## Karnataka 1st PUC Physics Question Bank Chapter 9 Mechanical Properties of Solids

### 1st PUC Physics Mechanical Properties of Solids Textbook Questions and Answers

Question 1.
A steel wire of length 4.7 m and cross-sectional 3.0 × 10-5 m2 stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 × 10-5 m2 under a given load. What is the ratio of the Young’s modulus of steel to that of copper?
Let the Young’s Modulus of steel and copper be Ys and Yc respectively.
Length of steel wire, Ls = 4.7 m
Length of copper wire, Lc = 3.5 m
Area of cross-section of steel wire,
As= 3 × 10-5 m2
Area of cross-section of copper wire,
Ac = 4 × 10-5 m2
Since change in lengths are equal.
∴ ∆Ls = ∆Lc = ∆ L.
As the load is same, Fs = Fc = F
We have, Y = $$\frac{F}{A} \times \frac{L}{\Delta L}$$
$$\therefore \frac{Y_{s}}{Y_{c}}=\frac{L_{s} A_{c}}{A_{s} L_{c}}=\frac{4 \times 4.7 \times 10^{-5}}{3 \times 3.5 \times 10^{-5}}$$ = 1.79
The ratio of Young’s Modulus of steel to that of copper is ≃ 1.8 :1.

Question 2.
Figure shows the strain – stress curve for a given material. What are
1. Young’s modulus and
2. approximate yield strength for this material?

1. F from the given graph,
For a stress of 150 × 106 Nm-2
Strain = 0.002
∴ Young’s modulus (Y) = $$\frac{\text { stress }}{\text { strain }}$$
$$=\frac{150 \times 10^{6} \mathrm{Nm}^{-2}}{0.002}$$ = 7.5 × 1010 Nm-2

2. 300 × 106 Nm-2 is the maximum stress that the material can sustain without crossing the elastic limit. Thus, the yield strength of the material is 3 × 108 Nm-2 or 300 × 106 Nm-2.

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

The graphs are drawn to the same scale.

1. Which of the materials has the greater Young’s modulus?
2. Which of the two is the stronger material?

1. Material A has greater Young’s modulus as the slope of the graph of material A is greater than material B.
Young’s Modulus = Y = Slope = $$\frac{\text { stress }}{\text { strain }}$$
∴ Material A has greater Young’s Modulus.
2. From the graph it can be approximated that the stress at the fracture point for material A is higher than material B, thus material A is stronger.

Question 4.
Read the following two statements below carefully and state, with reasons, if it is true or false.
1. The Young’s modulus of rubber is greater than that of steel;
2. The stretching of a coil is determined by its shear modulus.
1. The Young’s modulus of a -material defined as,
Y = $$\frac{\text { stress }}{\text { strain }}$$
Since, rubber stretches more than steel ∆ L for rubber is greater than that of steel, therefore ( $$\frac{\Delta L}{L}$$ 0f rubber) > ($$\frac{\Delta L}{L}$$ of steel) Since, Y $$\alpha \frac{1}{\text { strain }}$$ and, strain = $$\frac{\Delta L}{L}$$
Since the strain of rubber is greater, its Youngs modulus is less. Hence false,

2. The Stretching of a coil depends on the shear modulus of the material because for the coil to be stretched, a shear force has to applied to change the shape of the coil. Hence the statement is true.

Question 5.
Two wires of diameter 0.25 cm. one made of steel and the other made of brass are loaded as shown in Fig.9.13. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m. Compute the elongations of steel and the brass wires.

given, d = 0.25 cm
r = $$\frac{\mathrm{d}}{2}$$ = 0.125 cm.
As = AB = π × (0.125 × 10-2)2 m2
= 4.91 × 10-6 m2
As = AB because the diameters are same.
Fs = (4 × g + 6 × g) N
= (4 × 9.8 + 6 × 9.8) N
= 10 × 9.8 N =98 N
FB = (6 × 9.8) N = 58.8 N
∆Ls = ?
∆LB = ?
Ls = 1.5 m
LB =1.0m
Ys = 2 × 1011 N m-2 and
YB = 0.91 × 1011 Nm-2
∆Ls = $$\frac{F_{s} \cdot L_{s}}{A_{s} \cdot Y_{s}}=\frac{98 \times 1.5}{4.91 \times 10^{-6} \times 2 \times 10^{11}} \mathrm{m}$$
= 1.49 × 10-4m = 0.149 mm
To find the elongation of Brass,
∆LB = $$\frac{F_{B} L_{B}}{A_{B} Y_{B}}=\frac{(58.8)(1)}{\left(4.91 \times 10^{-6}\right)\left(0.91 \times 10^{11}\right)} \mathrm{m}$$
= 1.32 × 10-4m = 0.132 mm
The elongation of steel wire is 0.149 mm.
The elongation of Brass wire is 0.132 mm.

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 then attached to the opposite face of the cube. The shear modulus of aluminium is 25 GPa. What is the vertical deflection of this face?
Edge length, L = 10 cm = 0.1 m
Weight attached = 100 kg
F = (100 × 9.8) N = 980 N
A = L2 = 0.1 × 0.1 m2 = 0.01 m2

= 3.92 × 10-7 m
∴ the deflection is 3.92 × 10-7 m

Question 7.
Four identical hollow cylindrical columns of mild steel support a big structure of mass of 50,000 kg. The inner and outer radii of each column are 30 and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column.
Mass of big structure = 50,000 kg
Net force = (5 × 104 × 9.8) N
= 4.9 × 105 N = FB
Since it is supported by four cylinders the weight is equally distributed.
Force (F) on each cylinder is
F = $$\frac{F_{B}}{4}$$ = 122500 N
Young’s Modulus = Y = = $$\frac{\text { stress }}{\text { strain }}=\frac{F \times L}{A \times \Delta L}$$
∴ Compressional strain of each column is given by, $$\frac{\Delta L}{L}=\frac{F}{A \times Y}$$
Here, F = 122500 N
A = π(0.6)2 – π(0.3)2 = π(0.36 – 0.09)
= 0.27 π = 8.48 × 10-1 m2
Y = 2 × 1011 N m-2 (Standard Values)
∴ Strain = $$\frac{122500}{(0.27 \pi)\left(2 \times 10^{11}\right)}$$ = 7.22 × 10-7
Thus the compressional strain per cylinder is 7.22 × 10-7

Question 8.
A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain?
F = 44,500 N
Strain =?
Length = 19.1 mm, Breath = 15.2 mm
Area = A = 19.1 × 15.2 × 10-6m2
= 2.98 × 1 O-4 m2
G = 42 × 109 Nm-2
We have, G = $$\frac{F}{A \times \text { strain }}$$
∴ Strain = $$\frac{F}{G \times A}$$
$$=\frac{44,500}{2.98 \times 10^{-4} \times 42 \times 10^{9}}$$
= 3.65 × 10-3

Question 9.
A steel cable with a radius of 1.5 cm supports a chair lift at a ski area. If the maximum stress does not exceed 108 Nm-2, What is the maximum load the cable can support?
Maximum load = maximum stress x area of cross-section
= 108 n r2 =108 X (22/7) x (1.5 x 10-2)2 = 7.07 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 end are of copper and the middle one is of Iron. Determine the ratios of their diameters if each is to have the same tension.
Since the rigid bar is supported symmetrically by three wires each of length 2 m, the extension is same in all the three wires. Thus strain is same.
$$Y=\frac{F}{\pi r^{2} \times \text { strain }}$$ = $$\frac{4 \mathrm{F}}{\pi \mathrm{d}^{2} \times \text { strain }}$$
∴ Y $$\alpha \frac{1}{\mathrm{d}^{2}}$$
Yi = Young’s modulus of Iron
di = diameter of iron wire.
Yc = Young’s modulus of copper
dc = diameter of copper wire.
Yi = 190 × 109 N m-2
Yc = 110 × 109 N m-2
$$\frac{d_{c}}{d_{i}}=\sqrt{\frac{Y_{i}}{Y_{c}}}=\sqrt{\frac{19}{11}}=1.31: 1$$

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.
L = 1 m
W = 2 revolutions/s
A = 0.065 cm2 = 6.5 x 10-6 m2
F = mg + mlw2
= 14.5 × 9.8 +14.5 × 1 × (2)2
= 200.1 N
∆L = $$\frac{F l}{A Y}$$                [∵ Y = $$\frac{\mathrm{FI}}{\mathrm{A} \Delta \mathrm{I}}$$]
Young’s modulus of steel is 2 × 1011 Nm-2
∆L = $$\frac{200.1 \times 1}{6.5 \times 10^{-6} \times 2 \times 10^{11}} \mathrm{m}$$ = 1.539 × 10-4m

Question 12.
Compute the bulk modulus of water from the following data: Initial Volume = 100.0 litre. Pressure increase = 100.0 atm (1 atm = 1.013 × 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.
VF = final volume = 100.5 I
= 100.5 × 10-3 m3
Vi = initial volume = 100 × 10-3 m3
∆V = VF – Vi =0.5 × 10-3 m3
= 5 × 10-4 m3
∆ P = 100 atm = 100 × 1.013 × 105 Pa

= 2.03 × 109 Pa
Bulk modulus of air = 1 × 105 Pa
$$\frac{\text { Bulk modulus of water }}{\text { Bulk modulus of air }}=\frac{2.026 \times 10^{9}}{1 \times 10^{5}}$$
= 2.026 × 104
Since air is highly compressible compared to water, the ratio is so high.

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 1.03 × 103 kg m-3?
Density of water at surface
= $$\rho_{1}$$ = 1.03 × 103 kgm-3
density of water at depth h = $$\rho_{2}$$
Pressure at depth h = 80 atm
= 80 × 1.01 × 105 Pa.
V1 = Volume of water of mass m at surface.
V2 = Volume of water at depth h.

$$\frac{1}{\mathrm{B}}$$ Compressibility of water
= 45.8 × 10-11
∴ $$\frac{\Delta V}{V_{1}}$$ =80 × 1.013 × 105 × 45.8 × 10-11
= 3.71 × 10-3
Since,

= 1.034 × 103 kg m-3

Question 14.
Compute the fractional change in volume of a glass slab, when subjected to hydraulic pressure of 10 atm.
Hydraulic pressure exerted = 10 atm
= 10 × 1.013 × 105 Pa
B = $$\frac{P \cdot V}{\Delta V}$$
$$\frac{\Delta V}{V}$$ is the 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 × 10-5
∴ The fractional change in Volume is 2.73 × 10-5.

Question 15.
Determine the volume contraction of a solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of 7.0 × 106 Pa.
P = 7 × 106 Pa
I = 10 cm
V = I3= 10-3 m3
B = 140 × 109
B = $$\frac{P \times V}{\Delta V}$$
∆V = $$\frac{P}{B} \times V=\frac{7 \times 10^{6}}{140 \times 10^{9}} \times 10^{-3} \mathrm{m}^{3}$$
= 5 × 10-8 m3
= 0.05 cm3
∴ The contraction is 0.05 cm3

Question 16.
How much should the pressure on a litre of water be changed to compress it by 0.10%
P =?
V = 1L ,
B = 2.2 × 109Nm-2
$$\frac{\Delta v}{V}$$ × 100 = 0.1
⇒ $$\frac{\Delta v}{V}$$ = $$\frac{0.1}{100 \times 1}$$ = 10-3
Since, B = $$\frac{\mathrm{PV}}{\Delta \mathrm{V}}$$
P = B $$\frac{\Delta v}{V}$$
= 2.2 × 109 Pa × 10-3
= 2.2 × 106 Pa.
∴ The pressure is required is 2.2 × 106 Pa.

1st PUC Physics Mechanical Properties of Solids Additional Exercises Questions and Answers

Question 1.
Anvils made of single crystals of diamond, with the shape as shown in Fig. 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?

Diameter at narrow end
= d =0.5 mm
Compressional Force = F = 50,000 N

∴ The pressure is 2.546 × 1011 Pa

Question 2.
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 Fig.9.15. The cross-sectional areas of wires A and B are 1.0 mm2 and 2.0 mm2 respectively. At what point along the rod should a mass m be suspended in order to produce.

1. equal stresses.
2. equal in both steel and aluminium wires.

Let the initial lengths of the 2 wires be ‘L’
Given that the cross-sectional area of A
= Aa = 1 mm2 = 10-6
Cross-sectional area of
B = Ab = 2 mm2 = 2 × 10-6m2
Length of rod = 1.05 m.
Young’s modulus of wire A
= Ysteel = 2 × 1011 N/m2
Young’s modulus of wire B
YAl = 7 × 1010 N/m2
Let the mass ‘m’ be suspended at a distance ‘x’ from the end with wire A attached.

1. For equal stress on the two wires :
i.e., stress A = stress B

or Fb = 2 Fa …..(1)
where, Fa is force exerted on steel wire and Fσ is force exerted on aluminium wire. Since the total torque at the point when mass is suspended is zero, we can write
Fa × (x) = Fb × (1.05 – x)

⇒ 3x = 2 × 1.05 or x = 0.7 m
So, for equal stress on the wires the mass should be suspended at a distance 0.7m, from wire A.

2. For equal strain:
We know that, Strain = $$\frac{\text { Stress }}{Y}=\frac{F}{\text { AY }}$$
we require (strain)a = (strain)b

$$\frac{F_{a}}{F_{b}}$$ = $$\frac{10}{7}$$
……………(3) Since the torque is ‘0’ at the point of suspension equation (2) is valid.
⇒ $$\frac{F_{a}}{F_{b}}$$ = $$\frac{10}{7}$$ = $$\frac{1.05-x}{x}$$
⇒ 17 x = 7 × (1.05)
x = 0.432 m
So, for equal strain on the wires the mass should be suspended at a distance 0.432 m from wire A.

Question 3.
A mild steel wire of length 1.0 m and cross-sectional area 0.50 × 10-2cm2 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.
Given that length of the wire = 1.0 m
Area of cross-section = 0.50 × 10-2cm2
= 5.0 × 10-7m2
Mass suspended =100g
= 10-1 Kg
Ysteel = 2 × 1011 Pa
The mass causes the steel wire to bend as shown.
So new length of wire = AX + XB

Change in length is, ∆l = (AX + XB) – AB
Now, AX = $$\sqrt{\mathrm{AM}^{2}+\mathrm{MX}^{2}}$$ = XB
[ Pythogorus Theorem]

[Expanding and neglecting higher-order terms]
The tension in the steel wire is,
T cos θ + T cos θ = mg

$$\sqrt[3]{\frac{0.5 \times 0.1 \times 9.8}{4 \times 2 \times 10^{11} \times 5 \times 10^{-7}}}$$ = 0.0106 m
So, the depression at the centre is 0.0106 m or 1.06 cm.

Question 4.
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 × 107 Pa? Assume that each rivet is to carry one-quarter of the load.
Given diameter of metal strips
= 6.0 × 10-3 m
⇒ Area = $$\pi\left(\frac{\mathrm{d}}{2}\right)^{2}=\frac{\pi \times 36 \times 10^{-6} \mathrm{m}^{2}}{4}$$
= 2.827 × 10-5m2
For the maximum stress of 6.9 × 107 Pa
the maximum force is given by
Max Force = $$\frac{\text { Max Force }}{\text { Area }}$$
⇒ Max force = Max Stress × Area
= 6.9 × 107 × 2.827 × 10-5
= 1950.63 N
Since there are 4 rivets and each rivet carries 1/4th of the load,
Max tension that can be exerted by the rivet strip is 4 × 1.95063 KN
= 7.80252 KN.

Question 5.
The Marina trench is located in the Pacific Ocean, and at one place it is nearly eleven km beneath the surface of water. The water pressure at the bottom of the trench Is about 1.1 × 108 Pa. A steel ball of Initial volume 0.32 m3 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?
Pressure at bottom ’of the trench = ρgh = 1.1 × 108 Pa
Initial volume of the ball = 0.32 m3
Bulk modulus of steel = 1.6 × 1011 Pa
Let ∆v the change in volume of Ball when it falls down to the bottom.
We know that, B = $$\frac{P}{(\Delta V / V)}$$
⇒ ∆v = $$\frac{P V}{B}$$
$$=\frac{1.1 \times 10^{8} \times 0.32}{1.6 \times 10^{11}}$$ = 2.2 × 10-4 m3
Therefore, change in volume of the ball on reaching the bottom of the trench is 2.2 × 10-4 m3 or 220 cm3

### 1st PUC Physics Mechanical Properties of Solids One Mark Questions and Answers

Question 1.
What is elasticity?
The property of the body to restore the original shape or size on removal of the external force is called elasticity.

Question 2.
Define stress.
Stress is defined as the deforming force acting on a unit area of the body.

Question 3.
What is the S.l unit of stress?
Nm-2

Question 4.
Define strain.
Strain is the ratio of change in dimension of a body due to stress to the original dimension.

Question 5.
Define Shear Strain.
The angular tilt between different layers of a body due to tangential force (Shear Force) is called Shear Strain.

Question 6.
Define Modulus of elasticity.
Modulus of elasticity is defined as the ratio of stress to strain.

Question 7.
State Hooke’s law.
Within elastic limit, stress is directly proportional to strain.

Question 8.
Mention the expression for Young’s modulus in the case of a stretched string.
Young’s modulus in case of a stretched string is Y = $$\frac{\mathrm{FL}}{\pi r^{2}(\Delta \mathrm{L})}$$
F – Force acting on the wire,
L – Initial length of the wire
∆L – change in length of the wire,
r – radius of the wire.

Question 9.
Which is more elastic, rubber or iron?
Iron.

Question 10.
What are anisotropic materials?
The materials which have different physical properties along different directions are called anisotropic materials.

Question 11.
Name 2 main groups of solids.
Amorphous and Crystalline.

Question 12.
What are glass solids?
Glassy solids are amorphous solids.

Question 13.
What is the Bulk modulus of an incompressible object?
Since its incompressible strain is 0 and Bulk modulus is infinity.

Question 14.
Show graphically stress v/s strain curve for a Brittle material.

Question 15.
If the length of the wire Is made 2 times, what Is the effect on the Increase in length under same load?
∆l ∝ l. So, increase in length is also doubled.

### 1st PUC Physics Mechanical Properties of Solids Two Marks Questions and Answers

Question 1.
If a wire is replaced by another similar wire but of half the diameter. What is the effect on

1. Increase in the length for a given load,
2. Max load It can sustain.

We know that, $$\frac{\Delta L}{L}=\frac{F}{A Y}$$
1.

The increase in the length (∆L)is 4 times the original wire.

2.  We know that, F = F=A $$\left(\frac{\mathrm{Y} \Delta \mathrm{L}}{\mathrm{L}}\right)$$ Thus F ∝ A and hence maximum load it can sustain becomes 1/4th of the original wire.

Question 2.
What is a perfectly elastic and inelastic body?
If a body completely regains its natural shape or size after the removal of external forces, it is called a perfectly elastic body.
Eg: rubber.
If the body does not regain its original size or shape it is called inelastic body or plastic body.

Question 3.
Identical springs made of steel and aluminium are equally stretched. On which more work will have to be done? Why?
Work done for steel spring is more because steel is more elastic than aluminium, (alternatively: Spring constant of steel is more than that of aluminium).

Question 4.
Why does a cycle tube burst in summer?
In summer as the temperature increases, the pressure increases inside the tube This pressure would be enough to cross the breaking point of the tyre making it to burst.

Question 5.
When the pressure on a sphere increases by 50 atm then its volume decreases by 0.02%. Find Bulk modulus.
P = 50 atm
P = 50 atm ≅ 50 × 1.01 × 105
$$\frac{\Delta V}{V}=\frac{0.02}{100}=2 \times 10^{-4}$$
Bulk modulus= $$\frac{\mathrm{P}}{(\Delta \mathrm{V} / \mathrm{V})}=\frac{5.05 \times 10^{6}}{2 \times 10^{-4}}$$
= 2.525 × 1010N/m2

Question 6.
Which is more elastic, rubber or Steel? Why?
For a given force, rubber extends more than steel, i.e., Steel is more elastic than rubber. This is because for the same amount of extension. Steel requires more force. For the same force ∆l ∝ Y, and F ∝ Y, for same extension Ysteel > Yrubber.

Question 7.
What is elastic Limit?
Elastic limit is the value of applied force beyond which the material (body) does not come back to its original shape when the applied force is reduced.

Question 8.
Plot load v/s extension curve for a substance on the graph show,

1. Yield point
2. Breaking point
3. Elastic limit
4. Crushing point.

Question 9.
The Young ‘s modulus of a wire of length L and radius ‘r’ is Y. If length is reduced to L/3 and Radius r/4. What will be Its ratio of extension?
Young’s modulus is given by
Y = $$\frac{F}{A} \frac{l}{\Delta}$$ ⇒ ∆l = $$\frac{F}{A} \frac{l}{Y}$$
So, ∆l ∝ l and ∆l ∝ $$\frac{1}{A}$$ and A ∝ r2

Question 10.
The Young’s modulus of a material is Y. It has an area A and length ‘L’. Now if a is made $$\frac{A}{16}$$ and length L/2. What is the new Young’s modulus?
Even though the value of Young’s modulus is given by, Y = $$\frac{F}{A} \cdot \frac{l}{\Delta l}$$. It is only material dependent. It is constant for the given wire irrespective of its dimension. So, the new value of Young’s modulus is Y.

Question 11.
The length of a metal is L1 when a force is F1 is applied on it and its length is L2 when applied force is F2. Find the original length of wire.
We know that Y = $$\frac{\mathrm{F} l}{\mathrm{A} \Delta l}$$
Let original length be L. Then,
L1 = L + ∆l1
L2 = L + ∆l2
Since, Y is a constant,

Question 12.
Explain the terms Young’s modulus and elastic fatigue.
Young’s modulus is defined as the ratio of longitudinal stress to longitudinal strain.
Y = $$\frac{(F / A)}{(\Delta l / l)}$$
Elastic Fatigue is the time (delay) required for a material to regain its original position on removing the deforming force.

### 1st PUC Physics Mechanical Properties of Solids Three Marks Questions and Answers

Question 1.
When a wire is stretched by a certain force its elongation is ‘a’. If the second wire having double-radius and 4-time length. Find its elongation.

= 4 × 4 a
∆l2 = 16 a.

