 24 \times 60 \times 60)^{2}}\)
= \(\frac{133.24 \times 10^{33}}{6.64 \times 10^{4}}\) = 2.0 x 10³⁰kg
Hence, the mass of the Sun is 2.0 x 10³⁰ kg.

Question 14.
A Saáurn year is 29.5 times the Earth year. How far is the Saturn from the Sun if the Earth is 1.5 x 10⁸ km away from the Sun?
Solution:
Distance of the Earth from the Sun, r_e = 1.5x 10⁸ km= 1.5 x 10¹¹ m
Time period of the Earth = T_e
Time period of Saturn, T_s = 29.5 T_e
Distance of Saturn from the Sun = r_s
From Kepler’s third law of planetary motion, we have
T=\(\left(\frac{4 \pi^{2} r^{3}}{G M}\right)^{\frac{1}{2}}\)
For Saturn and Sun, we can write

= 1.43 x 10¹² m

Question 15.
A body weigths 63 N on the surface of the Earth. What is the gravitational force on it due to the Earth at a height equal to half the radius of the Earth?
Solution:
Weight of the body, W =63 N
Acceleration due to gravity at height h from the Earth’s surface is given by the relation:
g’ = \(\frac{g}{\left(1+\frac{h}{R_{e}}\right)^{2}}\)
where, g = Acceleration due to gravity on the Earth’s surface
R_e =Radius of the Earth
For h= \(\frac{R_{e}}{2}\)
g’ = \(\frac{g}{\left(1+\frac{R_{e}}{2 \times R_{e}}\right)^{2}}=\frac{g}{\left(1+\frac{1}{2}\right)^{2}}=\frac{4}{9} g\)

Weight of a body of mass m at height h is given as
W’ = mg’
= m x \(\frac{4}{9}\) g = \(\frac{4}{9}\) x mg
= \(\frac{4}{9}\) W
= \(\frac{4}{9}\) x 63 =28 N

Question 16.
Assuming the Earth to be a sphere of uniform mass density, how much would a body weigh halfway down to the centre of the Earth if it weighed 250 N on the surface?
Solution:
Weight of a body of mass m at the Earth’s surface, W = mg = 250 N
Body of mass m is located at depth, d = \(\frac{1}{2}\) R_e
where, R_e = Radius of the Earth
Acceleration due to gravity at depth d is given by the relation
g’ = \(\left(1-\frac{d}{R_{e}}\right) g=\left(1-\frac{R_{e}}{2 \times R_{e}}\right) g=\frac{1}{2} g\)
Weight of the body at depth d,
W’ = mg’
= m x \(\frac{1}{2} g=\frac{1}{2}\) mg = \(\frac{1}{2}\) W
= \(\frac{1}{2}\) x 250 = 125 N

Question 17.
A rocket is fired vertically with a speed of 5 km s^-1 from the Earth’s surface. How far from the Earth does the rocket ‘ go before returning to the earth? Mass of the Earth = 6.0 x 10²⁴ kg; mean radius of the Earth = 6.4 x 10⁶ m;
G = 6.67 x 10 ^-11 N-m² kg^-2.
Solution:
Velocity of the rocket, ν = 5 km/s = 5 x 10³ m/s
Mass of the Earth, M_e = 6.0 x 10²⁴ kg
Radius of the Earth, R_e = 6.4 x 10⁶ m
Height reached by rocket mass, m = h

At the surface of the Earth,
Total energy of the rocket = Kinetic energy + Potential energy
= \(\frac{1}{2} m v^{2}+\left(\frac{-G M_{e} m}{R_{e}}\right)\)
At highest point h,
ν = 0
and, Potential energy = – \(\frac{G M_{e} m}{R_{e}+h}\)
Total energy of the rocket = 0+ \(\left(-\frac{G M_{e} m}{R_{e}+h}\right)=-\frac{G M_{e} m}{R_{e}+h}\)
From the law of conservation of energy, we have
Total energy of the rocket at the Earth’s surface=Total energy at height h