Question 2.
Define stress and strain and derive their units. Write one limitation of Hooke’s Law.
Stress is the deforming force experience by the body per unit area.
Stress = $$\frac{F}{A}$$
Units = $$\frac{(N)}{\left(m^{2}\right)}$$ = Nm-2
Strain is the change in the considered quantity (l,v, A, θ…..) per unit of its original value.
Strain = $$\frac{\Delta x}{x}$$
It is unitless.
Hooke’s Law is valid only for small deformation i.e., in its limit.

### 1st PUC Physics Mechanical Properties of Solids Five Marks Questions and Answers

Question 1.
The Stress-Strain graph of a metal wire is shown in figure up to pint E. The wire returns to its original state 0 along EPO when it is gradually unloaded point B corresponds to the fracture of wire.

1. Up to what point of the curve Hooke’s Law is obeyed?
2. Which point on the curve corresponds to yield point or elastic limit.
3. Indicate the Elastic region and plastic region.
4. Describe what happens when the wire Is loaded up to the stress corresponding to point A and then unloaded gradually. In particular to the dotted line.
5. What is peculiar about portion of curve from C to B? Upto what stress can be applied without causing fracture?

1. Since the graph is linear till point E. Hooke’s Law is obeyed till point E.
2. The point E corresponds to the elastic limit because till there the body returns to its original state along EPO.
3. Elastic – O to E; Plastic – E to B
4. Up to point E an almost linear increase in strain with stress. After point E, if it is loaded till A, the strain increases faster for some increase in stress. But on unloading the body returns along line AO to a permanent strain (change in shape) of O’.
5. In the curve C to B, the body undergoes a strain even if it is being unloaded and then fractures. To present fracture the value of stress has to be just less than that of C.

Question 2.
Explain different types of strain.
There are 3 types of strain.
1. Longitudinal strain :
It is defined as the ratio of change in length produced (dl) to the original length (L) of a thin rod.
Longitudinal strain = $$\frac{d l}{L}$$

2. Shearing strain :
It is defined as the ratio of displacement of a surface on which stress is acting to the height of the surface.

Shearing strain θ = $$\frac{x}{h}$$
$$=\frac{\text { displacement }}{\text { heightof the surface }}$$

3. Bulk strain :
It is defined as the ratio of change in volume (dV) to the original volume (V).
Bulk strain = $$\frac{\mathrm{d} \mathrm{v}}{\mathrm{v}}$$.

Question 3.
Explain the different types of elastic constant.
1. Young’s modulus (y):
It is defined as the ratio of longitudinal stress to the longitudinal strain.

where F – Longitudinal force,
L – original length,
dl – change in length,
A – area of cross-section of the wire.

2. Rigidity modulus (η):
It is defined as the ratio of shearing stress to the shearing strain.

where F – tangential force,
A – area of the surface,
h – height of the surface,
x – displacement of the surface

3. Bulk modulus (K) :
It is defined as the ratio of bulk stress to the bulk strain. Bulk modulus is given by,

where F/A = P – pressure, V is the original volume and dV is the change in volume.

1st PUC Physics Mechanical Properties of Solids Numerical Problems Questions and Answers

Question 1.
What is the percentage increase in length of a wire of diameter 12.5 mm stretched by a weight of 1000 kg ωt ?
[Given Y = 12.5 × 1011 dyne/sq.cm]
Area = $$\frac{\pi d^{2}}{4}=\frac{\pi(1.25)^{2}}{4} \mathrm{cm}^{2}$$
F = 1000 kg ωt
= 1000 × 103 × 980
F = 9.8 × 108 dyne
We know that,

$$=\frac{9.8 \times 4 \times 10^{-3}}{\pi \times(12.5)(1.25)^{2}}$$
= 0.638 × 10-3
Percentage increase = 0.064%.

Question 2.
A cube of side 3 cm Is subjected to shearing force. The top of cube is sheared through 0.012 cm with respect to bottom face. If cube is made of a material of shear modulus 2.1 × 1010Nm-2 then find

1. Shearing strain
2. Shearing stress
3. Force applied.

1. Shearing strain
$$=\frac{\Delta l}{l}=\frac{0.012 \mathrm{cm}}{3 \mathrm{cm}}$$
= 0.004

2. Shearing stress
= Shearing strain × Shear modulus
Shearing stress = 0.004 × 2.1 × 1010Nm-1
= 8.4 × 107Nm-2

3. Shearing stress = $$\frac{\text { S.Force }}{\text { Area }}$$
Shearing force = Shearing stress × Area
= 8.4 × 107 × (3 × 10-1)2
= 7.56 × 104 N
= 75.6 KN.

Question 3.
A composite wire of materials A and B are of lengths 3.2 m and 2.4 m, and areas 1cm2 and 1cm2 respectively. If the extension observed is 1.2 mm. Find the force ‘F’. Given YA = 1.6 × 1011 Nm-2 and YB = 2.4 × 101 Nm-2

We know that Y = $$\frac{F}{A} \times \frac{l}{\Delta L}$$
Let the individual extension be ∆lA and ∆lB
∆LA = $$F\left(\frac{L_{A}}{A_{A} Y_{A}}\right)$$; ∆LB = $$F\left(\frac{L_{B}}{A_{B} Y_{B}}\right)$$
Given ∆LA + ∆LB = 1.2 mn

= 1.2 × 10-4
F(3 × 10-7) = 1.2 × 10-4
F = 400 N

Question 4.
A cube of 5 cm has its upper face displaced by 0.2 cm by a force of 8 N. Find Shear Modulus, Stress and Strain.
Shear stress = $$\frac{F}{A}=\frac{8 N}{25 \times 10^{-4} m^{2}}$$
= 3200 Nm-2
Shear Strain = $$\frac{\Delta l}{l}$$
But ∆l = 0.2 cm
l = 5 cm
Shear strain = $$\frac{0.2 \mathrm{cm}}{5 \mathrm{cm}}$$
= 0.04
Shear Modulus = $$\frac{\text { Shear Stress }}{\text { Shear Strain }}$$
$$=\frac{3200}{0.04}$$
= 80 × 103Nm-2 or 80 kPa.

Question 5.
A rubber string 10 m long is suspended from a rigid support at one end. Calculate extension of the string due to its own weight. The density of rubber is 1.5 × 103 kg/m3 and Yrubber = 5 × 106Nm-2.
The weight at any point x from below is, $$\frac{\mathrm{M}}{10}$$ (10 – x) where M is mass of string.
So, Force at any point dx is
F = $$\frac{\mathrm{Mg}}{l}$$ (l – x)dx
We know that,

Net extension =

So, Net extension of the rubber due to its own weight = $$\frac{1.5 \times 10^{3} \times 10 \times 100}{2 \times 5 \times 10^{6}}$$
= 0.15 m.

Question 6.
A metal rod is tied vertically. If the breaking stress is 5.4 × 109 Nm-2 and density 5.4 × 103 kg/m3. Find the maximum length of rod that can be held without breaking.
Stress = $$\frac{F}{A}$$
= $$\frac{\mathrm{Mg}}{\mathrm{A}}$$ [∵ $$\frac{M}{A L}$$ = ρ]
= ρgL
If ρgL > Maximum stress that it can sustain than the rod break.
So, ρgL < 5.4 × 109 Nm-2
l< $$\frac{5.4 \times 10^{9}}{5.4 \times 10^{3} \times\left(10 m s^{-1}\right)}$$
l < 105 m
Maximum length of wire is 105 m or 100 km..

Question 7.
Consider a tapered wire as shown. Find the extension if Y is its Young’s modulus.

Consider an element dx which is at a distance x from the end with r, as radius. Tfhe change in length in this portion is
dl = $$\left(\frac{F}{A}\right) \frac{d x}{Y}$$
From figure we can write radius at a distance x is, rx = r1 + xtanθ
⇒ A = π(rx)2
= π (r1 + xtanθ)2

Total extension = $$\frac{F\left(r_{2}-r_{1}\right)}{Y \cdot r_{1} \cdot r_{2} \tan \theta}$$
$$=\frac{\mathrm{FL} \tan \theta}{\mathrm{Y} \mathrm{r}_{1}-\mathrm{r}_{2} \tan \theta}$$
Total extension = $$\frac{F \cdot L}{Y r_{1} r_{2}}$$
[∵ tanθ = $$\frac{r_{2}-r_{1}}{L}$$]

## Karnataka 1st PUC Physics Question Bank Chapter 6 Work, Energy and Power

### 1st PUC Physics Work, Energy and Power TextBook Questions and Answers

Question 1.
State carefully
if the following quantities are positive or negative:

1. work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket.
2. work done by gravitational force ‘in the above case,
3. work done by friction on a body sliding down an Inclined plane,
4. work done by an applied force on a body moving on a rough horizontal plane with uniform velocity,
5. work done by the resistive force of air on a vibrating pendulum in bringing it to rest.

Solution:
W = FS cos θ, where θ is the angle between $$\vec { F }$$ and $$\vec { S }$$

1. Force on the bucket is in the direction of the displacement of the bucket i.e., θ = 0. Hence, work done is positive.
2. Gravitation force acts downwards and displacement of the bucket is in an upward direction i.e.,θ = 180; work done is negative.
3. Force of friction and displacement are in opposite direction i.e., θ= 180; work done is negative.
4. F and S are in the same direction e., θ = 0°; work done is positive.
5. The resistive force of air opposes the motion, hence work done is negative.

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 a coefficient of kinetic friction = 0.1. Compute the

1. work done by the applied force in 10 s
2. work done by friction in 10 s
3. work done by the net force on the body in 10 s
4. change in kinetic energy of the body in 10 s, and interpret your results.

Solution:
1. We know that: μk$$\frac{\text { Frictional force }}{\text { Normal force }}$$
Here, μk = 0.1 and normal force = mg
Frictional force = 0.1 × 2 × 9.8 N = 1.96 N
∴ Net force = (7 – 1.96) N = 5.04 N
acceleration = $$\frac{\text { Net force }}{\text { Mass }}$$ = $$\frac{5.04}{2}$$ m/s² = 2.52 m/s
To find the distance (d) moved in 10 s
d = $$\frac{1}{2}$$ × 2.52 × 10 × 10 = 126 m
∴ Work done by applied force in 10 s is work done = 7 × 126 J = 882 J.

2. Work done by the friction in 10 ‘s’ we know that frictional force = 1.96 N
∴ work done by frictional force = 1.96 × 126 = 246.96 J.
Since force is opposite to the direction of displacement (0 = 180°), work done is = 246.96 J.

3. Work done by the Net force in 10 ‘s’
we know that Net Force = 5.04 N.
∴ work done by the Net force = 5.04 × 126 = 635.04 J.

4. Change in the body’s kinetic energy is the net work done on the body = work done by the Net force in 10’s’.
∴ Change in kinetic energy of the body = 635.04 J

Question 3.

Given in Fig. 6 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 must have in each case. Think of simple physical contexts for which these potential energy shapes are relevant.
Solution:
(a) Total energy, E = P.E + K.E = constant. In the region x > a, the P.E. has a value more than E. Thus, K.E = E- P.E. i.e., K.E. has a negative value, so the particle can not exist in the portion x > a. (The particle exists in region in which its K.E. is positive).

(b) Here P.E. > E, and as such the kinetic energy of the particle would be negative. Thus, particle can not be present in any region of the graph.

(c) Here, in x = 0 to x = a and x > b, the P.E. is more than E, so K.E. is negative. The particle can not exist in these regions.

(d) The object can not exist in the region between
x =- $$\frac { b }{ 2 }$$ to x=- $$\frac { a }{ 2 }$$ and x=- $$\frac { a }{ 2 }$$ to x =- $$\frac { b }{ 2 }$$
Because in this region P.E > E. The minimum total energy = -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 Show that a particle of total energy 1 J moving under this potential must turn back when it reaches x = ± 2m

Solution:
Here force constant k = 0.5 Nm-1 and the total energy of particle E = 1 J. The particle can go up to a maximum distance of xm, at which its total energy transformed into elastic potential energy.
We know that , $$\frac{1}{2}$$ Kxm² = E

Question 5.
(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 a 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 mass of the casing of the rocket in flight is burnt by obtaining heat energy from the mechanical energy (K.E. + P.E.) of the rocket itself. It is obtained at the expense of its K.E and P.E.

(b) Work done by the conservative force in a closed path is zero. Since gravitational force is a conservative force so work done in the case of a given comet is zero.

(c) As satellite comes closer to earth, its P.E decreases. Since total energy (E = K.E. + P.E) is conserved, so its K.E increases. Hence, speed of satellite increases. But, total energy of satellite goes on decreasing due to the loss of energy against air friction.

(d) Using W = F S cos θ, where θ is the angle between F and S

θ = 90°; ∴ W = 0

Here θ=0°, when mass is lifted upward with some upward force given to the mass by man through the pulley. In (ii) case, work done is more.

Question 6.
Point out the correct alternative:

1. When a conservative force does positive work on a body, the potential energy of the body increases/decreases/remains the same.
2. Work done by a body against friction always results in a loss of its kinetic/potential energy.
3. 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.
4. 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:

1. Potential energy decreases
2. Kinetic energy
3. External force
4. Total linear momentum/total energy of two bodies.

Question 7.
State if each of the following statements is true or false. Give reasons for your answer.

1. In an elastic collision of two bodies, the momentum and energy of each body is conserved.
2. The total energy of a system is always conserved, no matter what internal and external forces on the body are present.
3. Work done in the motion of a body over a closed loop is zero for every force in nature.
4. In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system.

Solution:

1. False, total momentum, and total energy of the system remain conserved. However, the linear momentum and energy of each body of the system may not be conserved.
2. False, external force on the system can change the total energy of the system.
3. False, in the case of non-conservative forces like friction, the work done in a closed-loop is not zero.
4. True, there is atleast some loss of kinetic energy in an inelastic collision.

Question 8.

1. 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)?
2. Is the total linear momentum. conserved during the short time of an elastic collision of two balls?
3. What are the measures to (a) and (b) for an inelastic collision?
4. If the potential energy of two billiard balls depends only on the separation distance between their centres, is the collision elastic or Inelastic?

Solution:

1. In this case total kinetic energy is not conserved because when the bodies are in contact during elastic collision even the kinetic energy is converted to potential energy.
2. yes, because total momentum conserves as per the law of conservation of momentum.
3. The answers are the same for an inelastic collision also.
4. It is a case of elastic collision because in this case, the forces will be of conservative nature.

Question 9.
A body is initially at rest. It undergoes a one-dimensional motion with constant acceleration. The power delivered to it at time t is proportional to

1. t1/2
2. t
3. t3/2
4. t2

Solution:
we know that v = u + at & u = 0 v = at and P = F × V
∴But F = ma
hence P = ma × at = ma2t .
Thus power is proportional to ‘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

• t1/2
• t
• t3/2
• t2

Solution:
It is proportional ‘t3/2’
We know that P = F × V
[P] = [F] [V]
[P] = [MLT-2] [LT-1] since ‘P’ & ‘M’ are constant.
L2T-3 = constants ⇒ $$\frac{L^{2}}{T^{3}}$$= constant.
L2 & T3 ⇒ L & T3/2

Question 11.
A body constrained to move along the z-axis of a coordinate system is subject to a constant force ‘F’ given
by $$\vec{F}=-\hat{i}+2 \hat{j}+3 \hat{k} N$$.
What is the work done by the Force ‘F’ in moving the body 4m in the z-axis?
Solution:
$$\overrightarrow{\mathrm{S}}=4 \hat{\mathrm{k}}$$, $$\overrightarrow{\mathrm{F}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$$
We know that W = FS cos e = $$\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{S}}$$
W = $$[-\hat{i}+2 \hat{j}+3 \hat{k}]$$ . $$[4 \hat{k}]$$
= 1 Joules.

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 × 10-27kg. 1eV = 1.60 × 10-19J).
Solution:
Kelectron = 10 keV, Kproton = 10 keV
melectron = 9.11 × 10-31 kg and
mProton = 1.67 × 10-27 kg.
We know that K = $$\frac{1}{2}$$ mv², v = $$\sqrt{\frac{2 k}{m}}$$

= 13.54
∴ $$\frac{V_{\text {electron }}}{V_{\text {proton }}}$$ = 13.54
Electron is moving Faster.

Question 13.
A raindrop of radius 2 mm falls from a height of 500 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 by 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:
r = 2mm = 2 × 10-3 m
Distance moved half journey s = 250 m
we know that density of water,
ρ = 103 kg/m3
∴ mass of rain drop = volume × density
m = $$\frac{4 \pi r^{3} \rho}{3}$$ = $$\frac{4}{3}$$ × 3.14 × (2 × 10-3)3 × 103
m= 3.35 × 10-5 kg
∴ W = mgr = 3.35 × 10 5 × 9.8 × 250 = 0.082 J
∴ work done by gravitational force in first and second half is the same and is equal to 0.082 J. If there is no resistive forces, energy of the drop on reaching ground
E1 = mgr= 3.35 × 10-5 × 9.8 × 500
= 0.164 J
because of resistive force, the energy of the drop on reaching ground
E2 = $$\frac{1}{2}$$mv² = $$\frac{1}{2}$$ × 3.35 × 10-5 × 10²
= 1.675 × 10-3 J.
∴ work done by the resistive force is
W = E1 – E2 = 0.164 – 1.675 x 10-3
= 0.1623 J.

Question 14.
A molecule in a gas container hits a horizontal wall with a speed of 200 ms-1 and an angle of 30° with the normal, and rebounds with the same speed. Is momentum conserved in the collision? Is the collision elastic or inelastic?
Solution:
Let us consider the mass of the molecule be ‘m’ and that of the wall be ‘M’. The wall is at rest always. Resolving the momentum of the molecule along the x-axis and y-axis we get

The x-component of momentum of molecule
mu cos θ = – m 200 cos 30° = -m$$100 \sqrt{3}$$ kg m/s
y component of the molecule momentum is
= mu sin θ = m x 200 x sin 30° = m100kgm/s
Before collision:
x-component of total momentum = -100$$\sqrt{3}$$ m
Y component of momentum = 100 m
After collision:
x component of the momentum = m 200 cos 30° = 100 $$\sqrt{3}$$ m
Y component = m 100 sin 30° = 100 m
∴ It can be seen that momentum of system is conserved. The wall has a recoil momentum such that momentum of the wall + momentum of out going molecule equals the momentum of the incoming molecule.
Initial kinetic energy $$\left(\frac{1}{2} \mathrm{mu}^{2}\right)$$ is the same as final KE $$\left(\frac{1}{2} m v^{2}\right)$$ molecule.
∴ u = v = 200 m/s thus the collision is 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 water = 30m3
t = 15 min = 15 x 60 = 900 s
h = 40 m; η = 30%
we know that density of water
ρ =103 kgm-3
∴ mass of water pumped
m = volume x density = 30 x 103 kg
Actual power consumed
p0 =$$\frac{W}{t}$$ = $$\frac{m g h}{t}$$
p0 = $$\frac{30 \times 10^{3} \times 9.8 \times 40}{900}$$ = 13070 watt
If pi (required), then n = $$\frac{\mathrm{p}_{\mathrm{o}}}{\mathrm{p}_{\mathrm{i}}}$$
⇒ pi = $$\frac{\mathrm{p}_{\mathrm{o}}}{\mathrm{n}}$$ = $$\frac{13070}{0.3}$$ = 43567 w

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:
Let m be the mass of each ball Total KE of the system before collision = $$\frac{1}{2}$$mv²
Total KE of the system after collision is
case (i) E1 = $$\frac{1}{2}$$ (2m) $$\left(\frac{v}{2}\right)^{2}$$ = $$\frac{1}{4}$$mv²

case (ii) E2 = $$\frac{1}{2}$$mv²

case (iii) E3 = $$\frac{1}{2}$$(3m) $$\left(\frac{v}{2}\right)^{3}$$ = $$\frac{1}{6}$$mv²

∴ case (ii) is only possibility since KE of the system is conserved.

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

Solution:
Since the collision is elastic ‘A’ will come to rest and ‘B’ will begin to move with the velocity of ‘A’. The bob ‘A’ transfers its entire momentum to the bob ‘B’ on the table. Thus bob ‘A’ does not rise at all.

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:
On releasing the bob of the pendulum from a horizontal position, it falls vertically downward by a distance equal to the length of the pendulum i.e. h = I = 1.5.m. As 5% of loss is P E is dissipated against air resistance, the balance 95% energy is transformed into KE Hence,
$$\frac{1}{2}$$mv² = 0.95 x mgh
V = $$\sqrt{2 \times 0.95 \times 9.8 \times 1.5}$$ = 5.3 ms-1.

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, the sand starts leaking out of a hole 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 sandbag is empty?
Solution:
The system of trolley and sandbag is moving with a uniform speed. Clearly, the system is not being acted upon by an external force. If the sand leaks out, even then no external force acts. So there shall be no change in the speed of the trolley.

Question 20.
A particle of mass 0.5 kg travels in a straight line with velocity v = a x3/2 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:
Given m = 0.5 kg.
V = ax3/2, a = 5m-1/2 s-1
initial velocity at x = 0, v1= a × 0 = 0
Final velocity at x = 2n is,
V2 = a (2)3/3 = 5 × 23/2
work done = change in KE
$$\frac{1}{2}$$ m (v2² – v1²) = $$\frac{1}{2}$$ × 0.5 × 25 × 23
= 50 J .

Question 21.
The blades of a windmill sweep out a circle of area A.

1. If the wind flows at a velocity v perpendicular to the circle, what is the mass of the air passing through it in time?
2. What is the kinetic energy of the air?
3. Assume that the windmill converts 25% of the wind’s energy into electrical energy and that A = 30 m², v = 36 km/h and the density of air is 1.2 kg m-3. What is the electrical power produced?

Solution:
1. Volume of the wind flowing per sec = AV
Mass of wind flowing per sec = AVP
Mass of air passing in sec = AVPt

2. Kinetic energy of Air
= $$\frac{1}{2}$$ mv² = $$\frac{1}{2}$$ (Avpt)v²
= $$\frac{1}{2}$$Av3pt

3. Electrical energy produced =
$$\frac{25}{100}$$ × $$\frac{1}{2}$$Av3pt = $$\frac{\mathrm{Av}^{3} \mathrm{pt}}{8}$$
Electrical power = $$\frac{\mathrm{Av}^{3} \mathrm{p}}{8 \mathrm{t}}$$ = $$\frac{A V^{3} p}{8}$$
Now, A = 30m2,
v = 36 kmh-1 = $$\frac{36 \times 100}{3600}$$ m/s = 10ms-1
ρ = 1.2 Kgms-1
Electrical power =
$$\frac{30 \times 10^{3} \times 1.2}{8}$$ = 4500 W = 4.5kW.

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.

1. How much work does she do against the gravitational force?
2. Fat supplies 3.8 × 107J of energy per kilogram which is converted to mechanical energy with a 20% efficiency rate. How much fat will the dieter use up?

Solution:
Here, m = 10kg, h = 0.5m, n = 1000
1. Work done against gravitational force w = n (mgh) = 1000 × 10 × 9.8 × 0.5 = 49000 J.

2. Mechanical energy supplied by 1 kg of fat = 3.8 × 107 × $$\frac{20}{100}$$ = 0.7 × 107 J/kg
∴ Fat used up by the dieter = $$\frac{1 \mathrm{kg}}{0.7 \times 10^{7}}$$ × 49000 = 6.45 × 10-3kg

Question 23.
A family uses 8 kW of power.

1. 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?
2. Compare this area to that of the roof of a typical house.

Solution:
1. Power used by the family, p = 8kW= 8000 W
As only 20% of solar energy can be converted to useful electrical energy, the power to be supplied by solar energy =
$$\frac{8000 w}{20 \%}$$ = 40000 w.
Solar energy is incident at a rate of 200 w/m² hence Area needed.
A = $$\frac{4 \mathrm{0} \mathrm{00w}}{200 \mathrm{Wm}^{-2}}$$ = 200 m².

2. The area needed is comparable to the roof area of the large-sized house.

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 thin wires. Calculate the height to which the block rises. Also, estimate the amount of heat produced in the block.
Solution:
m1 = 0.012 kg, u1 = 70ms-1
m2 = 0.4 kg, u2 = 0.
After the collision, the bullet strikes in the block and both behave as a single body of mass m1 + m2 and moves with velocity V.
By conservation of momentum (m1 + m2) v = m1u1 + m2u2
= 2.04 ms-1
Let the block move to a height ‘h’
(m1 + m2)gh = $$\frac{1}{2}$$ (m1 + m2) v²
h = $$\frac{\mathrm{v}^{2}}{2 \mathrm{g}}$$ = $$\frac{2.04 \times 2.04}{2 \times 9.8}$$ = 0.212 m
To find the heat produced, calculate energy lost (w)
w = initial KE of bullet – final KE of combination
= $$\frac{1}{2}$$ m1 u1² – $$\frac{1}{2}$$ (m1 + m2) v²
= $$\frac{1}{2}$$ x 0.012(70)² – $$\frac{1}{2}$$ (0.412) (2.04)²
w = 29.4 – 0.86 = 28.5 J
∴ Heat produced
H = $$\frac{\mathrm{w}}{\mathrm{j}}$$ = $$\frac{28.54}{4.2}$$ = 68 cal.

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. 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 = 10 m, what are the speeds and times taken by the two stones?

Solution:
Assume the two stones are rest at the point ‘A’ & allowed to roll down. Since the height is same for the both the stones, the velocity at which they reach ground will be same given mass of both stones are equal.
i.e. $$\frac{1}{2}$$ mV² = mgh V = $$\sqrt{2 \mathrm{gh}}$$
VB = Vc = $$\sqrt{2 \times 10 \times 10}$$ = 14.14 ms-1
we know that H = l sin θ, l = $$\frac{\mathrm{h}}{\sin \theta}$$
and the acceleration at which they roll is a = g sin θ.

Question 26.

A 1 kg block situated on a rough incline is connected to a spring constant 100Nm-1. The block is released from rest with the spring in the unstretched position. The block moves 10cm down the incline before coming to rest. Find the coefficient of friction between the block and the incline.
Solution:
The normal reaction force of block on incline R = mg cos θ
Let the coefficient of friction be M
∴ F = MR
= μ mg cos θ
∴ Net force on block = mg sin θ – F
= mg (sin θ – μ cos θ )
we know that the block moves by distance x = 10cm = 0.1 m.
The work done by net force in moving block 0.1m is = Energy stored in the spring.
∴ mg (sin θ – μ cos θ) x = $$\frac{1}{2}$$ kx²
∴ 2 mg (sin 37° – μ cos 37°) = kx
⇒ 2 × 1 × 9.8 ms-2(sin37°- μ cos37°) = 100 × 0.1m
⇒ 19.6 (0.601 – μ × 0.798) = 10
0.601 – μ × 0.798 = 0.5102
⇒ μ × 0.798 = 0.09079
∴ μ = $$\frac{0.09079}{0.798}$$ = 0.1137.

Question 27.
A bolt of mass 0.3 kg falls the celling of an elevator moving down with a uniform speed of   7ms-1. It hits the floor of the elevator (length of 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:
P.E. of bolt = mgh = 0.3 x 9.8 x 3 = 8.82 J
The bolt does not rebound. So the whole of the energy is converted into heat.
Thus, heat produced = 8.82 J.
Since the value of acceleration due to gravity is the same in all inertial systems, therefore the answer will not change even if the elevator is stationary.

Question 28.
A trolley of mass 00 kg moves with a uniform speed of 36 km h-1 on a frictionless track. A child of mass 20 kg runs on a trolley from one and to the other (10m away) with a speed of 4ms-1 relative to the trolley in a direction opposite to the trolley’s 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:
Let there be an observer traveling parallel to the trolley at the same speed. He will observe the initial momentum of the trolley of mass M and the child of mass m as zero. When the child jumps in the opposite direction, he will observe the increase in the velocity of the trolley by ΔV.
Let u be the velocity of the child. He will observe the child landing at velocity (u – ΔV )
Therefore, initial momentum = 0
Final momentum = M ΔV – m (u – ΔV )
Hence, M ΔV – m (u – ΔV ) = 0
∴ ΔV = $$\frac{\mathrm{mu}}{\mathrm{M}+\mathrm{m}}$$ = $$\frac{4 \times 20}{20+220}$$ = 0.36 m/s
∴ final speed of trolley is 10.36 ms-1 and child take 2.5s to run on trolley.
∴ trolley moves a distance of 2.5 × 10.36 = 25.9 m.

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

Solution:
The potential energy of a system of two masses varies inversely as the distance (r) between them i.e.
V(r) < $$\frac{1}{r}$$ When the two balls touch each other, PE becomes zero i.e. at r – R + R = 2R , v (r) = 0. Out of the given graphs only the curve (V) satisfies these two conditions.

### 1st PUC Physics Work, Energy and Power One Mark Questions and Answers

Question 1.
What is the significance of the negative sign in w = – mgh.
A negative sign signifies that the work is done against the gravitational force.

Question 2.
Is friction a conservative force? Give reason.
No, because work done in a closed path against friction is not zero. Moreover, and the work done against fiction depends on the path.

Question 3.
Give the conditions under which a force is called a conservative force.
A force is called to be a conservative force,

• work done against the force has to be independent of the path.
• work done against the force in a closed path has to be zero.

Question 4.
A mass m collides with the other mass m and sticks to it. What Is the nature of collision?
Whenever a mass collides and sticks to other mass, the nature of collision is inelastic.

Question 5.
What is the relation between ergs and joule?
1 Joule = 107 ergs.

Question 6.
A mass is moving is a circular path with a constant speed. What is the work done on 1/2th of the rotation?
Work done is zero, since centripetal force and displacement are perpendicular to each other.

Question 7.
A light body and a heavy body have same momentum. Which one will have greater kinetic energy?
We know that KE = $$\frac{p^{2}}{2 m}$$,
given p is same ∴ k E α $$\frac{1}{2 m}$$
∴ Lighter body will have greater kinetic energy.

Question 8.
A spring is cut into equal halves. How is spring constant of each half affected?
We know that F α l
i.e. F = k l
∴ For a given force, k α $$\frac{1}{\ell}$$ since extension a length. If length is halved then the spring constant doubles.

Question 9.
Does a single external force on a particle necessarily change its kintetic energy?
Yes, a single external force cannot keep the particle is equilibrium.

Question 10.
What is the work done by earth’s gravitational force in keeping the moon in its orbit?
Work done is zero since force and displacement are in the perpendicular direction, i.e. w = o.

Question 11.
Define ‘work’.
Work is said to be done if the body is displaced by the application of force.

Question 12.
Write the dimensional formula of work.
ML2T-2

Question 13.
What is the work done when the displacement is perpendicular to the force applied.
Zero.

Question 14.
Define one Joule.
Work done is said to be 1 joule, if a force of 1 newton displaces a body through 1 meter in the direction of a force.

Question 15.
What does the area under the force-displacement curve indicate?
The area under the force-displacement curve indicates the work done.

Question 16.
Define energy.
The energy of a body is its capacity to do work.

Question 17.
State the law of conservation of energy.
Energy can neither be created nor be destroyed, but can only be converted from one form to another, i.e., The total energy of a closed system remains constant.

Question 18.
Define power.
Power is the rate of doing work.

Question 19.
What is the unit of power?
Watt.

Question 20.
Write the dimensional formula of power.
ML2T-3

Question 21.
Write an expression for instantaneous power in terms of force and velocity.
Instantaneous power = Force × uniform velocity.

Question 22.
State work-energy theorem.
The work-energy theorem states that the work done on a body is equal to the change in its kinetic energy.

Question 23.
What is a conservative force?
A force is said to be conservative if the work done by the force does not depend on the path followed by the body but depends only on the initial and final positions of the body.

Question 24.
What is an elastic collision?
An elastic collision is the one in which both kinetic energy and momentum are conserved.

Question 25.
Define coefficient of restitution. Write an expression for it?
Coefficient of restitution is defined as the ratio of relative velocity after collision to the relative velocity before collision. If u1 & u2 are the velocities of 2 bodies before collision & v1 & v2 are velocities after collision then coefficient of restitution
e = $$\frac{v_{1}-v_{2}}{u_{1}-u_{2}}$$

Question 26.
Write the S.l unit of work.
The S.I unit of work is the joule.

Question 27.
Give an example for conservative force
Gravitational force.

Question 28.
What is the value of the coefficient of restitution in case of a perfectly inelastic collision?
e = 0.

Question 29.
Name any one form of energy.
Kinetic energy.

### 1st PUC Physics Work, Energy and Power Two Marks Questions and Answers

Question 1.
What are conservative forces? Distinguish the conservative and nonconservative forces among the following:

1. Gravitational force
2. Frictional force
3. Air resistance
4. Electrostatic force

Conservative forces are those forces in which work done
1. in a closed path is zero and
2. is independent of path.

 Conservative Forces Non-Conservative Forces Gravitational force Electrostatic force Frictional force Air resistance

Question 2.
Two masses 10kg and 20 kg are connected by a massless spring. A force of 200 N acts on 20 kg mass. At the instant when the 10 kg mass has an acceleration 12 ms-2, what will be the energy stored in the spring? k = 2400 M/m

We know that F = ma
F = 10 × 12 = 120 N
But F = k x = 2400 x = 120 n
∴ x = $$\frac{1}{20}$$
∴ energy stored in spring
E = $$\frac{1}{2}$$ kx2 = $$\frac{1}{2}$$ × 2400 × $$\left(\frac{1}{2}\right)^{2}$$ = 3J

Question 3.
How are fast neutrons slowed down with the use of moderators?
Neutrons from stable nuclei with proton. Protons available in heavy water mix with a neutron. Since their masses are comparable, the neutrons come to rest in the collision.

Question 4.
A body of mass 4kg initially at rest is subjected to a force 16 N. What is the kinetic energy acquired by the body at the end of 10s?
Given m = 4kg, u = 0
F = 16 N, t = 10 s F = ma
16 = 4a, a = 4ms-2
We know that V = u + at = 0 + 4 × 10
V = 40ms-1

Question 5.
What is the amount of work done by

1. a weight lifter in holding a weight of 10 kg on his shoulder for 30 s, and
2. a locomotive against gravity, if it is traveling on a level plane?

1. zero
2. zero

Question 6.
A body of mass ‘M’ at rest is struck by a body of mass ‘m’. Show that the fraction of KE of mass m transferred to the struck particle is $$\frac{4 \mathrm{m} \mathrm{M}}{(\mathrm{m}+\mathrm{M})^{2}}$$
m1 = m, u1 = u
m2 = M, u2 = 0 V2 = ?
By conservation of momentum
m1 u1 = (m1 + m2) v2
∴ v2 = $$\frac{\mathrm{m}_{1} \mathrm{u}_{1}}{\mathrm{m}_{1}+\mathrm{m}_{2}}$$
k E of body struck after collision
E2 = $$\frac{1}{2}$$ m2v2² = $$\frac{1}{2}$$m $$\left(\frac{2 m u}{m+M}\right)^{2}$$
= $$\frac{2 \mathrm{Mm}^{2} \mathrm{U}^{2}}{(\mathrm{M}+\mathrm{m})^{2}}$$
Initial K E E1 = $$\frac{1}{2}$$ m1 u1² = $$\frac{1}{2}$$ mu²
∴ Fraction of initial K E transferred is

Question 7.
How high must a body be lifted to gain an amount of potential energy equal to the kinetic energy it has. When moving at sped 20ms-1. The value of acceleration due to gravity at that place is g = 9.8 ms-2
mgh = $$\frac{1}{2}$$ mv²
i.e. 9.8 h = $$\frac{1}{2}$$ × 20 × 20
h = $$\frac{200}{9.8}$$ = 20.4 m.

Question 8.
Define kinetic energy. Give an example of a body possessing kinetic energy.
The energy possessed by a body due to its motion is called kinetic energy. A bullet fired from a gun can penetrate a target due to its kinetic energy.

Question 9.
Explain potential energy with an example.
The energy possessed by a body due to its position or configuration is called potential energy. example: The potential energy of water in dams is used to run turbines in order to produce electrical energy.

Question 10.
Derive an expression for gravitational potential energy.
Consider a body of mass ‘m’ lying on the surface of earth. The force required to lift the body is equal to its weight mg. Let it be lifted through a height h. Work was done on the body W = Fs = mgh This work done is stored in the body as its potential energy
Ep = mgh.

Question 11.
What is a non-conservative force? Give an example.
A force is said to be non-conservative if the work done by the force depends on the path followed by the body, for example: Frictional force, Air resistance, Viscous force.

Question 12.
Give an example of a body possessing potential energy and kinetic energy.
A freely falling body possesses both potential & kinetic energy.

Question 13.
The momentum of a body is increased twice; How does its Kinetic energy change?
The Kinetic energy increases by four times.

Question 14.
Write the expression for the kinetic energy of a moving body and explain the symbols.
Kinetic energy E = $$\frac{1}{2}$$(mv2)
m → mass of the body
v → velocity of the body.

### 1st PUC Physics Work, Energy and Power Three Marks Questions and Answers

Question 1.
A stone is dropped from a height ‘h’ to prove that the energy at any point in its path is mgh.
Solution:

PE at ‘A’ = mgh
K E at ‘A’ = 0
∴ Total energy at A = mgh
As it starts falling at reaches the point ‘B’ it would have lost some PE and gained some KE.
we know that v² = u² + 2as
∴ V²B = 0 + 2g x
∴ VB = $$\sqrt{2 g x}$$
PE at B = mg (h – x)
KE at B = mgx
∴ Total energy at B = mgh – mgx + mgx = mgh.
On reaching the ground ‘C” the mass must have gained velocity of Vc =$$\sqrt{2 g h}$$
∴ PE at C =0
KE at C = mgh
∴ Total energy at C is mgh
Thus it is proved that the total energy at any point in its path is mgh.

Question 2.
A body of mass 3 kg makes an elastic collision with another body at rest and continues to move in the original direction with a speed equal to $$\frac{1}{3}$$ rd of
its original speed. Find the mass of the second body.
Solution:
Given,
m1 = 3 kg, u1 = xms-1, v1 = $$\frac{x}{3}$$ ms-1
m2 = mkg, u2 = 0
Since collision is elastic, both momentum and KE is conserved.
By conservation of momentum;
m1 u1 + m2 u2 = m1 v1 + m2 v2
3x + 0 = $$\frac{3 x}{3}$$ + mv2
∴ 2x = mv2 ……………. (1)
By law of conservation of KE
$$\frac{1}{2}$$ m1u1² + $$\frac{1}{2}$$ m2u2² = $$\frac{1}{2}$$ m1 v1² + $$\frac{1}{2}$$ m1 v1²
$$\frac{1}{2}$$ 3x² = $$\frac{1}{2}$$ × 3 × $$\frac{x^{2}}{9}$$ + $$\frac{1}{2}$$ mv2²
3x² – $$\frac{x^{2}}{3}$$ = m2v2²
m2v2² = $$\frac{8 x^{2}}{3}$$
Dividing 2 by 1, v2 = $$\frac{4 x}{3}$$
∴ from (1) m × $$\frac{4 x}{3}$$ = 2x
∴ m = $$\frac{3}{2}$$ kg = 1.5 kg

Question 3.
A block of mass 2kg is pulled up on a smooth incline of angle 30° with horizontal. If the block moves with an acceleration of 1ms-2, find the power delivered by the pulling force after 4 seconds. What is the average power delivered during these four seconds?
Solution:
Resolving the forces parallel to the incline.

F – mg sin θ = ma.
F = mg sin θ + ma
= 2 × 9.8 × sin 30 + 2 × 1
= 11.8 N.
The velocity after 4 seconds = u + at.
= 0 + 1 × 4
= 4 ms-1
Power delivered by force at t = 4 sec
= Force × velocity
= 11.8 N × 4ms-1
= 47 .2 W
The displacement during 4 sec
V² = u² + 2as
16 = 2S
S = 8M
Work done in 4 sec = F × s = 11.8 × 8 = 94.4 J
∴ Average power delivered = $$\frac{\text { work done }}{\text { time }}$$ = $$\frac{94.4}{4}$$ = 23.6 W.

Question 4.
The displacement x of a particle moving in one dimension under the action of a constant force is related to time by t = $$\sqrt{x}+3$$ Where x is in meter & t in sec. Calculate the work done by force in the first 6 seconds.
Solution:
Given t = $$\sqrt{x}$$ + 3
$$\sqrt{x}$$ = t – 3
∴ x = (t – 3)²
The velocity of the particle given by
V = $$\frac{d x}{d t}$$ = 2(t – 3) ms -1
∴ The acceleration of the particle given by
a = $$\frac{d v}{d t}$$ = 2 ms-2
Force required to produce this acceleration is given F = ma = m × 2 = 2m
Where ‘m’ is the mass of the particle. Distance travelled by the particle in first 6 sec.
x = (6 – 3)² = 9 m
∴ Hence work done = F × x = 2m × 9 = 18 m Joule.

Question 5.
An inclined plane (θ) has its topmost point at a height ii. Prove that the work done to bring a mass to the ground level either vertically or along the inclined surface is equal and is mgh. Also, prove that the velocity of the mass at the lowermost point is $$\sqrt{2 \mathrm{gh}}$$
Solution:
Consider path (a)
a = g sin θ

velocity on reaching ground is
V² = u² + zal
V² = 2 gsin θ l
where l = $$\frac{\mathrm{h}}{\sin \theta}$$
v = $$\sqrt{2 g \sin \theta \frac{h}{\sin \theta}}$$ = $$\sqrt{2 g h}$$
work done = change in K E
= $$\frac{1}{2}$$ mv² – 0 = $$\frac{1}{2}$$ x 2gh x m = mgh
Consider Path (b)
a = g
velocity on reaching ground is
v² = u² + 2gh
v² = 2gh
v = $$\sqrt{2 g h}$$
work done = change in K E.
$$\frac{1}{2}$$ mv² = mgh.

Question 6.
Define Elastic and Inelastic collisions. What are their basic characteristics?
Solution:
Elastic collision:
A collision in which there is absolutely no loss Of kinetic energy is called an elastic collision.
Characteristics:

1. The linear momentum is conserved
2. The total energy of the system is conserved
3. Kinetic energy is conserved
4. Forces involved during elastic collisions must be conservative forces.

Inelastic collision:
A collision in which there is some loss of kinetic energy is called an inelastic collision.

Characteristics:

1. Linear momentum is conserved
2. Total energy is conserved
3. K E is not conserved
4. Some or all forces involved may be nonconservative.

Question 7.
Derive an expression for the kinetic energy of a body.
Solution:
Consider a body of mass ‘m’ at rest. Let a constant force F act on the body producing an acceleration ‘a’ for a distance ‘s’. Let the velocity of the body be changed to v.
In this case, initial velocity = 0
final velocity = v.
Using, v² = u² + 2as
v² = 0 + 2as
v² = 2as
s = $$\frac{v^{2}}{2 a}$$
Work done F × s
= m a s = ma $$\frac{v^{2}}{2 a}$$
Work done = $$\frac{1}{2}$$ mv²
By definition, this work done is equal to kinetic energy.
∴ kinetic energy of the body, Ek = $$\frac{1}{2}$$ mv².

Question 8.
Distinguish between potential and b kinetic energy.
Solution:

1. Potential energy is the energy possessed by a body due to its position or configuration while kinetic energy is the energy possessed by a body due to its motion.
2. Gravitational potential energy is given by Ep = mgh, but kinetic energy is Ek = $$\frac{1}{2}$$ mv².
3. Potential energy is measured by the amount of work that a body can do when it returns to the standard position. Kinetic energy is measured by the work done on the body to change its velocity.
4. A wound spring has potential energy while a bullet fired from a gun has kinetic energy.

Question 9.
State and prove the work-energy theorem.
OR
Show that work done by a moving body is equal to the change in K.E. of the body.
Solution:
The work-energy theorem states that work done on a body is equal to the net change in its energy. (P.E or K.E)
Proof:
Consider a body of mass ‘m’ moving with an initial velocity u. Let a constant force F acting on a body changes its velocity to v. Let s is the distance traveled.
From the equation, v² = u² + 2as,
we get v² – u² = 2as
$$\frac{1}{2}$$ (v² – u²) = as
Multiplying both sides by m,we have
$$\frac{1}{2}$$ m(v² – u²) = mas
$$\frac{1}{2}$$ m(v² – u²) = F.s(∵ F = ma)
or $$\frac{1}{2}$$ mv² – $$\frac{1}{2}$$ mu² = W

Question 10.
Define potential energy. Derive an expression for gravitational potential energy. Mention its unit.
Solution:
The energy possessed by a body due to its position or configuration is called potential energy. Consider a body of mass ‘m’ lying on the surface of the earth. The force required to lift the body is equal to its weight
∴ F = mg

Let it be lifted through a height ‘h’
The work done on the body is W = FS
W = mgh (∵ S = h)
This work done is stored in the body as its potential energy
∴ Ep = mgh
i.e. Gravitational potential energy = mgh
Its unit is Joule.