Height achieved by the rocket with respect to the centre of the Earth = R_e +h
= 6.4 x 10⁶ +1.6 x 10⁶ = 8.0 x 10⁶ m

Question 18.
The escape speed of a projectile on the Earth’s surface is 11.2 km s^-1. A body is projected out with thrice this speed. What is the speed of the body far away from the Earth? Ignore the presence of the Sun and other planets.
Solution:
Escape velocity of a projectile from the Earth, υ_esc = 11.2 km/s
Projection velocity of the projectile, v_p = 3 v_esc
Mass of the projectile = m
Velocity of the projectile far away from the Earth = v_f
Total energy of the projectile on the Earth = \(\frac{1}{2} m v_{p}^{2}-\frac{1}{2} m v_{\mathrm{esc}}^{2}\)
The gravitational potential energy of the projectile far away from the Earth is zero.
Total energy of the projectile far away from the Earth = \(\frac{1}{2}\) mv²_f
From the law of conservation of energy, we have

Question 19.
A satellite orbits the Earth at a height of 400 km above the surface. How much energy must be expended to rocket the satellite out Of the Earth’s gravitational influence? Mass of the satellite = 200 kg mass of the Earth = 6.0 x 10²⁴ kg; radius of the Earth=6.4x 10⁶ m;G=6.67 x 10^-11 N-m² kg^-2.
Solution:
Mass of the Earth, M_e =6.0 x 10²⁴ kg
Mass of the satellite, m = 200 kg
Radius of the Earth, Re = 6.4 x 10⁶ m
Universal gravitational constant, G = 6.67 x 10^-11 N-m²kg²
Heightofthesatellite,h =400 km=4 x 10⁵ m=0.4 x 10⁶ m

Total energy of the satellite at height h = \(\frac{1}{2} m v^{2}+\left(\frac{-G M_{e} m}{R_{e}+h}\right)\)
Orbital velocity of the satellite, ν = \(\sqrt{\frac{G M_{e}}{R_{e}+h}}\)
Total energy of height, h = \(\frac{1}{2} m\left(\frac{G M_{e}}{R_{e}+h}\right)-\frac{G M_{e} m}{R_{e}+h}=-\frac{1}{2}\left(\frac{G M_{e} m}{R_{e}+h}\right) \)

The negative sign indicates that the satellite is bound to the Earth. This is called bound energy of the satellite.
Energy required to send the satellite out of its orbit = – (Bound energy)

Question 20.
Two stars each of one solar mass (= 2x 10³⁰ kg) are approaching each other for a head-on collision. When they are a distance 10⁹ km, their speeds are negligible. What is the speed with which they collide? The radius of each star is 10⁴ km. Assume the stars to remain undistorted until they collide. (Use the known value of G).
Solution:
Mass of each star, M = 2 x 10³⁰ kg
Radius of each star, R = 10⁴ km = 10⁷ m
Distance between the stars, r = 10⁹ km = 10¹² m
For negligible speeds, ν = 0 total energy of two stars separated at distance r
= \(\frac{-G M M}{r}+\frac{1}{2} m v^{2}\)
= \(\frac{-G M M}{r}+0\) = 0 ………………………… (i)

Now, consider the case when the stars are about to collide:
The velocity of the stars = ν
Distance between the centers of the stars = 2R
Total kinetic energy of both stars = \(\frac{1}{2} M v^{2}+\frac{1}{2} M v^{2}\) = Mv²
Total potential energy of both stars = \(\frac{-G M M}{2 R}\)
Total energy of the two stars = Mv² – \(\frac{-G M M}{2 R}\) ………………………. (ii)
Using the law of conservation of energy, we can write

Question 21.
Two heavy spheres each of mass 100 kg and radius 0.10 m are placed 1.0 m apart on a horizontal table. What is the gravitational force and potential at the midpoint of the line joining the centers of the spheres? Is an object placed at that point in equilibrium? If so, is the equilibrium stable or unstable?
Solution:
The situation is represented in the following figure :

Mass of each sphere, M = 100 kg
Separation between the spheres, r = 1 m
X is the mid-point between the spheres. Gravitational force at point X will be zero. This is because gravitational force exerted by each sphere will act in opposite directions. Gravitational potential at point X

Any object placed at point X will be in equilibrium state, but the equilibrium is unstable. This is because any change in the position of the object will change the effective force in that direction.