Question 11.
State law of conservation of energy. Explain the conservation of energy in the case of a body sliding down on an inclined plane.
Solution:
The law of conservation of energy states that “ Energy is neither created nor destroyed. But it can be transformed from one from to another’’

Consider a body of mass ‘m’ sliding down a smooth inclined plane of angle q and vertical height h. At the highest point A the body has only potential energy. Since it is at rest, kinetic energy is zero.
Total energy at A
= Potential Energy + Kinetic energy
= mgh ……………… (1)
Let the body slides down to a point D. Then it has both potential energy and kinetic energy. Let ‘x’ be the vertical distance travelled by the body.
∴ Potential energy at D is = mg (h – x) its velocity at D is v² = u² + 2as
v² = 0 + 2gsin θ. (AD)
(∵ u = 0, a = gsin θ, s = AD)
= 2g $$\frac{x}{A D}$$ × AD
∵ sin θ = $$\frac{x}{A D}$$ ∴ v² = 2gx
Kinetic energy of the body
= $$\frac{1}{2}$$ mv² = $$\frac{1}{2}$$ m. (2gx) = mgx
Total energy at D is = mg (h -x) + mgx
= mgh …………. (2)
At the lowest point B, potential energy = 0
Kinetic energy = $$\frac{1}{2}$$ mv² = $$\frac{1}{2}$$ m. (2gh)
∵ v² = u² + 2as = 0 + 2gsin θ. (AB)
= 2g $$\frac{h}{A B}$$ × AB = 2gh ∵ sin θ = $$\frac{h}{A B}$$
Kinetic energy = mgh
∴ Total energy = 0 + mgh = mgh …………… (3)
From (1), (2) and (3) it follows that total energy of the body remains constant while sliding down a inclined plane. Hence the principle.

### 1st PUC Physics Work, Energy and Power Five Marks Questions and Answers

Question 1.
A hammer of mass M drops from height ‘h’. It strikes a rail placed vertically on the ground and drives it. Into the ground through distance D. Calculate

1. The average resistance offered by the ground, assuming that the hammer and rail remain stuck together after impact,
2. The time for which the rail is in motion and
3. The loss in kinetic energy in input.

Solution:
The hammer of mass M falls freely under the gravity through a distance. Let v be the speed acquired by the hammer when it strikes the rail obviously.
v = $$\sqrt{\mathrm{gh}}$$ ………… (1)
On impact, the hammer and rail are stuck together. Let v be the speed of the combination after impact. The law of conservation of momentum gives.
Mv = (M + m) v1 …………….. (2)
Where m is the mass of the rail.
Let FG be the average resistance (or resistive force) exerted by the ground. The net upward force F on the hammer rail combination is
F = FG – (M + m)g.
The combination moves through a distance ‘d’ against this net upward force. Obviously, the work done against the force F in a distance ‘d’ must equal to the kinetic energy the combination had just after impact.

= [FG – (M + m) g] × d …………… (3)
By equations (1) & (2), we can rewrite equation (3) as

2. Let the hammer rail combination be moving for a time Δt, before coming to rest. If ΔP is the change in momentum,
$$\frac{\Delta p}{\Delta t}$$ = FG – (M + m)g
$$|\Delta p|$$ = (M + m) v1 – 0

3. The kinetic energy of the hammer rail combination just before the impact is $$\frac{1}{2}$$
Mv² and after impact, it is
Loss in kinetic energy = $$\frac{1}{2}$$ Mv² – $$\frac{1}{2}$$

Question 2.
A small block of mass m slides along the frictionless loop-to-loop track shown in the figure

1. If it starts from rest at P, what is the resultant force acting on it at Q?
2. At what height above the bottom of the loop should the block be released so that the force it exerts against the track at the top of the loop equal its weight?

Solution:
1. Point Q is at a height R above the ground. Thus, the difference in height between points P & Q is 4 R. Hence the difference in gravitational potential energy of the block between these points is 4 mg R. Since the block starts from rest at P its kinetic energy at Q is equal to its change in potential energy. By the conservation of energy.
$$\frac{1}{2}$$ mv² = 4 mg R
v² = 8g R
At Q, the only forces acting on the block are its weight ring acting downward and the force M of the track on block acting in radial direction. Since the block is moving in a circular path, the normal reaction provides the centripetal force for circular motion.
N = $$\frac{m v^{2}}{R}$$ = $$\frac{\mathrm{m} \times 8 \mathrm{g} \mathrm{R}}{\mathrm{R}}$$ = 8 mg
The loop must exert a force on block equal to eight times the blocks weight.

2. For the block to exert a force equal to its weight against the track at the top of the loop,
$$\frac{m v^{2}}{R}$$ = 2 mg
or v2 = 2gR

The block must be released at a height of 3 R above the bottom of the loop.

Question 3.
A bullet of mass ‘m’ moving with a velocity V is embedded into a block of mass ‘x’ suspended by a thread. As a result of this collision the block along with the bullet rises to a height ‘h’ Prove that velocity of bullet was
$$\left(\frac{m+M}{m}\right) \sqrt{2 g h}$$
Solution:
let v1 be velocity of the system of block and the bullet. Applying principle of conservation of linear momentum, (M + m) v1 = mv
v1 = $$\frac{\mathrm{mv}}{(\mathrm{M}+\mathrm{m})}$$ ………. (1)
As K E of the system = P E of the system at height ‘h’
$$\frac{1}{2}$$ (m + M) v1² = (m + M) gh
V1 = $$\sqrt{2 \mathrm{gh}}$$
From (1)
v = $$\frac{(m+m) v_{1}}{m}$$ = $$\frac{(m+M) \sqrt{2 g h}}{m}$$

Question 4.
A block of mass 200 g is released from ‘P’. if it slides down without friction till it reaches a point Q of traversing a circular path of radius 2.0 m.

1. Find the velocity of the block at point Q and
2. The coefficient of friction if the block comes to rest 2.0 m. from Q, Assuming the horizontal part of the path is rough. Take g – 10ms-2.

Solution:
1. P E of the block at P = kintetic energy of block at Q
∴ mgh = $$\frac{1}{2}$$ mv²
v = $$\sqrt{2 g h}$$ = $$\sqrt{2 \times 10 \times 2}$$ = $$\sqrt{40 \mu}$$
∴ θ = 6.32 ms -1.

2. When the block comes to rest after travelling a distance, r = 2m from then, work done against friction = change in K E
μ mgs = $$\frac{1}{2}$$ mv²
∴ μ = $$\frac{v^{2}}{2 g r}$$ = $$\frac{40}{2 \times 10 \times 2}$$ = 1

Question 5.
A mass ‘m’ moving with a speed u colloids with a similar mass m at rest, elastically and obliquely. Prove that they will move in directions making an angle $$\frac{\pi}{2}$$ with each other.
Solution:
Since momentum is a vector, the component along the x-axis and y-axis must separately balance each other.
On considering the components along the x-axis we have,
mu = mv1 cosθ1 + mv2cosθ2
i.e. u = v1 cosθ1 + v2 cos θ2 …………………. (1)
Similarly on considering the components along the y-axis we have,
0 = v1 sin θ1 + v2 cos θ2 ……………… (2)

Being an elastic collision, kinetic energy is also conserved.
$$\frac{1}{2}$$mu² = $$\frac{1}{2}$$ mv1² + $$\frac{1}{2}$$ mv2²
u² = v1² + v2² ………………. (3)
squaring & adding (1) and (2) we get
u² = v1² + v2² + 2 v1v2 + (cos θ1cos θ2 – sin θ1sin θ2)
∴ u² = v1² + v2² + 2 v1v2 cos (θ1 + θ2)
using (3) in this equation, we get
2v1v2cos (θ1 + θ2) = 0
cos (θ1 + θ2) = 0
∴ θ1 + θ2 = $$\frac{\pi}{2}$$
i.e., the masses move at right angles after the collision.

Question 6.
An object of mass 0.4 kg moving with a velocity of 4ms-1 collides with another object of mass 0.6 kg moving in the same direction with a velocity of 2ms-1. If the collision is perfectly inelastic, what is the loss in K E due to impact?
Solution:
m1 = 0.4 kg, u1 = 4 ms-1 m2 = 0.6kg, u2 = 2 ms-1
∴ Total K E of system before collision
ki = $$\frac{1}{2}$$ m1u1² + $$\frac{1}{2}$$ m2u2²
$$\frac{1}{2}$$ × 0.4 × 4² + $$\frac{1}{2}$$ × 0.6 × 2²
= 3.2 + 1.2 = 4.4J
∴ As collision is perfectly inelastic, the common velocity after collision v is given by
v = $$\frac{m_{1} u_{1}+m_{2} u_{2}}{m_{1}+m_{2}}$$ = $$\frac{0.4 \times 4+0.6 \times 2}{0.4+0.6}$$
= 2.8 ms-1
∴ Total K E of system after collision
kf = $$\frac{1}{2}$$ (m1 + m2) v² = $$\frac{1}{2}$$ (1) × (2.8)²
= 3.92 J.
∴ Loss in K E = Δ K E = Δ K = Ki – Kf
= 4.4 – 3.92 J
= 0.48 J.

1st PUC Physics Work, Energy and Power Additional Marks Questions and Answers

Question 1.
A force of 600 N displaces an object through 2 m. Find the work done if the force and displacement are

1. Parallel
2. At right angles.

Solution:
Work done is given by W = Fs cos θ
1. when F and s are parallel, θ = 0 and hence cos θ = 1
∴ W = Fs = 600 × 2 = 1200 J.

2. when F and s are at right angles,
θ = 90° and hence cos θ =0
∴ Work done, W = 0.

Question 2.
A sphere of radius 2mm and of density 7.5 × 103 kg m-3 falls from a height of 500m above the ground. Due to air resistance, its acceleration gradually decreases and becomes zero when it drops to half its original height. The drop then falls with a uniform terminal velocity. What is the work done by the gravitational force during the first and second half of its Journey?
Solution:
Even though the resultant force acting on the sphere is different during the first and second half of its journey, the force due to gravity remains constant throughout and is equal to the weight of the sphere. The amount of work done during the first half is given by,
W = mgh where m is the mass of the body.
But m = $$\frac{4}{3}$$πr3d and h = $$\frac{500}{2}$$ = 250m
W = $$\frac{4 \pi}{3}$$(2 x 10-3)3 × 7.5 × 103 × 9.8 × 250
= 4$$\frac{4 \pi}{3}$$ (2)3 × 7.5 × 9.8 × 250 × 10-6 J
= 615.4 × 10-3 = 0.615J
The work done remains the same during both halves.

Question 3.
A pump is used to fill a tank 6m × 3m × 2m in half an hour. If the average height to which water to be lifted from the well is 110m, find the horsepower of the engine.
Solution:
Volume of the tank = 6 × 3 × 2 = 36m3
∴ mass of water to be lifted
= V × d = 36 × 103 kg
∴ Work done = mg × h
= 36 × 103 × 9.8 × 110
Power of the pump = $$\frac{W}{t}$$
= $$\frac{36 \times 9.8 \times 1.1 \times 10^{5}}{30 \times 60}$$ = 0.216 × 105W
= $$\frac{0.216 \times 10^{5}}{746}$$ = $$\frac{21600}{746}$$ = 28.9 HP.

Question 4.
A circular well of radius 5m is filled with water to a height of 25m. It is emptied using a pump. If the efficiency of the pump is 60% and the initial level of water from the surface of the ground Is 10 m, calculate the time taken to empty it.
Solution:
If h1 and h2 are the heights of the initial and final levels of water in the well, then the average height through which water is lifted to empty it is,
h = $$\frac{h_{1}+h_{2}}{2}$$ = $$\frac{10+35}{2}$$ = 22.5 m
mass of water to be lifted
= volume × density = πr²Ld
Here, r = 5 m, L = 25m, d = 103 kgm-3
∴ total work to be done = mgh
= π × 5² × 25 × 103 × 9.8 × 25
= 1.73 × 107 ………………. (1)
Power of the pump used = 15 HP = 15 × 746 W
Total work done in lifting the water = 60% of the power of the pump.
If t is the time taken to empty it, the total amount of work done by the pump is
W = 15 × 746 × t.
Only 60% is used to lift the water.
∴ Amount of work done by the pump to lift water
W= $$\frac{60}{100}$$ × 15 × 746 × t ………… (2)
Equating equation (1) and (2), we have,
1.73 × 107 = $$\frac{60}{100}$$ × 15 × 746 × t
or t = $$\frac{1.73 \times 100 \times 10^{7}}{60 \times 15 \times 746}$$
= 2.5 × 103 S.

Question 5.
A lorry of mass 5 tons runs on a level track with a speed of 80km/hr. If the resistance due to air and friction is 80 N per ton, find the H.P. of the engine. Solution:
Total frictional force acting on the lorry is,
F = 5 × 80 = 400 N.
When a constant force is acting, then the power
P = x force × velocity
= 400 × $$\frac{80 \times 1000}{60 \times 60}$$ W
= 8.89 × 103 W
= $$\frac{8.89 \times 10^{3}}{746}$$ HP = 11.9 HP

Question 6.
An engine pulls a car of mass 1500 kg on a level road at a constant velocity of 5ms-1. If the frictional force is 500 N, what power does the engine generate? What extra power must the engine develop to maintain the same speed up a plane inclined to have gradient in 10?
Solution:

Since the car moves with a constant velocity, the power generated by the engine is just sufficient to overcome the frictional force.
P = F × V = 500 × 5
= 2500 W
= 2.5 kW
When the car moves over an inclined road, in addition to the friction, the engine must over come the component of the weight of the car along the inclined road. The component of the weight along the AB is, F = mg Sin θ
= 1500 × 9.8 × $$\frac{1}{10}$$ (since Sin θ = $$\frac{1}{10}$$)
∴ Extra power required = F × v
= 1500 × $$\frac{9.8 \times 5}{10}$$
= 7350 W = 7.35 kW

Question 7.
A gun of mass 5 kg fires a bullet of mass 10-2 kg with a speed of 5 x 10²ms-1. Find the kinetic energy of the bullet and the gun. Also, find the ratio of the distance moved by the gun due to recoil to that moved by the bullet.
Solution:
If m1 and m2 are the mass of the gun and the bullet and v1 and v2 are their corresponding velocities, then from the law of conservation of linear momentum,
Pi = Pf where Pi is the total initial momentum of bullet and the gun which is zero.
But P1 = m1v1 + m2 v2
m1v1 + m2 v2 = 0 or
v1 = $$\frac{m_{2} v_{2}}{m_{1}}$$ = $$\frac{10^{-2} \times 5 \times 10^{2}}{5}$$
∴ Recoil velocity of the gun,
v1 = – 1 ms-1
∴ Kinetic energy of the bullet.
= $$\frac{1}{2}$$ m2v2²
= $$\frac{1}{2}$$ × 10-2 × (5 × 10²)²
= $$\frac{25 \times 10^{2}}{2}$$ =12.5 × 10² J

Question 8.
A labourer weighing 70kgs climbs 11 step of a stair case each 0.21 m height in 1.4. Find the power in watt and the Horse Power (g = 9.8ms 2)
Solution:
Mass of labourer m = 70kg
Total height of stair case = 11 × 0.21
h = 2.31m
Time taken to climb = 1.4seconds, g = 9.8ms-2
∴ Work done, W = mgh
= 70 × 9.8 × 2.31
= 1585 J.
∴ Power P = $$\frac{W}{t}$$ = $$\frac{1585}{1.4}$$ = 1132 W
But, 1 HP =746 Watts
∴ Power in Horsepower = 1.517 HP.

Question 9.
An engine is used to lift water from a well to a height of 40m to fill a tank 2m × 2m × 3m in 5 minutes. Find the power of the engine.
Solution:
Volume of the tank V = 2 × 2 × 3 = 12m3
∴ Mass of water to be lifted = v × d
= 12 × 103kg
∴ work done, W = mgh
= 12 × 103 × 9.8 × 40
∴ W = 4704 × 103J
∴ Power of the engine P = $$\frac{W}{t}$$
= $$\frac{4704 \times 10^{3}}{5 \times 60}$$
= 15.68 × 103W.

Question 10.
Find the height from which body of a mass 50kg should fall in order to have the K.E of a car of mass 200kg travelling at 10ms-1 Given g = 9.8ms-2
Solution:
Mass of a body, m1 = 50 kg
Mass of car m2 = 200kg
velocity of the car v = 10ms-1 g = 9.8ms-2
P.E of the body, P.E = mgh
= 50 × 9.8 × h
= 490h
K.E of the car K.E = $$\frac{1}{2}$$ m2
= $$\frac{1}{2}$$ × 200 × 10²
= 10,000J
But P.E of body = K.E of car
490h = 10,000
h = $$\frac{10,000}{490}$$ = 20.41 m

Question 11.
A man fires 10 bullets, each of mass 20 × 10-3 Kg in every second with a gun of mass 10 Kg. If the velocity of each bullet is 300 ms-1, what force must be applied by the man to the gun to hold it in position? If the number of bullets fired per second doubles what will be a new force?
Solution:
No of bullets, N = 10
Mass of each bullet, M = 20 × 10-3kg
∴ Total mass of 10 bullets,
m = N × M
= 10 × 20 × 10-3
= 200 × 10-3kg
Velocity of each bullet = 300ms-1
From, the equation, F = ma
= $$m\left(\frac{v-u}{t}\right)$$
= 200 × $$10^{-3}\left(\frac{300-0}{1}\right)$$
= 60 N
∴ Force required by the man to the gun to hold it in position is 60N.
If the number of bullets fired per sec doubles, then force required also doubles.

Question 12.
A bullet of mass 20gm strikes a wooden plank with a velocity of 60ms-1 and emerges out in 0.01 seconds with a velocity of 30ms-1. Calculate
(i) the work done by the bullet and
(ii) the thickness of the plank.
Solution:
Given m = 20 × 10-3kg
u = 60ms-1 & v = 30 ms-1
work done = decrease in kinetic energy
= $$\frac{1}{2}$$ m (u² – v²)
= $$\frac{1}{2}$$ × 0.02 (60² – 30²)
= 27J
From, v = u + at
30 = 60 + a (0.01)
a = $$\frac{-30}{0.01}$$ = – 3000 ms-2
Retarding force, F = ma
= 0.02 × 3000 = 60N
work done, W = F × S
27 = 60 × S
∴ Thickness of the plank, S = $$\frac{27}{60}$$
= 0.45m.

Question 13.
A body of mass 0.1 kg falling freely under gravity takes 10second to reach the ground. Calculate kinetic energy & potential energy of the body, when the body has travelled for 6 seconds.
Solution:
Given: m = 0.1 kgm, u = 0,
t = 6 sec, a = g
∴ From v = u + at
v = 0 + 9.8 × 6
v = 58.8 ms-1
∴ kinetic energy of the body, when the body has travelled for 6 seconds, is
K.E = $$\frac{1}{2}$$ mv²
= $$\frac{1}{2}$$ × 0.1 × (58.8)²
K.E = 172.9 J
To Find Potential Energy (PE):
Distance travelled by the body in 10 sec, is
S10 = ut + $$\frac{1}{2}$$ gt²
= 0 + $$\frac{1}{2}$$ × 9.8 × 10²
S10 = 490m
Distance travelled by the body in 6 sec is
S6 = ut + $$\frac{1}{2}$$gt²
= 0 +$$\frac{1}{2}$$ × 9.8 × 6²
S6 = 176 .4m
∴ h = S10 – S6 = 490 – 176 .4 = 313.6m
∴ P.E = mgh
= 0.1 × 9.8 × 313.6
= 307.3 J.

## Karnataka 1st PUC Physics Question Bank Chapter 10 Mechanical Properties Of Fluids

### 1st PUC Physics Mechanical Properties Of Fluids Textbook Questions and Answers

Question 1.
Explain why
(a) The blood pressure in humans is greater at the feet than at the brain.
(b) Atmospheric pressure at a height of about 6 km decreases to nearly half its value at the sea level, though the height of the atmosphere is more than 100 km.
(c) Hydrostatic pressure is a scalar quantity even though the pressure is force divided by area, and force is a vector.
(a) P = h pg, where h is the height of the liquid column. Since the height of blood tolumn at the feet is greater than at the brain, therefore, the blood pressure in humans is greater at the feet than at the brain.
(b) The density of air in the atmosphere does not decrease linearly with height. It decreases exponentially with height. For this reason, atmospheric pressure at a height of about 6 km decreases to nearly half of its value at sea level.
(c) Hydrostatic pressure is transmitted equally in all directions. Hence no definite direction is associated with pressure. Therefore, the pressure is a scalar quantity.

Question 2.
Explain why
1. The angle of contact of mercury with glass Is obtuse, while that of water with glass is acute.
When a small quantity of liquid is poured on solid, three interfaces, namely liquid – air, solid – air, and solid-liquid are formed. The surface tension corresponding to these three layers are SLA, SSA, and SSL respectively. Let θ be the angle of contact between the liquid and solid.

The molecules in the region, where the three interfaces meet is in equilibrium. It means that net force acting on them is zero. For the molecule at K to be in equilibrium, we have
SSL + SLA cos θ = SSA or cos θ = $$\frac{\mathrm{S}_{\mathrm{SA}}-\mathrm{S}_{\mathrm{SL}}}{\mathrm{S}_{\mathrm{LA}}}$$ In case of mercury-glass, SSA < SSL, therefore cos θ is negative or θ > 90° i.e. (obtuse). In case of water-glass SSA > SSL, therefore cos θ is positive or θ < 90° (acute).

2. Water on a clean glass surface tends to spread out while mercury on the same surface tends to form drops.
In the case of mercury the mercury glass, angle of contact is obtuse. To obtain this obtuse value of angle of contact, mercury tends to form a drop. But in the case of water, the water-glass angle of contact is acute. Thus the water tends to spread out to achieve this acute angle of contact.

3. The surface tension of a liquid is independent of the area of the surface.
The surface tension of a liquid is the force acting per unit length on a line drawn tangentially to the liquid surface which is at rest. As this force is independent of the area of the liquid surface, the surface tension is also independent of the area of the liquid surface.

4. Water with detergent dissolved in ‘ it should have small angles of contact.
The cloth has narrow spaces in the form of capillares. The rise of liquid in a capillary tube is directly proportional to cos θ. If θ is small cos θ will be large. Hence capillary rise will be more so that the detergent will penetrate more in cloth and hence gets dissolved.

5. A drop of liquid under no external forces is always spherical in shape.
In the absence of external forces, the surface of the liquid drop tends to acquire the minimum surface area due to surface tension. For a given volume, the surface area of the sphere is the least and hence the liquid drop takes the spherical shapes.

Question 3.
Fill in the blanks using the word(s) from the list appended with each statement:
1. Surface tension of liquids generally …….with temperatures (increases /decreases)
decreases

2. Viscosity of gases ……with temperature, whereas Viscosity of liquids …….with temperature (increases/decreases).
increases, decreases

3. For solids with elastic modulus of rigidity, the shearing force is proportional to……, while for fluids it is proportional to …..(shear strain /rate of shear strain).
Shear strain, rate of shear strain.

4. For a fluid in a steady flow, the increase in flow speed at a constriction follows (conservation of mass / Bernoulli’s principle).
Both conservation of mass and Bernoulli’s principle.

5. For the model of a plane in a wind tunnel, turbulence occurs at a ….. speed for turbulence for an actual plane (greater / smaller).
Greater

Question 4.
Explain why.
1. To keep a piece of paper horizontal, you should blow over, not under, it.
If we blow over a piece, velocity of air above the paper becomes more than that below the paper. This reduces the pressure of the air above the paper in accordance with Bernoulli’s theorem. But the pressure of air below the paper is still atmospheric and is higher than the pressure above. Hence the paper remains horizontal and does not fall.

2. When we try to close a water tap with our fingers, fast jets of water gush through the openings between our fingers.
When we try to close the water top with our fingers, the area of outlet of water eject gets reduced. Therefore in accordance with the principle of Continuity (AV = constant), the velocity of the water will increase creating fast jets of water.

3. The size of the needle of a syringe controls flow rate better than the thumb pressure exerted by a doctor while administering an injection.
The size of the needle controls the velocity of flow and the thumb pressure controls the pressure. According to Bernoulli’s equation:
P + ρgh+ $$\frac{1}{2} \rho$$ v2 = constant
we can note that the equation varies linearly with pressure, P but varies as the square of the velocity. Hence the contribution of the velocity of flow is greater, due to which the needle has better control over the flow.

4. A fluid flowing out of a small hole in a vessel results in a backward thrust on the vessel.
When a fluid flows out of a small hole in a vessel, it acquires a large velocity and hence has a large momentum. Since no external force is acting on the system, a backward velocity must be acquired by the vessel (By-law of conservation of momentum). As a result backward thrust is experienced by the vessel.

5. A spinning cricket ball In air does not follow a parabolic trajectory.
A spinning ball displaces air. The ball moves formed and relative to it, the air moves backward. Hence, the velocity of air above the ball relative to it is larger and below it is smaller. This difference in the velocities of air results in a pressure difference between the lower and upper faces leading to a net upward force on the ball. Therefore the ball does not follow a parabolic trajectory.

Question 5.
A 50 kg girl wearing high heel shoes balances on a single heel. The heel Is circular with a diameter of 1.0 cm. What is the pressure exerted by the heel on the horizontal floor?
Force exerted by the heel due to the weight of the girl, F = mg = 50 × 10 = 500N
Diameter of the heel’s circular are, d = 1 cm = 10-2m
∴ Area of the heel $$\frac{\pi \mathrm{d}^{2}}{4}$$
= $$\frac{\pi \times\left(10^{-2}\right)^{2}}{4}=7.85 \times 10^{-5} \mathrm{m}^{2}$$
∴ Pressure exerted by the heel = $$\frac{F}{A}$$
$$=\frac{500}{7.85 \times 10^{-5}}$$
= 6.366 × 106 Pa.

Question 6.
Toricelli’s barometer used mercury. Pascal duplicated it using French wine of density 984 kg m-3. Determine the height of the wine column for normal atmospheric pressure. (Assume g = 9.8 m/s2)
Normal atmospheric pressure = 1 atm
⇒ gauge pressure of wine = 1 atm
ρgh = 1.013 × 105 Pa (Since 1 atm = 1.013 × 105pa)
⇒ h = $$\frac{1.013 \times 10^{5}}{984 \times 9.8}$$
⇒ h = 10.5 m.

Question 7.
A vertical off-shore structure is built to withstand maximum stress of 109 Pa. Is the structure suitable for putting up on top of an oil well in the ocean? Take the depth of the ocean to be roughly 3 km, and ignore ocean currents.
The depth of ocean, h = 3 km = 3000 m.
Density of water a 1000 kg m 3 (assumed) ; g = 9.8 ms’2.
∴ Pressure, P = ρgh = 1000 x 9.8 x 3000 = 2.94 x 107 Pa.
This value of pressure is less than the stress of 109 Pa which the structure can withstand. Therefore, the structure is suitable.

Question 8.
A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000kg. The area of cross¬section of the piston carrying the load is 425 cm2. What maximum pressure would the smaller piston have to bear.
Let Fmax be the maximum force experienced by the Bigger piston.
Fmax = mg = 3000 × 9.8 = 29400 N
Area of the bigger piston = 425 cm2
= 425 x 10-4m2
Maximum pressure on the bigger piston
pmax = $$\frac{F_{\max }}{A}$$
$$P=\frac{29400}{425 \times 10^{-4}}$$ = 6.92 × 105 Pa
Since pressure is transmitted uniformly throughout the liquid the smaller piston will also bear a pressure of 6.92 × 105Pa.

Question 9.
A U-tube contains water and methylated spirit separated by mercury. The mercury columns in the two arms are In level with 10.0 cm of water in one arm and 12.5 cm of spirit In the other. What is the specific gravity of spirit?

$$\rho_{1}$$ → density of water
$$\rho_{2}$$ → density of spirit
h1 → height of water column
h2 → height of spirit column
Let Po be the atmospheric pressure
⇒ PA = Po + $$\rho_{1}$$ gh1 ….(1)
PB = Po + $$\rho_{2}$$ gh2 …..(2)
Since the mercury levels are same at point A and B we can say that,
PA = PB
⇒ Po + $$\rho_{1}$$ gh1 = Po + $$\rho_{2}$$ gh2
from (1) and (2)
⇒ $$\rho_{1}$$ gh1 = $$\rho_{2}$$ gh2
⇒ $$\frac{\rho_{2}}{\rho_{1}}=\frac{h_{1}}{h_{2}}=\frac{10}{12.5}=0.8$$
∴ specific gravity of spirit
$$=\frac{\text { density of spirit }}{\text { density of water }}=\frac{\rho_{2}}{\rho_{1}}=0.8$$

Question 10.
In the previous problem, if 15.0 cm of water and spirit each are further poured into the respective arms of the tube, what is the difference in the levels of mercury in the two arms ? (Specific gravity of mercury = 13.6)
Since we are putting equal amount of water and spirit and specific gravity of spirit is < 1 we can assume that the mercury level will rise on the spirit side,
$$\rho_{1}$$ → density of water,
$$\rho_{2}$$ → density of spirit,
$$\rho_{3}$$ → density of mecury,

Since A and B are at the same height,
PA = PB
we know that, PA = $$\rho_{1}$$gh1
and PB = $$\rho_{2}$$ gh2 + $$\rho_{3}$$ gx
⇒ $$\rho_{1}$$ gh1 = $$\rho_{2}$$ gh2 + $$\rho_{3}$$ gx
⇒ x = $$\frac{\rho_{1} h_{1}-\rho_{2} h_{2}}{\rho_{3}}$$
x = $$\frac{\mathrm{h}_{1}-\frac{\rho_{2}}{\rho_{1}} \mathrm{h}_{2}}{\frac{\rho_{3}}{\rho_{1}}}$$
where $$\frac{\rho_{2}}{\rho_{1}}$$ ⇒ specific gravity of spirit and $$\frac{\rho_{3}}{\rho_{1}}$$ ⇒ specific gravity of mercury.
Thus x = $$\frac{25-0.8 \times 27.5}{13.6}$$ = 0.221 cm
Therefore the difference in the mercury level in the arms is 0.221 cm.

Question 11.
Can Bernoulli’s equation be used to describe the flow of water through a rapid in a river? Explain.
Bernoulli’s theorem is applicable only for the ideal fluids in streamlined motion. Since the flow of water in a river in a rapid way (i.e., turbulent) can not be treated as streamlined motion and hence the theorem can not be used.

Question 12.
Does it matter if one uses gauge Instead of absolute pressures in applying Bernoulli’s equation? Explain.
No, unless the atmospheric pressure at the two points where Bernoulli’s equation is applied are significantly different.

Question 13.
Glycerine flows steadily through a horizontal tube of length 1.5 m and radius 1cm. If the amount of glycerine collected per second at one end is 4.0 x 10-3 kg s-1, what is the pressure difference between the two ends of the tube? (Density of glycerine = 1.3 x 103 kg m-3 and viscosity of glycerine = 0.83 Pas). [You may also like to check if the assumption of laminar flow in the tube is correct].
The volume of liquid floawing out per second is given by, $$Q=\frac{\pi R^{4}\left(p_{1}-p_{2}\right)}{8 h l}$$
⇒ pressure difference at tube ends, $$\mathrm{P}_{1}-\mathrm{P}_{2}=\frac{8 \mathrm{Q} \eta}{\pi \mathrm{r}^{4}}$$
where Q is the volume of liquid flowing per second = $$\frac{\text { mass of liquid flowing per second }}{\text { density of liquid }}$$
$$=\frac{4 \times 10^{-3} \mathrm{kg} / \mathrm{s}}{1.3 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3}}$$
= 3.077 x 10-6 m3/s
Therefore,
P1 – P2 = $$=\frac{8 \times 0.83 \times 1.5 \times 3.077 \times 10^{-6}}{3.14 \times(1 \times 10)^{3}}$$
= 9.76 x 102 N/m2
A pressure difference of 9.76x 102 N/m2 is obtained.
Reynolds Number,
R = $$\frac{4 \rho V}{\pi d n}$$ = $$\frac{4 \times 1.3 \times 10^{3} \times 3.08 \times 10^{-6}}{\pi \times(0.02) \times 0.83}$$ = 0.3
Since the Reynolds number is about 0.3, the flaw is laminar.

Question 14.
In a test experiment on a model aeroplane in a wind tunnel, the flow speeds on the upper and lower surfaces of the wing are 70 m s-1 and 63 m s-1 respectively. What is the lift on the wing if its area is 2.5 m2? Take the density of air to be 1.3 kg m-3.
P1 = pressure at the lower surface of wings
P2 = pressure at the lower surface of wings
ρ = Density of air
V1 = Speed of wind at the lower surface = 63 m/s.
V2 = speed of wind at the upper surface = 70 m/s
According to Bernoulli’s theorem,
P1 + $$\frac{1}{2} \rho V_{1}^{2}$$ = P2 + $$\frac{1}{2} \rho V_{2}^{2}$$
or P1 – P2 = $$\frac{1}{2} \rho\left(V_{1}^{2}-V_{2}^{2}\right)$$
P1 – p2 = $$\frac{1}{2}$$ × 1.3 × (702 – 632) = 605.15 pa
Force on wings = (P1 – P2) × area
= 605.15 × 2.5 = 1512.87 N
∴ The lift on the wings is about 1.512 × 103N

Question 15.
Figures 10.23(a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect? Why?

Figure (a) is incorrect. From the equation of continuity AV = constant. Thus, when the area of constriction is narrow, velocity of streamline is high. If the velocity is high, we can say that the pressure head is low at that point (from Bernoulli’s equation). Correspondingly the height of the liquid column should be lesser at the narrow construction than that of the wide construction. Hence figure (a) is incorrect.

Question 16.
The cylindrical tube of a spray pump has a cross-section of 8.0 cm2 one end of which has 40 fine holes each of diameter 1.0 mm. If the liquid flow inside the. tube is 1.5 m min-1, what is the speed of ejection of the liquid through the holes?
Number of holes = 40
Diameter of each hole, D = 10-3m
Area of cross section of each hole
$$=\frac{\pi D^{2}}{4}$$
$$=\frac{\pi \times 10^{-6}}{4}$$
Total area of cross section of 40 holes,
a2 = $$\frac{40 \times \pi \times 10^{-6}}{4}$$ = π × 10-5m2
speed of liquid inside the tube,
V1 = 1.5 m/min = $$\frac{1.5}{60}$$ m/s
Area of cross section of tube,
a1 = 8.0cm2 = 8 × 10-4m2
If V2 is the velocity of ejection of liquid through the holes then from equation of
continuity,
a1 v1 = a2 v2
⇒ v2 = $$\frac{a_{1} v_{1}}{a_{2}}=\frac{8 \times 10^{-4} \times 1.5}{60 \times \pi \times 10.5}$$
⇒ v2 = 0.637 m/s
∴ the speed of ejection is 0.637 m/s.

Question 17.
A U-shaped wire is dipped in a soap solution and removed. The thin soap film formed between the wire and the light slider supports a weight of 1.5 x 10-2N (which includes the small weight of the slider). The length of the slider is 30 cm. What is the surface tension of the film?
The soap film has two free surfaces.
∴ Total length of the film to he supported,
l = 2 × 30 = 60 cm = 0.6
Total force on the slider due to surface tension will be, F = S × 2l = S × 0.6N
At equilibrium , the force F on slider due to surface tension should balance the weight mg
∴ F = mg = 1.5 × 10-2 N
⇒ S × 0.6 = 1.5 × 10-2
⇒ S = 2.5 × 10-2 N/m.

Question 18.
Figure 10.24 (a) shows a thin liquid film supporting a small weight = 4.5 × 10-2 N. What is the weight supported by a film of the same liquid at the same temperature in Fig. (b) and (c)? Explain your answer physically.

In all three figures, the liquid is at the same. Temperature is also the same for each case. Hence, the surface tension in figure (b) and (c) is the same as in figure.
1. Since the length of the film in all the cases is 40 cm, the weight supported in each case is 4.5 × 10-2 N.

Question 19.
What is the pressure inside the drop of mercury of radius 3.00 mm at room temperature? The surface tension of mercury at that temperature (20 °C) is 4.65 × 10-1 N m-1. The atmospheric pressure is 1.01 × 105 Pa. Also, give the excess pressure inside the drop.
r = 3mm = 3 × 10-3m Surface tension of mercury, S = 4.65 × 10-1 Nm-1. Atmospheric pressure, Po = 1.01 × 105 Pa
Total pressure inside the mercury drop = Excess pressure inside mercury + Atmospheric pressure
= $$\frac{2 \mathrm{S}}{\mathrm{r}}$$ + Po
= $$\frac{2 \times 4.65 \times 10^{-1}}{3 \times 10^{-3}}$$ + 1.01 × 105
Total Pressure = 1.013 × 105 Pa
Excess pressure = $$\frac{2 \mathrm{S}}{\mathrm{r}}$$
$$=\frac{2 \times 4.65 \times 10^{-1}}{3 \times 10^{-3}}$$
= 310 Pa.

Question 20.
What is the excess pressure inside a bubble of soap solution of radius 5 mm, given that the surface tension of soap solution at the temperature (20 °C) is 2.50 × 10-2 N/m? If an air bubble of the same dimension were formed at a depth of 40.0 cm inside a container containing the soap solution (of relative density 1.20), what would be the pressure inside the bubble? (1 atmospheric pressure is 1.01 × 105 Pa).
Radius of the soap bubble, r = 5mm
= 5 × 103m
Surface tension of the soap solution,
S = 2.5 × 10-2 N/m
Since the soap Bubble has two surfaces,
Excess pressure, P = $$\frac{4 \mathrm{S}}{\mathrm{r}}$$
$$=\frac{4 \times 2.5 \times 10^{-2}}{5 \times 10^{-3}}$$
P = 20 Pa
Excess pressure inside the soap bubble is 20 Pa
Radius of air bubble, r = 5 mm = 5 × 10-3
Depth, h = 40 c = 0.4 m
Density of soap solution, ρ = 1.2 × 10-3 kg/m2
Atmospheric pressure, Po = 1.01 × 105 Pa
g = 9.8 m/s2
Excess pressure inside air bubble,
P1 = $$\frac{25}{r}$$
⇒ P1 = $$\frac{2 \times 2.5 \times 10^{-2}}{5 \times 10^{-3}}$$
P1 = 10Pa
At a depth of 0.4m, the total pressure inside the air bubble,
PTotal = Po + ρgh + P1
PTotal = 1.01 × 105 + 1.2 × 103 × 9.8 × 0.4 + 10
⇒ PTotal = 1.057 × 105Pa
∴ The pressure inside the air bubble is 1.06 × 105Pa.

Question 21.
A tank with a square base of area 1.0 m2 is divided by a vertical partition in the middle. The bottom of the partition has a small-hinged door of area 20 cm2. The tank is filled with water in one compartment, and an acid (of relative density 1.7) in the other, both to a height of 4 m. compute the force necessary to keep the door closed.
Assume g = 9.8 m/s2
Density of water, $$\rho_{1}$$ = 103 kg/m3
height of water column, h1 = 4m
∴ Pressure due to water, P1 = $$\rho_{1}$$ gh1
= 103 × 9.8 × 4
= 3.92 × 104 Pa
Density of acid
= Relative, densityx density of water = 1.7 × 103 kg/m3
height of acid column, h2 = 4m
∴ Pressure due to acid, P2 = $$\rho_{2}$$ gh2
= 1.7 × 103 × 9.8 × 4
⇒ P2 = 6.664 × 104Pa
Pressure difference between the water and acid columns :
∆ P = P2 – P1
∆ P = 6.664 × 104 – 3.92 × 104
⇒ ∆ P = 2.744 × 104 Pa
Area of the door, a = 20 cm2
= 20 × 10-4m2
∴ Force exerted on the door, F = ∆ P × a
⇒ F = 2.744 × 104 × 20 × 10-4
F = 54.88 N
Therefore, the force necessary to keep the door closed is 54.88 N.

Question 22.
A manometer reads the pressure of a gas in an enclosure as shown in Fig. 10.25 (a) When a pump removes some of the gas, the manometer reads as in Fig. 10.25 (b) The liquid used in the manometers is mercury and the atmospheric pressure is 76 cm of mercury,

1. Give the absolute and gauge pressure of the gas in the enclosure for cases (a) and (b), In units of cm of mercury.
2. How would the levels change in case of (b) if 13.6 cm of water (immiscible with mercury) are poured into the right limb of the manometer? (Ignore the small change in the volume of the gas).

1. Difference between the mercury levels in the two limbs gives gauge pressure
∴ Gauge pressure, Pa = 20 cm of mercury
Atmospheric pressure, Po = 76 cm of mercury
∴ Absolute pressure = Atmospheric pressure + Gauge pressure
= 76 cm + 20 cm
= 96 cm of mercury

2. Difference between the levels of mercury in the two limbs = – 18 cm
∴ Gauge pressure = – 18cm of mercury
∴ Absolute pressure = Atmospheric pressure + Gauge pressure
= 76cm – 18 cm
= 58 cm of mercury.

Question 23.
Two vessels have the same base area but different shapes. The first vessel takes twice the volume of water that the second vessel requires to fill up to a particular common height. Is the force exerted by the water on the base of the vessel the same in the two cases? If so, why do the vessels filled with water to that same height give different readings on a weighing scale?
Two vessels having water-filled up to the same height, exert equal pressure on their bases. Also since the base area is the same, they exert an equal amount of force on the base. (F = PA). Force exerted on the sides of the vessels has non-zero vertical components. When their vertical components are added, the total force experienced by one vessel will be different than the other vessel depending on the shape of the vessels. That is why the two vessels filled with water to the same vertical height show different readings on a weighing machine.

Question 24.
During a blood transfusion, the needle is inserted in a vein where the gauge pressure is 2000 Pa. At what height must the blood container be placed so that blood may just enter the vein?
Gauge pressure, P = 2000 Pa
Density of blood, ρ = 1.06 × 103 kg/m3
g = 9.8 m/s2
Let the height of the blood container be h
∴ pressure of the blood container, P = ρgh
⇒  ρgh = 2000
h = $$\frac{2000}{1.06 \times 10^{3} \times 9.8}$$
⇒ h = 0.1925 m
The blood may just enter the vein if the blood container is kept at a height slightly greater than 0.1925 m.

Question 25.
In deriving Bernoulli’s equation, we equated the work done on the fluid in the tube to its change in the potential and kinetic energy.

1. What is the largest average velocity of blood flow in an artery of diameter 2 × 10-3 m if the flow must remain laminar?
2. How does the pressure change as the fluid moves along the tube if dissipative forces are present?
3. Do the dissipative forces become more important as the fluid velocity Increases? Discuss qualitatively.

1. Diameter of artery = 2 × 10-3m
Viscosity of blood, = 2.084 × 10-3Pa s
Density of blood, ρ = 1.06 × 103kg/m3
Reynolds number for laminar flow,
NR = 2000
The largest average velocity of blood is
$$\mathrm{V}_{\mathrm{avg}}=\frac{\mathrm{N}_{\mathrm{R}} \eta}{\rho \mathrm{d}}$$
$$=\frac{2000 \times 2.084 \times 10^{-3}}{1.06 \times 10^{3} \times 2 \times 10^{-3}}$$
= 1.966 m/S
The largest average velocity of blood is 1.966 m/s.

2. If dissipative forces are present, then some forces in liquid flow due to pressure difference is spent against dissipative forces due to which the pressure drop becomes large.

3. The dissipative forces become more important with increasing flow velocity. This is because of the rise of turbulence. Turbulent flow causes dissipative losses in fluid.

Question 26.

1. What is the largest average velocity of blood flow in an artery of radius 2 × 10-3m if the flow must remain laminar?
2. What is the corresponding flow rate? (Take viscosity of blood to be 2.084 × 10-3 Pas).

1. Diameter of artery, d = 2 × 2 × 10-3 m
= 4 × 10-3 m
Viscosity of blood, h = 2.084 × 10-3 m Pas
Density of blood, ρ = 1.06x 10-3 kg/m3
Reynold’s number for laminar flow,
NR = 2000
The largest average velocity of blood is,
$$\mathrm{V}_{\mathrm{avg}}=\frac{\mathrm{N}_{\mathrm{R}} \eta}{\rho \mathrm{d}}$$
$$V_{\text {avg }}=\frac{2000 \times 2.084 \times 10^{-3}}{1.06 \times 10^{3} \times 2 \times 10^{-3}}$$
V = 0.983 m/s
Hence, the largest average velocity of blood is 0.983 m/s.