Additional Exercises

Question 22.
As you have learnt in the text, a geostationary satellite orbits the Earth at a height of nearly 36,000 km from the surface of the Earth. What is the potential due to Earth’s gravity at the site of this satellite? (Take the potential energy at infinity to be zero). Mass of the Earth = 6.0 x 10²⁴ kg, radius = 6400 km.
Solution:
Mass of the Earth, M = 6.0 x 10²⁴ kg
Radius of the Earth, R = 6400 km = 6.4 x 10⁶ m
Height of a geostationary satellite from the surface of the Earth,
h = 36000 km = 3.6 x 10⁷m
Gravitational potential energy due to Earth’s gravity at height h,

Question 23.
A star 2.5 times the mass of the Sun and collapsed to a size of 12 km rotates with a speed of 1.2 revs. per second. (Extremely compact stars of this kind are known as neutron stars. Certain stellar objects called pulsars belong to this category). Will an object placed on its equator remain stuck to its surface due to gravity? (Mass of the Sun = 2 x 10³⁰ kg).
Solution:
Yes, A body gets stuck to the surface of a star if the inward gravitational force is greater than the outward centrifugal force caused by the rotation of the star.
Gravitational force, fg = \(\frac{G M m}{R^{2}}\)
where,M =Mass of the star=2.5 x 2 x 10³⁰ = 5 x 10³⁰ kg
m = Mass of the body

R=Radiusofthestar=12 km=1.2 x 10⁴ m
∴ f_g = \(\frac{6.67 \times 10^{-11} \times 5 \times 10^{30} \times m}{\left(1.2 \times 10^{4}\right)^{2}}\)
= 2.31 x 10¹² mN
Centrifugal force, f_c = mrω²

where, ω = Angular speed = 2 πv
ν = Angular frequency = 1.2 rev s^-1
f _c=mR(2πv)²
= m x (1.2 x 10⁴) x 4 x(3.14)² x (1.2)² = 1.7 x 10⁵ mN
Since f_g > f_c, the body will remain stuck to the surface of the star.

Question 24.
A spaceship is stationed on Mars. How much energy must be expended on the spaceship to launch it out of the solar system? Mass of the spaceship = 1000 kg; mass of the Sun = 2 x 10³⁰ kg; mass of mars = 6.4 x 10²³ kg; radius of mars = 3395 km; radius of the orbit of mars = 2.28 x 10⁸ km;
G = 6.67 x 10^-11 N-m²kg^-2.
Solution:
Given, mass of the Sun, M = 2 x 10³⁰ kg

Mass of the mars, m = 6.4 x 10²³ kg
Mass of spaceship, Δm = 1000 kg
Radius of orbit of the mars, r₀ = 2.28 x 10¹¹
Radius of the mars, r = 3.395 x 10⁶ m
If v is the orbital velocity of mars, then

Since the velocity of the spaceship is the same as that of the mars, Kinetic energy of the spaceship,


Total potential energy of the spaceship,
U = Potential energy of the spaceship due to its being in the gravitational
field of the mars + potential energy of the spaceship due to its being in the gravitational field of the Sun.

Total energy of the spaceship E=K +U = 2.925 x 10¹¹ J – 5.977 x 10¹¹J
= – 3.025 x 10¹¹ J = -3.1 x 10¹¹ J
Negative energy denotes that the spaceship is bound to the solar system. Thus, the energy needed by the spaceship to escape from the solar system = 3.1 x 10¹¹ J.