2. Flow rate is given by
R= πr2Vavg
R = 3.14 × (2 × 10-3)2 × 0.983
⇒ R = 1.235 × 10-5m3/s
Therefore, the corresponding flow rate is 1.235 × 10-5m3/s

Question 27.
A plane is in level flight at constant speed and each of Its two wings has an area of 25 m2. If the speed of the air is 180 km/h over the lower wing and 234 km/h over the upper wing surface, determine the plane’s mass. (Take air density to be 1 kg m-3).
Total area of wings, A = 2 × 25
= 50m2
speed of air lover the longer wing,
V1 = 180 km/h = 50 m/s
speed of air over the upper wings,
V2 = 243 km/h = 65 m/s
Density of air, ρ =1kg/m3
Let P1 and P2be the pressure of air over the lower wing and upper wing respectively.
From Bimoull’s equation,
P1 + $$\frac{1}{2} \rho V_{1}^{2}$$ = P2 + $$\frac{1}{2} \rho V_{2}^{2}$$
P1 – P2 = $$\frac{1}{2} \rho\left(V_{2}^{2}-V_{1}^{2}\right)$$
∴ pressure difference, ∆ P
= $$\frac{1}{2} \rho\left(V_{2}^{2}-V_{1}^{2}\right)$$
= $$\frac{1}{2}$$ × 1 × (652 – 502)
= 862.5 Pa
The upward force on the plane, F = ( ∆ P) A
= 862.5 × 50
=43125 N
We know that the upward force balances the weight of the plane
∴ F = mg
43125 = m × 9.8
⇒ m = 4400.51 kg.

Question 28.
In Millikan’s oil drop experiment, what is the terminal speed of an uncharged drop of radius 2.0 × 10-5 m and density 1.2 × 103 kg m-3? Take the viscosity of air at the temperature of the experiment to be 1.8 × 10-5 Pa s. How much is the viscous force on the drop at that speed? Neglect buoyancy of the drop due to air.(Take g = 9.8 ms2)
Radius of the given unchanged drop, r = 2 × 10-5 m
Density of the uncharged drop,
ρ = 1.2 × 103 kg/m-3
Viscosity of air, $$\eta$$ = 1.8 × 10-5 Pa S
The density of air ($$\rho_{0}$$) can be taken as zero in order to neglect the buoyancy of air.
Terminal velocity (v) is given by the relation:

V = 5.807 × 10-2 m/s
V = 5.8 cm/s
The terminal speed of the drop is 5.8 cm/s
Viscous force, F = π 6 hrv
∴ F = 6 × 3.14 × 1.8 × 10-5 × 2 × 105 × 5.8 × 102
⇒ F = 3.9 × 10-10N
The viscous force on the drop is
= 3.9 × 10-10 N

Question 29.
Mercury has an angle of contact equal to 140° with soda-lime glass. A narrow tube of radius 1.00 mm made of this glass is dipped in a trough containing mercury. By what amount does the mercury dip down in the tube relative to the liquid surface outside? The surface tension of mercury at the temperature of the experiment is 0.465 N m. Density of mercury = 13.6 × 103 kg m-3.
Angle of contact 1 θ = 140°
Radius of the tube, r = 1 mm = 10-3m
Surface tepsion of mercury S = 0.465 N/m
Density of mercury, ρ = 13.6 × 103 kg/m3
g = 9.8 m/s2
Let the Dip in height of mercury be h Surface tension is given by,
S= $$\rho_{o} e^{-y / y_{o}}$$
⇒ h = $$\frac{2 \mathrm{S} \cos \theta}{\rho \mathrm{gr}}$$
⇒ h = $$\frac{2 \times 0.465 \times \cos 140^{\circ}}{13.6 \times 10^{3} \times 9.8 \times 10^{-3}}$$
h = – 0.00534 m = -5.34 mm

Question 30.
Two narrow bores of diameters 3.0 mm and 6.0 mm are Joined together to form a U-tube open at both ends. If the U-tube contains water, what Is the difference In its levels in the two limbs of the tube? The surface tension of water at the temperature of the experiment is 7.3 × 10-2 N/m. Take the angle of contact to be zero and the density of water to be 1.0 × 103 kg m-3 (g= 9.8 m s-2).
Radius of the first bore, r1 = $$\frac{3 \mathrm{mm}}{2}$$ = 1.5 × 10-3 m
Radius of the second bore, r2 = $$\frac{6 \mathrm{mm}}{2}$$ = 3 × 10-3 m
Surface tension of water, S = 7.3 × 10-2 N/m
Angle of contact, θ = 0°
(give in the question) Density of water, ρ = 1 × 103 kg/m3
Let h1 and h2 he the heights to which the water rises in the 3 mm and 6 mm diameter bores respectively.We have,
h1 = $$\frac{2 \mathrm{S} \cos \theta}{\mathrm{r}_{1} \rho \mathrm{g}}$$
h2 = $$\frac{2 \mathrm{S} \cos \dot{\theta}}{\mathrm{r}_{2} \rho \mathrm{g}}$$
Difference in water level = h1 – h2

= 4.966 × 10-3m
= 4.97 mm
The difference in the water levels in the cores is 4.97 mm.

Question 31.
1. It is known that density P of air decreases with height y as$$\rho_{\mathrm{o}} \mathrm{e}^{-y / \mathrm{y}_{0}}$$ where $$\rho_{0}$$ = 1.25 kg/m-3 is the density at sea level, and yo is a constant. This density variation is called the law of atmospheres. Obtain this law assuming that the temperature of the atmosphere remains constant (isothermal conditions). Also, assume that the value of g remains constant.
2. A large He balloon of volume 1425 m3 is used to lift a payload of 400 kg. Assume that the balloon maintains a constant radius as It rises. How high does it rise?
[Take y0 = 8000m and $$\rho_{\mathrm{He}}$$ = 0.18kg/m3].
1. Volume of the balloon, V = 1425m3
Payload mass, m = 400 kg
g = 9.8 m/s2
Give that, yo = 8000m
$$\rho_{\mathrm{He}}$$ =0.18 kg/m3
$$\rho_{\mathrm{o}}$$ = 1.25 kg/m3
Density of the balloon = ρ
Let the height to which the balloon rises be y Density (ρ) of air decreases with height (y) as,

we can infer from (1), that rate of decrease of density with height is directly proportional to ρ,i.e.,

$$\frac{\mathrm{d} \rho}{\rho}=-\mathrm{kdy}$$
where, K is the constant of proportionality Height changes from o to y, while density changes from $$\rho_{\mathrm{o}}$$
to ρ by integrating.
we get:

from (1) and (2),
yo = 1/K
⇒ K = $$\frac{1}{y_{0}}$$
From (2) and (3) we get,

2. Density,

⇒ ρ = 0.46 kg/m3
we have proved that,

⇒ y = – 8000 × loge $$\frac{0.46}{1.25}$$
⇒ y = 8000m = 8km .
Hence balloon will rise to a height of 8 km.

### 1st PUC Physics Mechanical Properties Of Fluids One Mark Questions and Answers

Question 1.
Pascal’s law is valid for dynamic fluids. True/False.
False. Valid only for static fluid.

Question 2.
What is the relation between absolute pressure and gauge pressure?
Absolute pressure = atmospheric pressure ± gauge pressure.

Question 3.
Angle of contact for a glass tube dipped in mercury is obtuse. True / False.
True.

Question 4.
Water is an ideal fluid. True/False.
False. It is real fluid.

Question 5.
Define thrust.
The total normal force exerted by the fluid at rest on a surface is called thrust.

Question 6.
What is the dimensional formula for pressure?
ML-1T-2

Question 7.
What happens to the viscosity of a gas when the temperature is increased?
Viscosity of gases increases as the temperature is increased.

Question 8.
How does the coefficient of viscosity of liquid vary with temperature?
The coefficient of viscosity of liquids decreases as temperature is increased.

### 1st PUC Physics Mechanical Properties Of Fluids Two Marks Questions and Answers

Question 1.
Name two applications of Pascal’s law.
Hydraulic lifts, hydraulic breaks.

Question 2.
What is streamline? What are its properties?
The path taken by a fluid particle under a steady flow is a streamline.
Properties:

1. The tangent at any point in streamline gives direction of fluid velocity at that point.
2. No two streamlines can cross.

Question 3.
Define the specific gravity of a fluid. What is its unit?
Specific gravity is defined as a ratio of a density of a fluid to the density of water at 4°C. It has no unit.

Question 4.
Which are the factors on which viscous force depends?
Viscous force acting between two layers of liquids depends on,

1. area of the layers
2. relative velocity of the two layers and
3. distance between two layers.

Question 5.
A force 6f 1 kN is applied on a road surface by the tyre of a car. If the contact area of the tyre is 10 cm2, find the pressure on the contact area.
Pressure = $$\frac{\text { Force }}{\text { area }}$$
$$\frac{1000 N}{10 \times 10^{-4} M^{2}}$$ = 106 Pa

Question 6.
State whether true or false :

1. An ideal liquid has non zero viscosity but zero compressibility.
2. With increase in temperature, the viscosity of liquid decreases but the viscosity of gases increases.

1. False
2. True.

Question 7.
A bigger raindrop falls faster than a smaller one (True/False) Explain.
True.
The raindrop moves with terminal velocity due to viscous drag of the air. The terminal velocity of drop varies as the square of its radius. Hence, a bigger drop has a higher terminal velocity than a smaller one.

Question 8.
Machine parts get jammed during winter. Why?
In winter, due to low temperature, the viscosity of oil between the machine parts increases considerably resulting in jamming of machine parts.

Question 9.
Discuss the effect of temperature on the viscosity of gases and liquids.
Viscosity of liquid decreases with increase in temperature and viscosity of gases increases with increase in temperature.

Question 10.
What is an ideal liquid?

1. An ideal liquid is incompressible meaning density of liquid remains irrespective of the pressure.
2. An ideal liquid is non-viscous. No tangential forces between layers of liquid in relative motion.
3. Ideal liquid cannot withstand any shearing stress.

Question 11.
Explain the effect of density and pressure on the viscosity of liquids/gases.
Liquids:
1. Viscosity increases with increase in density of the liquid. Viscosity increases with increase in pressure (except water).

2. Gases:
Viscosity decreases with increase in density. Viscosity decreases with increase in pressure.

Question 12.
The excess pressure inside a soap bubble is four times the excess pressure inside a moving soap bubble. What is the ratio of the first and second bubble?
Let r1 and r2 be radius of soap bubbles. Excess pressure in them are
$$\frac{4 \sigma}{r_{1}}$$ and $$\frac{4 \sigma}{r_{2}}$$ where σ is surface tension.
Given P1 = 4P2 $$\frac{4 \sigma}{r_{1}}$$ = 4 $$\frac{4 \sigma}{r_{2}}$$
ratio of volume of I to II
$$=\frac{r_{1}^{3}}{r_{2}^{3}}=\frac{r_{1}^{3}}{64 r_{1}^{3}}=\frac{1}{64}$$

Question 13.
Explain why water with detergent dissolved in it should have small angles of contact.
Cloth has narrow spaces inform of capillaries. The rise of liquid in a capillary tube is directly proportional to cos e.
θ ∝ $$\frac{1}{\cos \theta}$$. Hence e should be less in order that detergent penetrates more in clothes.

Question 14.
What is a venturi meter? Name two applications.
Venturi meter is a device to measure the flow speed of incompressible fluid. The principle is used in filter pumps, Bunsen burner, atomiser.

Question 15.
Water is filled as shown In following containers. In which case highest force is exerted on the base of the container? Why?

Water exerts same pressure at the bottom of the container, regardless of its shape. Since height, his constant, equal pressure is exerted.
P = ρgh (h = constant)

Question 16.
Is it better to wash clothes in hot soap solution? Why?
Yes. Surface tension decreases with temperature. Hence, in hot soap solution, spreading of solution over the surface of clothes happens easily. Hot soap solution can penetrate and clean clothes better.

Question 17.
Explain why a drop of liquid under no external force is always spherical in shape.
Due to surface tension, a drop of liquid tries to acquire a shape with minimum surface area. For a given volume, surface area of sphere is least. Hence, liquid drop takes a spherical shape.

Question 18.
Point out any two properties required for a fluid to be used in a barometer.

1. Density of fluid should be high so as to avoid use of long barometer,
2. Temperature variation of fluid should be minirial.

Question 19.
Why does water flow faster that honey?
The coefficient of viscosity of water is less compared to that of honey. This makes water to move faster.

Question 20.
What are the factors affecting viscosity?

1. Increase in temperature decreases viscosity.
2. Increases in pressure increases viscosity in liquids. In water, it decreases whereas, in gases, it remains the same.

Question 21.
What do you mean by angle of contact of a liquid with a solid surface. What are the factors that effect it?
The angle made by the tangent drawn to the meniscus from the point of contact with the walls of the container measured from within the liquid is called angle of contact. It depends on the atmosphere pressure, adhesive and cohesive forces of the liquid in the tube.

Question 22.
Bernoulli’s equation can be used to describe the flow of a river rapidly. True/False. Explain why?
False. Bernoulli’s equation can be applied only to streamline flow.

Question 23.
Archimedes’s Principle holds good in a vessel at free fall. True/False. Give reason.
False
The principle does not hold good in this particular case as the vessel in free fall is in a condition of weightlessness, where the buoyant force accounting for the Archimedes’s principle does not exist.

Question 24.
Why are cars and aeroplanes streamlined?
The shape of the aeroplane is streamlined because when it flies in air. The velocity on the top surface is more than the bottom surface. Hence the pressure on the top surface decreases. This causes an upward thrust on the wings of the place which gives uplift to the aeroplane.

Question 25
Why do some liquids rise and some liquids dip in a capillary tube? Explain
The rise or fall of liquid in a capillary tube is due to surface tension. If the liquid wets the glass then there is rise in the liquid inside the capillary tube.

Question 26.
Define cohesive force and adhesive, force of molecules.
Force of attraction between molecules of the same substance is called cohesive force. Force of attraction between molecules of the different substances is called adhesive force.

Question 27.
Diameter of a ball A is thrice that of B. What will be the ratio of their terminal velocities in water?
Terminal velocity × (rad. of ball)2 Required ratio is $$\frac{3^{2}}{1}$$ = 9

Question 28.
A hole of area 2 mm opens near bottom of a large water storage tank and stream of water shoots from it. If top of the water in the tank is to be kept 30 m above leaking point, how much water in litres Is should be added to reservoir tank to keep this level? Take g = 10 m/s2
Velocity of the outflowing water
v = $$\sqrt{2 g h}=\sqrt{2 \times 10 \times 30}$$ = 24.49 m/s
Quantity of water flowing out per sec.
= av = 10-5 × 4 × 24.49
= 97.96 × 10-6
= 9.796 × 10-5 × 103 litres/s
= 97.96 ml/s.

Question 29.
A small ball of mass m and density ρ is dropped in a viscous liquid having density $$\rho_{0}$$. After some time, ball falls with a constant velocity. Calculate viscous force acting on the ball.
Volume of the ball, v = m/ρ; mass of liquid displaced by the ball.
$$\mathrm{m}^{\prime}=\left(\frac{\mathrm{m}}{\rho}\right) \rho_{0}$$
ρ = density of ball
When the ball falls with a constant velocity
⇒ viscous force = effective weight of the ball or F = mg = m’g = (m – m’)g
$$=\left(\mathrm{m}-\frac{\mathrm{m} \rho_{0}}{\rho}\right) \mathrm{g}=\mathrm{mg}\left(1-\frac{\rho_{0}}{\rho}\right)$$

Question 30.
What is Reynold’s number? What is its significance?
Turbulent flow is less likely for viscous fluid flowing at low rates. A dimensionless number, whose value gives one an approximate idea whether the flow would be turbulent. This number is called the Reynold’s Re.
$$R_{e}=\rho v d / \eta$$
ρ → density of fluid
speed v,d stands for the dimension of the pipe.
$$\eta$$ → viscosity of the fluid. Significance is that,
flow is lamina’s/streamline for Re <1000 flow is turbulent for Re > 2000.

Question 31.
A wooden log 100 mm × 100 mm × 5 m hangs vertically from a vertical string so that 3 m length of log Is submerged in water. Find the tension in string. Take specific gravity of wood = 0.65.

FB + T = W
(0.1 × 0.1 × 3) × 9810 + T
= (0.1 × 0.1 × 5) × 0.65 × 9810
T = 24.52 N
(0.1 × 0.1 × 3) × 9810 + T
= (0.1 × 0.1 × 5) × 0.65 × 9810
∴ T = 24.52 N

Question 32.
A piston of small cross-section A1 is used to exert a force F1 on liquid to transfer it from the existing cylinder to a larger cylinder attached with a larger piston of area A2. What is the weight, say if a truck that can be placed on a platform supported by the larger piston due to the force exerted on the liquid in existing cylinder?
F2 = $$\frac{\mathrm{F}_{1} \mathrm{A}_{2}}{\mathrm{A}_{1}}=\frac{100 \mathrm{kN} \times 1 \mathrm{m}^{2}}{10 \times 10^{-4} \mathrm{m}^{2}}$$
= 105 kN
∴ Weight of truck = 105 kN.

Question 33.
The pressure inside a droplet of water is 10.3 N/m2 in excess of atmospheric pressure. Its diameter is 50 mm. Find surface tension of the water film.
P = $$\frac{2 \lambda}{r}$$
λ = $$\frac{P r}{2}=\frac{10.3 \times 25 \times 10^{3}}{2}$$
= 128.75 N/m

### 1st PUC Physics Mechanical Properties Of Fluids Four/Five Marks Questions and Answers

Question 1.
Define viscosity. Discuss the cause the viscosity.
Viscosity is the property of a fluid, (liquid or gas) by virtue of which an internal frictional force comes into play when the fluid is in motion and opposes the relative motion of its different layers. Viscosity is due to the intermolecular forces which are effective when the different layers of the liquid are moving with different velocities.

These forces are of Vander waal type. Due to these forces, every fast-moving liquid layers tends to accelerate the adjoining slow-moving layers and every slow-moving layer tends to retard the adjoining fast-moving layer of liquid. As a result a backward dragging force called viscous drag comes into play which accounts for viscosity of liquid.

Question 2.
Derive the dimensional formula for co-efficient of viscosity and hence its unit.

Units:
We have $$\eta=\frac{F}{A d v / d n}$$
in cgs system, the unit of $$\eta$$ is called poise.
1 poise $$=\frac{1 \mathrm{dyne}}{1 \mathrm{cm}^{2} \times(1(\mathrm{cm} / \mathrm{s}) / \mathrm{cm})}$$
= dyne cm-2 sec.
The S.I.unit of $$\eta$$ is Pa.s (Pascal second) or deca poise.
1 dec poise = $$\frac{1 \mathrm{N}}{1 \mathrm{m}^{2}\left(1 \mathrm{ms}^{-1} / \mathrm{m}\right)}$$ = 1Nsm-2

Question 3.
State Stoke’s law? What are the factors on which viscous drag depends?
Stoke’s law states the backward dragging force (F) acting on a small spherical body of radius r, moving through a medium having coefficient of viscosity $$\eta$$ and velocity v is given by
F = 6π$$\eta$$rv
viscous drag (F) depends on

1. coefficient of viscosity of the medium $$\eta$$
2. velocity of the body (v)
3. radius of the spherical body (r)

Question 4.
How is the rise of liquid affected if top of the capillary is closed?
There will be a small rise in the capillary tube if top of capillary tube is closed. Because the rise of liquid in capillary tube due to surface tension will be opposed by the downward force exerted by the compressed air above the liquid in tube. This downward force increases with increase in height of liquid column. Therefore only a small rise of liquid column is possible in a capillary tube with a closed top.

Question 5.
What is terminal velocity? What are the factors on which terminal velocity depends?
The terminal velocity of an object is the maximum constant velocity acquired by the object while falling freely in a viscous medium.
Terminal velocity (v) of an object of radius (r) density $$(\rho)$$ moving through a viscous medius of viscosity $$\eta$$ and density $$\rho_{0}$$ is given by
$$\mathrm{v}=\frac{2}{9} \frac{\mathrm{r}^{2}}{\eta}\left(\rho-\rho_{0}\right) \mathrm{g}$$
Hence, terminal velocity depends on

2. coefficient of viscosity of the medium
3. density of the object.
4. density of the medium

Question 6.
Derive expression for capillary rise in a tube in terms of surface tension, radius of tube angle of contact and density of fluid.
Consider a tube of radius ‘a’ dipped in a container filled with fluid of density of ρ as shown.

h = capillary rise
s = surface tension
θ = angle of contact
The weight of column of liquid of height h in the tube is balanced by vertical component of the surface tensile force of fluid. Vertical component of the surface tensile force
= (S cos θ) × (length over which it acts)
= (S cos θ) × circumference of tube
= S cos θ × 2πa ……..(1)
Weight of column of fluid balanced by surface tension = volume of fluid × density × g
= (πa2h) × ρg ……(2)
Equating (1)and(2),
S cosθ × 2πa = πa2h ρg
h $$=\frac{S \cos \theta \times 2}{a \rho g}$$
$$=\frac{2 \mathrm{S} \cos \theta}{\mathrm{a} \rho \mathrm{g}}$$

Question 7.
State and prove the Archimedes principle?
Archimedes principle states that when a body is wholly partially immersed in a liquid at rest. The loss of weight of the body in the liquid is equal to the weight of the liquid displaced by the immersed part of the body.
Proof:

Let A be the cross-sectional area of the top or bottom face of the object.
Volume of liquid displaced, v = volume of the object = Ah
∴ mass of liquid displaced, m = vρ = Ahρ
ρ is the density of the liquid.
Liquid pressure on the top of the body
P1 = xρg
Vertical downward thrust (Force f1) on top face of the object.
F1 = P1A = xgAρ
Liquid pressure on the bottom face of body P2 = (x + h) ρg
Vertical upward thrust (Force F2) on bottom force
F2 = P2 A(X + h) ρgA.
Since F2 >F1, the net upward force acting is F = F2 – F = (x + h) ρ gA – x ρgA
= hρgA = hA(ρg)
= mg
weight of liquid displaced.
True weight of the object = Mg
Upward thrust (Force F) on the body F = mg
Apparent weight of the body in liquid.
= W-F = Mg – mg
This means that observed weight at the body (object) immersed in a liquid is less than its true weight by an amount equal to weight of the liquid displaced by body.

Question 8.
State and prove Bernoulli’s theorem.
Bernoulli theorem states that for streamlined flow of an ideal liquid, the total energy (Pressure energy + potential energy + kinetic energy) per unit mass remains constant at every cross-section throughout the flow.
Proof:
Consider a tube AB of varying cross-section through which an ideal liquid is in streamlined flow.

Let P1 be the pressure applied on the liquid at A and P2 be the pressure against which liquid has to move out at B
a1, a2 be the cross-sectional area at tube at A and B
h1 and h2 are the mean height at section A and B from the reference level.
v1 and v2 are the velocity of liquid flow at A and B
ρ is the density of the ideal liquid flowing through the tube.
The liquid flows from A to B hence,
P1 > P2
The mass m of the liquid crossing per second through any section at the tube with respect to equation of continuity is,
a1v1 ρ = a2v2 = ρm
a1v1 = a2v2 = V (say)
Force on liquid at A = P1a1
Force on liquid at B = P2a2
Work done per second by liquid at A = P1a1 × v1 = P1V
Work done per second by liquid at B = P2a2 × v2 = P2V
Net work done per second on liquid by pressure energy moving from A to B is
= P1V- P2V = v (P1-P2)
Increase in potential energy per second as the liquid flows from A to B
= mgh2 – mgh,
(since v2 is greater than v1)
Increase in kinetic energy per second of liquid from A to B = 1/2 m$$v_{2}^{2}$$ – 1/2 m$$v_{1}^{2}$$
According to work – energy principle, work done per second by pressure energy
= increase in P.E per second + increase in K.E per second
or P1v – P2v = (mgh2 – mgh1) + $$\left(\frac{1}{2} m v_{2}^{2}-\frac{1}{2} m v_{1}^{2}\right)$$
i.e., P1v + mgh1 + $$\frac{1}{2} m v_{1}^{2}$$
= P2v + mgh2 + $$\frac{1}{2} m v_{2}^{2}$$
Dividing by m1

Hence, $$\frac{\mathrm{P}}{\rho}+\mathrm{gh}+\frac{1}{2} \mathrm{v}^{2}$$ = constant.
But, $$\frac{P}{\rho}$$ is the pressure energy per unit mass, gh is the potential energy per unit mass and $$\frac{1}{2} v^{2}$$ is the kinetic energy
per unit mass Hence,
Pressure Energy + Potential Energy + Kinetic Energy remains constant.
Hence proved.

Question 9.
Derive the equation at continuity.
Consider a non-viscous liquid in streamlined flow through a tube AB at varying cross-section.

Let a1, a2 be the area of cross-section of the tube at A and B.
v1, v2 = velocity of flow of liquid at A and B.
Volume of the liuid entering per second at A = a1v1
Mass of liquid entering per second at A = a1v1$$\rho_{1}$$
Similarly mass of liquid leaving per second at B = a2v2 $$\rho_{2}$$
Assuming, no loss of liquid in the tube and
steady flow, then mass of liquid entering at A/sec = mass of liquid leaving at B/sec.
⇒ a1v1$$\rho_{1}$$ = a2v2$$\rho_{2}$$
Assuming liquid is incompressible,
$$\rho_{1}$$ = $$\rho_{2}$$
⇒ a1v1 = a2v2
av = constant
This is the equation of continuity.

Question 10.
Explain the limitations of Bernoulli theorem.

1. While deriving the Bernoulli theorem, it is assumed that the velocity of every particle of liquid across any crosssection is uniform. Practically this is not possible.
2. Viscous drag of the liquid which comes into play when the liquid is in motion has not been considered.
3. While deriving the equation, it is assumed that there is no loss of energy when liquid is in motion. In fact, some KE is converted to heat energy.
4. If a liquid is flowing along a curved path, energy due to centrifugal force should also be considered.

Question 11.
Explain why the angle of contact of water with glass is acute while that mercury with glass is obtuse?

SSL + SLA cos θ = SSA
or cos θ = $$\frac{S_{S A}-S_{S L}}{S_{L A}}$$
In case of mercury – glass, SSA < SSL
∴ cos θ is negative or θ>90° i.e., obtuse.
In case of water – glass, SSA > SSL
cos θ is positive or θ<90° i.e., acute.
Note:
SSL:
Surface tension corresponding to solid-liquid interface.
SLA:
Surface tension corresponding to liquid – air interface.
SSA:
Surface tension corresponding to solid-air interface.

Question 12.
Derive an expression for velocity of fluid at wide neck in a venturi meter.
Speed of fluid flowing through tube at broad neck = V1
Speed of fluid flowing through tube at narrow neck = V2
By equation of continuity,
A1V1 = A2V2
V2 = $$\frac{A_{1} V_{1}}{A_{2}}=\frac{A V_{1}}{a}$$
From Bernoulli’s equation,

Question 13.
What is the Magnus effect? Explain.
A ball Which is spinning drags air along with it. If the surface is rough, more air will be dragged.

Ball is moving forward and relative to it air is moving backwards. This results in a larger velocity of air relative to the ball above it and a smaller velocity below it. This difference in velocities of air results in pressure difference between lower and upper faces and there is a net upward force on the ball. This dynamic lift due to spinning is called the Magnus effect.

Question 14.
Define terminal velocity. Give reasons why a sphere attains this velocity. Derive a relation for the terminal velocity.
The constant velocity with which a body drops down after initial acceleration in a dense liquid or fluid is called terminal velocity. This is attained when the apparent weight is compensated by the viscous force. It is given by
v = $$\frac{2}{9} \frac{r^{2} g}{\eta}(\rho-\sigma)$$, where ρ and σ are densities of the body and liquid respectively, $$\eta$$ is the coefficient of viscosity of liquid and r is the radius of the spherical body.
The net force on the sphere becomes zero as the viscous force equals the apparent weight.
Consider a long column of dense liquid-like glycerine. As the ball is dropped in it, the forces experienced are;

1. weight = mg = $$\frac{4}{3} \pi r^{3} \rho g$$, where ρ is the density of the ball.
2. upthrust = U = $$\frac{4}{3} \pi r^{3} \rho_{1} g$$, where$$\rho_{1}$$ is the density of the liquid.
3. viscous force Fv = $$6 \pi \eta \rho v$$, where v is the terminal velocity.

Net force and the acceleration should be 0.
∴ mg – U – Fv = 0

Question 15.
Write the differences between laminas/streamline flow and turbulent flow.
1. Laminar Flow

• Layered flow with one layer of fluid sliding smoothly over the other.
• Reynold’s No. < 1000
• Observed usually in flow through porous materials.

2. Turbulent

• Haphazard flow with rapid and continuous mixing of fluid resulting in momentum transfer.
• Reynold’s No. > 2000
• Pipeline flow, river flow, etc.

Question 16.
A razor blade can be made to float on water. What forces act on this blade? Is Archimedes’s principle applicable?
Three forces are acting on the blade when it is made to float on water:

1. Weight of blade acting vertically downwards.
2. Reaction in blade exerted by the liquid surface acting vertically upwards.
3. Force of surface tension on circumference, of the blade acting tangentially to the liquid surface.
In this case, as no portion of razor blade is immersed in water, hence Archimedes’s principle is not applicable.

1st PUC Physics Mechanical Properties Of Fluids Numerical Problems Questions and Answers

Question 1.
A capillary tube of diameter 1.5 mm is dipped in

1. mercury.
2. water.

Find capillary rise for each case. Surface tension for water and mercury can be taken as 0.07 N/m and 0.52 N/m respectively. The contact angle for water and mercury can be taken as 0° and 130°.
1. h = $$\frac{4 \lambda \cos \theta}{\mathrm{dr}}$$
$$=\frac{4 \times 0.52 \times \cos (130)}{1.5 \times 10^{-3} \times 13.65 \times 9810}$$
= 6.66 mm (fall)

2. h = $$\frac{4 \lambda \cos \theta}{\mathrm{dr}}$$
$$=\frac{4 \times 0.07 \times 1}{1.5 \times 10^{-3} \times 9810}$$
= 19 mm (rise).

Question 2.
Calculate the energy evolved when 8 droplets of water (surface tension 0.72 N/m) of radius 0.5 mm combine into one.
Here, s = 0.072 N/m
r = 0.5 mm = 0.5 × 10-3 m.
Let R be the radius of big drop formed.
Volume of the big drop = volume of 8 small drops.
$$\frac{4}{3}$$ π R3 = 8 × $$\frac{4}{3}$$ πr3
or R = 2 r = 2 × 0.5 × 10-3= 10-3 m
Surface area of big drop :
= 4π R2 = 4π × (10-3)2
= 4π × 10-6m2
Surface area of 8 small drops:
= 8 × 4π × r2
= 8 × 4π × (0.5 × 10-3)2
= 8π × 10-6m2
Energy evolved = S.T × decrease in area.
= 0.072 × 4π × 10-6 = 9.05 × 107J.

Question 3.
A plate 0.025 mm distant from a fixed plate moves at 60 cm/s and requires a force of 2 N per meter square to maintain its speed. Determine fluid viscosity between the plates.
dy = 0.025 mm
= 0.025 × 10-3m
Velocity of upper plate .
= 60 cm/s = 0.6 m/s
Force on upper plate
= 2$$\frac{\mathrm{N}}{\mathrm{m}^{2}}$$ = Shear stress , τ
du = change of velocity = u – 0 = 0.6 m/s
dy = change of distance = 0.025 × 10-3m
Viscosity,
µ = $$\frac{\tau}{\mathrm{du} / \mathrm{dy}}=\frac{2}{\left(0.6 / 0.25 \times 10^{-3}\right)}$$
8.33 × 10-5 $$\frac{\mathrm{Ns}}{\mathrm{m}^{2}}$$ = 8.33 × 10-5 Pas

Question 4.

In the above figure, if area of plate is A, viscosity of fluid is µ, distance of plate from top and bottom plane surfaces are y1 and y2 respectively, find the expression for force F required to drag the plate at a velocity v.

Required force F = F1 + F2
$$=\frac{\mu_{1} A_{1} v_{1}}{y_{1}}+\frac{\mu_{2} A_{2} v_{2}}{y_{2}}$$
$$=\mu \mathrm{Av}\left(\frac{1}{\mathrm{y}_{1}}+\frac{1}{\mathrm{y}_{2}}\right)$$

Question 5.
An open tank contains water up to a depth of 3m and above it an oil of sp.gr 0.9 for a depth of 1m. Find pressure intensity

1. at the interface of two liquids
2. at the bottom of the tank.

Height of water, z1 = 3 m
Height of oil, z2 = 1 m
Density of oil
= Sp.gr of oil × Density of water
= 0.9 × 1000 kg/m3 = 9000 kg/m3
= $$\rho_{2}$$
Density of water = $$\rho_{1}$$ = 1000 kg/m3

1. At the interface, i.e., at A,
P= $$\rho_{2}$$ g × z2 = 900 × 9.81 × 1
= 8829 N/m2

2. At the bottom i.e., at B,
p =$$\rho_{2}$$gz2 + $$\rho_{1}$$gz1
= 900 × 9.81 × 1 + 1000 × 9.81 × 3
= 8829 + 29430 = 38259 N/m2

Question 6.
Determine pressure difference between two points A and B for the setup as shown in figure.

G = specific gravity,
PA = pressure at A,
PB = Pressure at B
Pressure at any point is the same horizontal line x – x should be same.
Hence, pressure at C in left column = pressure at D in right column.
Pressure at C
= (2 × 1000 × g × 2) + (1 × 1000 × g × 4) + PA …..(1)
pressure at D
= (4 × 1000 × g × 2) + PB ……(2)
Equating (1) and (2),
8000 g + PA = 8000 g + PB
PA – PB = 0

Question 7.
A closed tank contains 1m of mercury, 2m of water and 3 m of oil of sp.gravity 0.6. There is an unknown fluid in space above oil. If gauge pressure at the bottom of the tank is 200 kPa. What is the pressure on the top surface of oil?

Let pressure on top of oil be PA
Total pressure = PA + Poil +Pwater + Pmercury
P= ρgh
200 × 103 =PA + 0.6 × 1000 × g × 3 + 1000 × g × 2 + 13.6 × 1000 × g × 1
PA = 200 × 103 – 170694
= 29.306 kPa

Question 8.
The flow rate of water from a tap of diameter 1.2 cm is 0.48 L/min. The coefficient of viscosity of H20 is 10-3 Pa s. After some time the flow rate is increased to 4 L/min. Characterise the flow for both the flow rates.
Let the speed of the flow be v and the area of tap be d = 1.2 cm. The volume of the water flowing out per second is
Q = v × πd2 /4
v = 4Q/d2
Reynold’s number is
Re = 4P2 / πd$$\eta$$
= 4 × 103kg/m3× Q/(3.14 × 1.2 × 10-2m × 10-1Pa s)
= 1.061 × 108 m-3s   Q = 1.061 × 108Q
Initially,
Q = 0.48 L/min = 8.0 cm3/s
= 8.00 × 10-6m3/s
we obtain
R =484.8
Since this is below 1000, the flow is steady.
After sometime when Q = 4L/min,
= 66.67 cm3/s
= 6.67 × 10-5 m3s-1
we obtain
R = 1.061 × 108 × 6.67 × 10-5
= 7076.9
The flow now is turbulent.

Question 9.
Calculate the total energy possessed by one kg of water at a point where pressure is 30 gm wt/sq.mm. Velocity of 0.1 ms-1 and height is 60 cm above the ground.
Given, P = 30 g wt/sq.mm
= $$\frac{30}{1000}$$ × 9.8 × 106N/m2
v = 0.1ms-1 h = 0.6 m
Total energy per unit mass
$$\frac{P}{\rho}$$ + gh + 1/2 v2
= $$\frac{30 \times 9.8 \times 10^{3}}{10^{3}}$$ + 9.8 × 0.6 + 1/2 × (0.1)2
= 294 + 5.88 + 0.005
= 299.885 J.

Question 10.
Water flows at the rate of 4 litres per second through an orifice at the bottom of the tank which contains water 720 cm deep. Find the rate of escape of water If additional pressure of 20 kg/cm2 is applied to the surface of water.
Given, h = 720 cm v = $$\sqrt{2 g h}$$
v = $$\sqrt{2 \times 980 \times 720}$$ cm/s ;
= 1187.93 cm/s v
v = 4 litres
= 20000 g/cm2
$$\frac{20000 \times 980}{980}$$ Cm 0f Water
= 20,000 cm of water column.
Now, pressure head (h1) = 20,000 + 720
= 20,720 cm
New velocity v1 = $$\sqrt{2 \mathrm{gh}_{1}}$$
$$\sqrt{2 \times 980 \times 20,720}$$ cm/s
= 6372.69 cm/s
As v = av and V1 = av1
v1 = V$$\frac{v_{1}}{v}$$ = 4 × $$\sqrt{\frac{20720}{720}}$$
= 21.45 litres/s

Question 11.
A venturi meter is connected to two points in the mains where its radii are 20 cm and 10 cm. and the levels of water column in the tubes differ by 10 cms. How much water flows through the pipe per minute?
Volume of water flowing per second
v = a1a2$$\sqrt{\frac{2 g h}{a_{1}^{2}-a_{2}^{2}}}$$
where a1 = $$\pi r_{1}^{2}$$ = 22/2 × 2012cm2
a2 = $$\pi r_{2}^{2}$$ 22/7 × 1012cm2
g = 980 cm/s2;
h = 10 cm .
Volume of water flowing per minute

= 2726.58 litres/minute.

Question 12.
When a drop of mercury of radius R is split into n similar drops, What is the change in surface energy? Assume σ as surface tension of mercury.
Volume of initial mercury drop = $$\frac{4}{3}$$ πR3
r = radius of smaller drops,
volume conservation
⇒ $$\frac{4}{3}$$ πR3 = n $$\frac{4}{3}$$ πr3
r = Rn-1/3
Initial Surface Energy = σ × surface area
= σ × 4πR2
Final Surface Energy = n × σ × 4πr2
= nσ 4π[Rn-1/3]2
= nσ 4πR2n-2/3
= n1/34σ πR2
Change in surface energy
= 4σ πR2[1 – n1/3]

Question 13.

1. The diameter of pipe at section (1) and (2) are 10 and 15 cm respectively. Find discharge through pipe if the velocity of water flowing through pipe at section (1) is 5 m/s. Find also velocity at section (2)
2. Find the work done in blowing a soup bubble of surface tension 0.06 N/m from a 2 cm radius to a 5 cm radius.

1.

Q = A1V1 = π × $$\frac{0.1^{2}}{4}$$ × 5
= 0.0393 m3/s
V2 = $$\frac{A_{1} V_{1}}{A_{2}}$$ (continuity equation)
= $$\frac{0.0393}{\frac{\pi}{4} \times 0.15^{2}}$$ = 2.22 m/s

2. Here, s = 0.06 N/m
r1 = 2 cm = 0.02 m ;
r2 = 5 cm = 0.05 m
Since bubble has 2 surfaces, initial surface area of the bubble.
= 2 × 4πr12= 2 × 4π(0.02)2
= 32π × 10-4m2
Final surface area of the bubble
= 2 × 4πr22 = 2 × 4π(0.05)2
= 200π × 10-4m2
Increase in surface area
= 200π × 10-4 – 32π × 10-4
= 168π × 10-4m2
∴ work done = S × increase in surface area
= 0.06 × 168π × 10-4 = 0.003168 J

Question 14.
A tank is filled with water to a height H. At a depth ‘h’ from the free surface a hole is made so that the water comes out of it. What is the velocity of efflux? Also, what is the maximum range and the position of hole for the same? Find the time taken by a water molecule to reach the ground. Determine the horizontal length covered by the molecule.

Take two points at the same height (H- h) from ground one inside and one outside the hole. Applying Bernoulli’s theorem for the points, we have
$$\frac{P_{0}+h p g}{\rho g}$$ + 0 + (H – h) = (H – h) + $$\frac{v^{2}}{2 g}+\frac{P_{0}}{\rho g}$$
On Solving, we get, v = $$\sqrt{2 \mathrm{gh}}$$
The velocity of efflux thus depends on the depth of which the hole is made from the surface of the liquid. Time taken to reach
ground = t = $$\sqrt{\frac{2(\mathrm{H}-\mathrm{h})}{\mathrm{g}}}$$,
Since v = $$\sqrt{2 \mathrm{gh}}$$ is horizontal and ax = 0;
R = vt = $$\sqrt{2 \mathrm{gh}}$$ $$\sqrt{\frac{2(\mathrm{H}-\mathrm{h})}{\mathrm{g}}}$$
R = $$2 \sqrt{\mathrm{h}(\mathrm{H}-\mathrm{h})}$$
To maximise Range, $$\frac{\mathrm{d} \mathrm{R}}{\mathrm{dh}}$$ = 0
i.e., h = H/2
Maximum Range,
$$R_{\max }=2 \sqrt{\frac{H}{2}(H-H / 2)}=H$$.

Question 15.
Water flows through a horizontal pipe of varying cross-section. If the pressure is 1 cm of mercury when the velocity is 0.35 m/s. Find the pressure at a point where velocity is 0.65 m/s.
At 1st point, P1 = 1 cm of mercury
= 0.01m of Hg.
= 0.01 × (13.6 × 103) × 9.8 P a;
v1 = 0.35 m/s
At the second nd point,
p2 = ?
v2 = 0.65 m/s and
ρ = 103 hg/m3.
According to Bernoulli’s theorem,
p1 + 1/2 $$\rho \mathrm{V}_{1}^{2}$$ = p2 + 1/2 $$\rho \mathrm{V}_{2}^{2}$$
or P2 = P1 + 1/2 $$\rho\left(v_{2}^{2}-v_{1}^{2}\right)$$
= 0.01 × 13.6 × 103 × 9.8 – 1/2 × 103 × [(0.65)2 – (0.35)2]
= 1182.8 Pa ⇒ $$\frac{1182.8}{9.8\left(13.6 \times 10^{3}\right)}$$
= 0.00887 m of Hg.

Question 16.
If excess pressure inside a soap bubble of diameter 2 cm is balanced by that due to a column of oil of sp.gravity 0.8, 2mm high, find the surface tension of the soap bubble.
R = 1 cm = 10-2 m
ρ of oil = 0.8 × 103kg/m3
h = 2 mm = 2 × 10-3m
Pressure due to 2 mm column of oil,
P = hρg – (2 × 10-3)(0.8 × 10-3) × 9.8
= 2 × 0.8 × 9.8 Pa
In case of soap bubble, P = $$\frac{4 \mathrm{s}}{r}$$
or s = $$\frac{P r}{4}=\frac{2 \times 0.8 \times 9.8 \times 10^{-2}}{4}$$
= 3.92 × 10-2N/m

Question 17.
A spray pump having a cylindrical tube of cross-section 8 cm2 has 50 fine holes of radius 0.5 mm. Consider the flow of liquid inside the tube to be 1.5 m/min. What is the speed of ejection of the liquid through the whole?
Area of cross-section of the tube:
a1 = 8 cm2 = 8 × 10-4m2
Number of holes = 50
Diameter of each hole = 0.5 × 2 = 1 mm
= 10-3
= 0.5 mm = 5 × 10-4m
Area of cross-section of each hole
= π r2 = π (5 × 10-4)2 m2
Total area of cross-section of 50 holes,
a2= 50 × π(5 × 10-4) m2
Speed of the liquid inside the tube,
v1 = 1.5 m/min = $$\frac{1.5}{60}$$ m/s
If v2 is the velocity of ejection of the liquid through the holes,
then, a1v1 = a2v2 or v2 = a1v1/a2
∴ v2 = $$\frac{\left(8 \times 10^{-4}\right) \times 1.5}{60 \times 50 \times \pi \times\left(5 \times 10^{-4}\right)^{2}}$$
= 0.5096 m/s.