Question 25.
A rocket is fired ‘vertically’ from file surface of mars with a speed of 2 kms^-1. If 20% of its initial energy is lost due to martian atmospheric resistance, how far will the rocket go from the surface of Mars before returning to it?
Mass of mars = 6.4x 10²³ kg; radius of mars = 3395 km;
G = 6.67 x 10^-11 N-m² kg².
Solution:
Initial velocity of the rocket, ν = 2 km/s = 2 x 10³ m/s
Mass of Mars, M = 6.4 x 10²³ kg .
Radius of Mars, R = 3395 km = 3.395 x 10⁶ m
Universal gravitational constant, G = 6.67 x 10^-11 N-m² kg^-2
Mass of the rocket = m
Initial kinetic energy of the rocket = \(\frac{1}{2} m v^{2}\)
Initial potential energy of the rocket = \(\frac{-G M m}{R}\)
Total initial energy = \(\frac{1}{2} m v^{2}-\frac{G M m}{R}\)

If 20 % of initial kinetic energy is lost due to martian atmospheric resistance, then only 80 % of its kinetic energy helps in reaching a height.
Total initial energy available = \(\frac{80}{100} \times \frac{1}{2} m v^{2}-\frac{G M m}{R}=0.4 m v^{2}-\frac{G M m}{R} \)
Maximum height reached by the rocket = h
At this height, the velocity and hence, the kinetic energy of the rocket will become zero.
Total energy of the rocket at height h = \(-\frac{G M m}{(R+h)}\)
Applying the law of conservation of energy for the rocket, we can write

= 495 x 10³ m = 495 km

Punjab State Board PSEB 11th Class Biology Book Solutions Chapter 1 The Living World Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Biology Chapter 1 The Living World

PSEB 11th Class Biology Guide The Living World Textbook Questions and Answers

Question 1.
Why are living organisms classified?
Answer:
Living organisms are classified because:

• there are millions of organisms on the earth, which need a proper system of classification for identification.
• a number of new organisms are discovered each year. They require a particular system to be identified and to find out their correct position in a group.

Question 2.
Why are the classification systems changing every now and then?
Answer:
Evolution is the major factor responsible for the change in classification systems. Since, evolution still continues, so many different species of plants and animals are added in the already existed biodiversity. These newly discovered plant and animal specimens are then identified, classified and named according to the already existing classification systems. Due to evolution, animal and plant species keep on changing, so necessary changes in the already existed classification systems are necessary to place every newly discovered plant and animal in their respective ranks.

Question 3.
What different criteria would you choose to classify people that you meet often?
Answer:
The different scientific criteria to classify people that we meet often would be :
(i) Nomenclature: It is the science of providing distinct and proper names to the organisms. It is the determination of correct name as per established universal practices and rules.

(ii) Classification: It is the arrangement of organisms into categories based on systematic planning. In classification various categories used are class, order, family, genus and species.

(iii) Identification: It is the determination of correct name and place of an organism. Identification is used to tell that a particular species is similar to other organism of known identity. This includes assigning an organism to a particular taxonomic group.
The same criteria can be applied to the people we meet daily. We can identify them will their names classify them according to their living areas, profession, etc.

Question 4.
What do we learn from identification of individuals and populations?
Answer:
Identification of individuals and population categorized it into a species. Each species has unique characteristic features. On the basis of these features, it can be distinguished from other closely related species, e. g.,

Question 5.
Given below is the scientific name of Mango. Identify the correctly written name.
(i) Mangifera Indica
(ii) Mangifera indica.
Answer:
(ii) Mangifera indica (the name of species can never begins with a capital letter).

Question 6.
Define a taxon. Give some examples of taxa at different hierarchical levels.
Answer:
Taxon is a grouping of organisms of any level in hierarchical classification, which is based on some common characteristics, e.g., insects represent a class of phylum – Arthropoda. All the insects possess common characters of three pairs of jointed legs. The term ‘taxon’ was introduced by ICBN in 1956. Examples of taxa are kingdom, phylum or division, class, order, family, genus and species. These taxa form taxonomic hierarchy, e.g., taxa for human :
Phylum – Chordata
Class – Mammalia
Order – Primata
Family – Hominidae
Genus – Homo
Species – sapiens