Question 18.
Calculate the total energy/unit mass possessed by water at a point, where the pressures is 10gm f/sq mm, velocity is 0.1 m/s and height of water level from the ground is 0.2 m (g = 9.8 m/s2).
Here, P = 10 gm f/sq mm
= $$\frac{10}{1000}$$ × 9.8 × (103)2
= 98 × 103Nm-2
v = 0.1 m/s ;
ρ = 103 kg/m3;
h = 0.20 m
1. Pressure energy / unit mass
= $$\frac{P}{\rho}=\frac{9.8 \times 10^{3}}{10^{3}}$$ = 98 J/kg

2. Gravitational potential energy / unit mass = gh = 9.8 × 0.2 = 1.96 J/kg

3. Kinetic energy / unit mass
= 1/2 v2= $$\frac{1}{2}$$(0.1)2 = 0.0053/kg
∴ Total energy / unit mass
= $$\frac{P}{\rho}$$ + gh + 1/2 v2 = 98 + 1.96 + 0.005
= 99.965 J/kg

Question 19.
A mass of 15 kg is placed on the wider tube of a U tube as shown in the figure. Given that the area of the wider tube is 5 m2, find the difference in the water levels in the two tubes, (density of water = 1 kg/m3).

Area of the wider tube, A = 500 cm2
= 5 × 10-2 m2
The pressure applied by the mass at A is transmitted through the liquid and leads to a difference in the water levels.
Since the water level is same a A and B,
PA = PB ……(1)
But PA = $$\frac{F}{A}=\frac{m g}{A}=\frac{15 \times 9.8}{5}$$ = 29.4 Pa
Also we know that PB = ρgh = 1 × 9.8 × h
From (1) PB = PA
9.8 × h = 29.4
h = 3m
∴ The difference in water level obtained is 3 m.

Question 20.
If two liquids of mass m1 and m2 and density $$\rho_{1}$$, and $$\rho_{2}$$ are mixed together, then what is the density of the resulting mixture?
Volume occupied by first liquid,
V1 = $$\frac{m_{1}}{\rho_{1}}$$
Volume occupied by second liquid,
V2 = $$\frac{m_{2}}{\rho_{2}}$$
∴ density of the mixture,

Question 21.
The container shown in the figure below is filled with a liquid of density ρ. Note that the container has a height h1 and area of cross-section A1 for the upper half height h2 and area of cross-section A2 for the lower half. Find:

1. The pressure at the base of the container.
2. Force exerted by the liquid on the base of the container.

1. The pressure at the base of the container is due to the liquids both at the upper and lower half.
∴ PTotal = P1 + P2
PTotal = ρgh1 + ρgh2
PTotal = ρg(h1 + h2)

2. Force exerted on the bottom of the container,
P = PTotal × A2
F = ρg(h1 + h2) × A2.

## Karnataka 1st PUC Physics Question Bank Chapter 11 Thermal Properties of Matter

### 1st PUC Physics Thermal Properties of Matter Textbook Questions and Answers

Question 1.
The triple points of neon and carbon dioxide are 24.57 K and 216.55 K respectively. Express these temperatures on the Celcius and Fahrenheit scales.
Tc = T – 273.15
∴ Tc for neon = 24.57 – 273.15
= – 248.58°C
and Tc for carbon dioxide
= 216.55 – 273.15
= – 56.60°C
tF = $$\frac{9}{5}$$ tc + 32
∴ tF for neon = $$\frac{9}{5}$$ × (-248.58) + 32
= – 415.44°C
tF for carbon dioxide
= $$\frac{9}{5}$$ × (-56.60) + 32
= – 69.88°F

Question 2.
Two absolute scales A and B have triple points of water defined to be 200A and 350B. What is the relation between TA and TB?
Triple point of water
= 273.16 K = 200A
= 350B
∴ Size of 1K on scale A = $$\frac{273.16}{200}$$
and size of 1K on scale B = $$\frac{273.16}{350}$$
Value of temperature TA on scale A = $$\frac{273.16}{200}$$ TA
Value of temperature TB on scale B = $$\frac{273.16}{350}$$ TB
Since the two temperatures are equal,
$$\left(\frac{273.16}{200}\right) \mathrm{T}_{\mathrm{A}}=\left(\frac{273.16}{350}\right) \mathrm{T}_{\mathrm{B}}$$
⇒ TA = $$\frac{4}{7}$$ TB

Question 3.
The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law: R = R0 [1 + α (T – T0)] The resistance is 101.6Ω at the triple point of water 273.16K, and 165.5 Ω at the normal melting point of lead (600.5K). What Is the temperature when the resistance is 123.4 Ω?
At T0 = 273.16K, R0= 101.6 Ω
At T1 = 600.5K, R1 = 165.5 Ω
Let T2 be the temperature at which the resistance is R2 = 123.4 Ω
R = R0 [1 + α (T – T0)]
⇒ T – T0 = $$\frac{\mathrm{R}-\mathrm{R}_{0}}{\alpha \mathrm{R}_{0}}$$
∴ T1 – T0 = $$\frac{\mathrm{R}_{1}-\mathrm{R}_{0}}{\alpha \mathrm{R}_{0}}$$→ (1)
T2 – T0 = $$\frac{\mathrm{R}_{2}-\mathrm{R}_{0}}{\alpha \mathrm{R}_{0}}$$→ (2)
Dividing (2) by (1) gives,
$$\frac{\mathrm{T}_{2}-\mathrm{T}_{0}}{\mathrm{T}_{1}-\mathrm{T}_{0}}=\frac{\mathrm{R}_{2}-\mathrm{R}_{0}}{\mathrm{R}_{1}-\mathrm{R}_{0}}$$
$$\mathrm{T}_{2}=\mathrm{T}_{0}+\left(\mathrm{T}_{1}-\mathrm{T}_{0}\right)\left(\frac{\mathrm{R}_{2}-\mathrm{R}_{0}}{\mathrm{R}_{1}-\mathrm{R}_{0}}\right)$$
= 293.16 + (600.5-273.16) $$\left(\frac{123.4-101.6}{165.5-101.6}\right)$$
⇒ T2 = 384.83 K

Question 4.
1. The triple point of water is a standard fixed point in modern thermometry. Why? What is wrong in taking the melting point of ice and the
boiling point of water as standard fixed points (as was originally done in the Celcius scale)?

2. There were two fixed points in the original Celcius scale as mentioned above which were assigned the number 0°C and 100°C respectively. On the absolute scale, one of the fixed points is the triple point of water, which on the Kelvin absolute scale is assigned the number 273.16K. What is the other fixed point on this (Kelvin) scale?

3. The absolute temperature (Kelvin scale) T is related to the temperature te on the Celcius scale by tc = T – 273.15 Why do we have 273.15 in this relation and not 273.16?

4. What is the temperature of the triple point of water on an absolute scale whose unit interval size is equal to that of the Fahrenheit scale?
1. The triple point is used as a standard because of the unique conditions of temperature and pressure at this point. The melting point of ice and boiling point of water depend on pressure. So they cannot be used as reference points.

2. The other fixed point is absolute zero, the theoretically lowest possible temperature.

3. We have 273.15 in the relation and not 273.16 because the triple point of water is 0.01 °C and not 0°C.
In other words, 0.01 °C → 273.16 K
⇒ 0°C → 273.15 K
∴ tc = T – 273.15

4. The unit interval size in the Fahrenheit scale = $$\frac{212-32}{100}$$ =1.8
∴ The triple point of water = 273.16 × 1.8
= 491.69

Question 5.
Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made:

1. What is the absolute temperature of normal melting point of sulphur as read by thermometers A and B?
2. What do you think Is the reason behind the slight difference in the answers of thermometers A and B? (The thermometers are not faulty) What further procedure is needed In the experiment to reduce the discrepancy between the two readings?
1. Let P be the pressure in the pressure thermometer and T be the corresponding absolute temperature.
Let P0 and T0 be the values for the triple point of water.
Then $$\frac{T}{T_{0}}=\frac{P}{P_{0}}$$
$$\therefore \frac{T_{A}}{273.16}=\frac{1.797 \times 10^{5}}{1.250 \times 10^{5}}$$
⇒ TA = 392.69 K
$$\frac{T_{B}}{273.16}=\frac{0.287 \times 10^{5}}{0.200 \times 10^{5}}$$
⇒ TB= 391.98 K

2. The discrepancy is due to the non¬ideal behaviour of the gases. To reduce this discrepancy, measurements at lower pressures should be taken.

Question 6.
A steel tape 1m long Is correctly calibrated for a temperature of 27.0°C. The length of a steel rod measured by this tape Is found to be 63.0 cm on a hot day when the temperature is 45.0°C. What is the actual length of the steel rod on that day? What is the length of the same steel rod on a day when the temperature is 27.0° C? Coefficient of linear expansion of steel = 1.20 × 10-5K-1.
l0 = 63 cm
∆T = 45°C – 27°C .
= 18°C = 18 K
α = 1.2 × 10-5K-1
l = l0 (1 + α ∆T)
= 63(1 + 1.2 × 10-5 × 18)
l = 63.0136 cm.
∴ Actual length of the steel rod at 45°C =63.0136 cm
At 27°C, the steel tape is correctly calibrated.
∴ the length of the steel rod = 63.0 cm

Question 7.
A large steel wheel Is to be fitted on to a shaft of the same material. At 27°C, the outer diameter of the shaft is 8.70 cm and the diameter of the central hole In the wheel is 8.69 cm. The shaft Is cooled using ‘dry ice’. At what temperature of the shaft does the wheel slip on the shaft? Assume the coefficient of linear expansion of the steel to be constant over the required temperature range: αsteel = 1.2 × 10-5K-1.
Given αsteel = 1.2 × 10-5K-1
l1 = 8.70 cm
l2 = 8.69 cm (because the shaft is cooled to help it slip on the wheel)
T1 = 27°C = 300 K
T2 =?
l2 = l1 [1 + αsteel(T2 – T1)]
⇒ 8.69 = 8.70 [1 + 1.2 × 10-5(T2 – 300)]
⇒ 1 + 1.2 × 10-5(T2 – 300) = $$\frac{8.69}{8.70}-1$$
⇒ T2 = 300 + $$\frac{-0.01}{8.70} \times \frac{1}{1.2 \times 10^{-5}}$$
= 204.21 K
= – 68.94°C
= – 69°C
∴ The required temperature = – 69°C

Question 8.
A hole drilled in a copper sheet. The diameter of the hole is 4.24cm at 27.0°C. What is the change in the diameter of the hole when the sheet is heated to 227° C? Coefficient of linear expansion of copper = 1.70 × 10-5K-1.
Here, L = diameter of hole = 4.24 cm
ΔT = 227 – 27 = 220 °C and α = 1.70 x 10-5 C-1
Change in diameter,
ΔL = aLΔT = 1.70 x 10-5 x 4.24 x 200 = 1.44 x 10-2 cm.

Question 9.
A brass wire 1.8 m long at 27°C Is held taut with little tension between two rigid supports. If the wire Is cooled to a temperature of -39°C, what Is the tension developed In the wire, if its diameter is 2.0 mm?
Coefficient of linear expansion of brass = 2.0 × 10-5 K-1, Young’s modulus Of brass = 0.91 × 1011 Pa.
l = 1.8 m,
T1 =27°C = 300 K,
T2 = – 39°C =234 K,
d = 2mm = 2 × 10-3
αbrass = 2.0 × 10-5 K-1,
Ybrass = 0.91 × 1011 Pa
$$\frac{\Delta \ell}{\ell}=\alpha \Delta \mathrm{t}$$
$$Y=\frac{\Delta F / A}{\Delta \ell / \ell}=Y\left(\frac{\Delta \ell}{\ell}\right) A$$
= Y(α ∆T) $$\pi\left(\frac{\mathrm{d}}{2}\right)^{2}$$
= 0.91 × 1011 × 2.0 × 10-5 α (234 – 300) × π × $$\left(\frac{2 \times 10^{-3}}{2}\right)^{2}$$
= – 377.37 N
Tension developed ∆F = 380N (positive value, rounded to 2 significant figures)

Question 10.
A brass rod of length 50cm and diameter 3.0 mm is Joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at 250°C, if the original lengths are at 40.0°C? Is there a ‘thermal stress’ developed at the junction? The ends of the rod are free to expand (Coefficient of linear expansion of brass = 2.0 × 10-5K-1, steel = 1.2 × 10-5K-1).
Given:  lbrass = 50cm, lsteel = 50cm
T1 = 40°C = 313K
T2 = 250°C = 523 K
αbrass = 2.0 × 10-5K-1
αsteel = 1.2 × 10-5K-1
∆lbrass = lbrass × αbrass × ∆T
= 50 × 2.0 × 10-5 × (523 – 313)
∆lbrass = 0.21 cm
∆lsteel = lsteel × αsteel × ∆T
= 50 × 1.2 × 10-5 × (523 – 313)
= 0.126 cm
∆lsteel = 0.13cm
(round off to two significant figures) Since the rods are free to expand, no thermal stress is developed at the junction.

Question 11.
The coefficient of volume expansion of glycerin is 49 × 10-5K-1. What is the fractional change in its density for a 30°C rise In temperature?
Given, rglycerin = 49 × 10-5K-1
∆T = 30°C = 30 K
Let the volume be V1 at temperature and V2 at temperature T2
Using V2 = V1(1 + r ∆T), we get
$$\frac{V_{1}}{V_{2}}=\frac{1}{1+r \Delta T}$$
$$=\frac{1}{1+49 \times 10^{-5} \times 30}=0.9855$$
Fractional change in density

= 0.9855 – 1
= – 0.0145
= – 1.5 × 10-2
(The negative sign denotes a decrease in density).

Question 12.
A 10 kW drilling machine is used to drill a bore In a small aluminium block of mass 8.0 kg. How much is the rise in temperature of the blocks in 2.5 minutes, assuming 50% of power is used up in heating the machine itself or lost to the surroundings? Specific heat of aluminium = 0.91 J g-1 K-1
Given,
P = 10kW,
t = 2.5 minutes = 150s,
m = 8 kg,
saluminium = 0.91 Jg-1 K-1 = 910 Jkg-1 K-1
Energy transferred to the aluminium
= 50% of p × t
This energy heats up the aluminium.
So, $$\frac{50}{100}$$ × pt = m saluminum ∆T
⇒ $$\frac{1}{2}$$ × 10 × 103 × 150 = 8 × 910 × ∆T
⇒ ∆T = 103 K
or ∆T = 103°C

Question 13.
A copper block of mass 2.5kg is heated in a furnace to a temperature of 500°C and then placed on a large ice block. What is the maximum amount of ice that can melt? (Specific heat of copper = 0.39 Jg-1 K-1 heat of fusion of water = 335 Jg-1)
Given mcopper = 2.5 kg,
T1= 500°C,
T2 = 0°C,
Scopper = 0-39 Jg-1 K-1
Lice = 335 Jg-1
Let the mass of ice that melts be mice
Heat dissipated by the copper block = Heat absorbed by the ice.
⇒ mcopper Scopper ∆T = mice Lice
⇒ 2500 × 0.39 × (500 – 0) = mice × 335
⇒ mice = 1455 g
⇒ mice = 1.5 kg
(rounding off to 2 significant figures)

Question 14.
In an experiment on the specific heat of a metal, a 0.20 kg block of the metal at 150°C Is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150 cm3 of water at 27°C. The final temperature is 40°C. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value of the specific heat of the metal?
Given, mass of metal block,
m1 = 0.20 kg
Initial temperature of the metal block,
T1 = 150°C
Initial temperature of the calorimeter and water,
T2 = 27°C
Final temperature of the mixture,
T3 = 40°C
Volume of water in the calorimeter,
V =150 cm3
= 150 × 10-6 m3
Mass of the water = Density × Volume
= 1000 × 150 × 10-6
= 150 × 10-3 kg
Let s1 be the specific heat of the metal.
s2 is the specific heat of water.
Water equivalent of the calorimeter,
W = 0.025 kg
Heat lost by the metal block = Heat gained by the calorimeter and water.
⇒ m1 s1 ∆T1 = (m2 + W) s2 ∆T2
⇒ 0.2 × s1 × (150 – 40)
= (150 × 10-3 + 0.025) × 4200 × (40 – 27)
⇒ s1 = 434.3 J kg-1K-1
= 430 J kg-1K-1
or 0.43 Jg-1K-1
If the heat losses to the surroundings are not negligible, the left-hand side of the expression would have had an extra term added to it to account for this loss. Upon solving for s2 we would get a smaller value than the actual value.

Question 15.
Given below are observations on molar specific heats at room temperature of some common gases.
Gas                         Molar specific heat (cv) (cal mol’1 K’1)
Hydrogen               4.87
Nitrogen                 4.97
Oxygen                   5.02
Nitric oxide             4.99
Carbon monoxide   5.01
Chlorine                  6.17
The measured molar specific heats of these gases are markedly different from those for monoatomic gases. Typically, the molar specific heat of a monoatomic gas is 2.92 cal/mol K. Explain this difference. What can you Infer from the somewhat larger (than the rest) value for chlorine?
The measured molar specific heats of the gases mentioned in the table are markedly different because all these gases are diatomic gases. A monoatomic gas has only translational motion. In a diatomic gas, apart from translational motion, vibrational as well as rotational motion is also possible. Therefore, to increase the temperature of this gas through 1°C, an extra amount of heat is required to increase the average energy of different modes of motion. Under ordinary temperature conditions (say room temperature) diatomic gases usually have rotational motion apart from their translational motion. But for chlorine, the vibrational motion also occurs. This is the reason that chlorine has a somewhat larger value of molar-specific heat.

Question 16.
Answer the following questions based on the P-T diagram of carbon dioxide:
1. At what temperature and pressure can the solid, liquid and vapour phases of Co2, exist in equilibrium?
2. what is the effect of a decrease of pressure on the fusion and boiling point of Co2?
3. What are the critical temperature and pressure for Co2? What is their significance?
4. Is CO2 solid, liquid or gas at

• – 70°C under 1 atm,
• – 60°C under 10 atm,
• 15°C under 56 atm?

1. At -56.6°C and 5.11 atm, which is the triple point of Co2, all 3 .phases can coexist in equilibrium.
2. From the fusion and vaporisation curves, we see that reducing the pressure lowers the fusion and boiling point.
3. The critical temperature and pressure are 31.1°C and 73.0 atm respectively. Above this temperature, Co2 will not liquefy even if compressed to high pressures.
4.

• Gas
• Solid
• Liquid

Question 17.
Answer the following questions based on the P-T phase diagram of Co2:

1. CO2 at 1 atm pressure and temperature – 60°C is compressed isothermally. Does it go through a liquid phase?
2. What happens when Co2 at 4 atm pressure is cooled from room temperature at constant pressure?
3. Describe qualitatively the changes In a given mass of solid Co2 at 10 atm pressure and temperature -65°C as it is heated up to room temperature at constant pressure.
4. CO2 is heated to a temperature of 700C and compressed isothermally. What changes in its properties do you expect to observe?

(a) Since – 60° lies to left of – 56.6 °C on the curve, the vapour will condense directly into the solid.

(b) Since 4 atm pressure is less than 5-11 atm, the vapour will condense directly into the solid without entering into the liquid region.

(c) When the solid CO2 at – 65 °C is heated at 10 atm pressure, it is first converted into a liquid. A further increase in its temperature brings it into the vapour phase. If a horizontal line at P = 10 atm is drawn parallel to the T­axis, then the points of intersection of this line with the fusion and vaporization curve give the fusion and boiling points at 10 atm.

(d) Above 31.1°C, the gas can not be liquefied. Therefore, on being compressed isothermally at 70°C, there will be no transition to the liquid region. However, the gas will depart more and more from its perfect gas behaviour with the increase in pressure.

Question 18.
A child running a temperature of 101°F is given an antipyrin (i.e., a medicine that lowers fever) which causes an increase in the rate of evaporation of sweat from his body. If the fever is brought down to 98°F in 20 min, what is the average rate of extra evaporation caused by the drug? Assume the evaporation mechanism to be the only way by which heat is lost. The mass of the child is 30 kg. The specific heat of the human body is approximately the same as that of water, and the latent heat of evaporation of water at that temperature is about 580 cal g-1.
Mass of the child, m = 30 kg
Change in temperature ∆T ,
= 101°F – 98°F
= 3°F
= 3 × 5/9 = 5/3°c
Specific heat of water,
s = 4.2 × 103 J kg-1°C-1
Latent heat of vaporisation,
L = 580 cal g-1
= 580 × 4.2 × 103 J kg-1
Time taken for the sweat to evaporate,
t = 20 min.
Let m1 be the mass of sweat that evaporates.
Now, heat lost by the child during evaporation
= Heat needed for the sweat to evaporate.
⇒ ms ∆T = m1L
⇒ 30 × 4.2 × 103 × 5/3
= m1 × 580 × 4.2 × 103
⇒ m1 = 0.0862 kg
m1 = 86.2 g
Rate of extra evaporation caused by the drug = Rate of evaporation of sweat
$$=\frac{m_{1}}{t}=\frac{86.2 g}{20 m i n}$$
= 4.3 g min-1

Question 19.
A ‘thermacole’ icebox is a cheap and efficient method for storing small quantities of cooked food In summer in particular. A cubical icebox of side 30 cm has a thickness of 5.0 cm. If 4.0 kg pt Ice is put in the box, estimate the amount of Ice remaining after 6h. The outside temperature is 45°C, and the coefficient of thermal conductivity of thermacole is 0.01 J s-1 m-1K-1. [Heat of fusion of water = 835 × 103 J kg-1]
Total surface area of the cube,
A = 6 × (30 cm)2 = 0.54 m2
Mass of ice in the box M = 4 kg
Thickness of the box, L = 5 cm = 0.05 m
Thermal conductivity of thermacole,
k = 0.01 Js-1m-1K-1.
Outside temperature T1 = 45°C
The temperature of ice T2 = 0°C
Let the mass of ice that melts be m kg.
Heat of fusion of water,
Lf ice = 335 × 103 J kg-1
Time t = 6h = 6 × 3600 s = 21600 s
Heat gained by ice during fusion = Heat conducted through the walls of the box
⇒ m Lf, ice = $$\frac{\mathrm{KA}\left(\mathrm{T}_{1}-\mathrm{T}_{2}\right) \mathrm{t}}{\mathrm{L}}$$
⇒ m × 335 × 103
$$=\frac{0.01 \times 0.54 \times(45-0)}{0.05} \times 21600$$
⇒ m = 0.31 kg
∴ Mass of ice left in the box = M – m
= 4 – 0.31
= 3.69
= 3.7 kg.

Question 20.
A brass boiler has a base area of 0.15 m2 and thickness of 1.0 cm. It bolls water at the rate of 6.0 kg/min when placed on a gas stove. Estimate the temperature of the part of the flame in contact with the boiler. Thermal conductivity of brass = 109 J s-1m