Question 7.
Can you identify the correct sequence of taxonomical categories?
(a) Species → Order → Phylum → Kingdom
(b) Genus → Species → Order → Kingdom
(c) Species → Genus → Order → Phylum
Answer:
The correct sequence of taxonomical categories is as follows :
Species → Genus → Order → Phylum

Question 8.
Try to collect all the currently accepted meanings for the word ‘species’. Discuss with your teacher the meanings of species in case of higher plants and animals on one hand, and bacteria on the other hand.
Answer:
A group of individual organisms with fundamental similarities is called species. It can be distinguished from other closely related species on the basis of distinct morphological differences.
In case of higher plants and animals, one genus may have one or more than one species, e.g.,Panthera leo (lion) and Panthera tigris (tiger).
In this example, Panthera is genus, which includes leo (lion) and tigris (tiger) as species.
In case of bacteria, different categories are present on the basis of shape. These are spherical, coccus, rod-shaped, comma and spiral-shaped. Thus, meaning of species in case of higher organisms and bacteria are different.

Question 9.
Define and understand the following terms:
(i) Phylum,
(ii) Class,
(iii) Family,
(iv) Order,
(v) Genus.
Answer:
(i) Phylum: Phylum comes next to Kingdom in the taxonomical hierarchy. All broad characteristics of an animal or plant are defined in a phylum. For example all chordates have a notochord and gill at some stage of life cycle. Similarly all arthropods have jointed legs made of chitin.

(ii) Class: The category class includes related orders. It is higher than order and lower than phylum. For example, class – Mammalia has order – Carnivora, Primata, etc.

(iii) Family: It is the category higher than genus and lower than order, which has one or more related genera having some common features. For example, Felidae, Canidae, etc.

(iv) Order: Order further zeroes down on characteristics and includes related genus. For example humans and monkeys belong to the order primates. Both humans and monkeys can use their hands to manipulate objects and can walk on their hind legs.

(v) Genus: It comprises a group of related species which has more characters in common in comparison to species of other genera. We can say that genera are aggregates of closely related species. For example, j potato, tomato and brinjal are three different species but all belong to the genus Solatium. Lion (Panthera leo), leopard (P. pardus) and tiger (P. tigris) with several common features, are all species of the genus Panthera. This genus differs from another genus Felis which includes cats.

Question 10.
How is a key helpful in the identification and classification of an
organism?
Answer:
Key is a device (scheme) of diagnostic alternate (contrasting) characters, which provide an easy means for the identification of unknown organism. The keys are taxonomic literature based on the contrasting characters generally a pair called couplet. Each statement in the key is called a lead. Separate taxonomic keys are required for each taxonomic category such as family, genus and species for identification purposes. Being analytical ‘ in nature, two types of keys are commonly used-indented key and bracketed key.

(i) Indented key provides sequence of choice between two or more statements of characters of species. The user has to make correct choise for identification.

(ii) Bracketed key (1) are used for contrasting characters like indented key but they are not repeated by intervening sub-dividing character and each character is given a number in brackets.

Question 11.
Illustrate the taxonomical hierarchy with suitable examples of a plant and an animal.
Answer:

Taxonomical hierarchy is the system of arrangement of taxonomic categories in a descending order depending upon their relative dimensions. It was introduced by Linnaeus (1751) and is therefore, also called Linnaeus hierarchy. Each category, referred to as a unit of classification commonly called as taxon (pi. taxa), e.g., taxonomic categories and hierarchy can be illustrated by a group of organisms, i.e., insects. The common features of insects is ‘three pair of jointed legs’. It means insects are recognisable objects which can be classified, so given a rank or category.
Category further denotes a rank. Each rank or taxon, represents a unit of classification taxonomic studies of all plants and animals led to the development of common categories such as kingdom, phylum or division (for plants), class, order, family, genus and species. All organisms, including those in the plant and animal, kingdoms have ‘species’ as the lowest category.

To place an organism in various categories is to have the knowledge of characters of an individual or group of organism. This helps to identify similarities and dissimilarities among the individual of the same kind of organisms as well as of other kinds of organism. Some organisms with their taxonomical categories are given in following table: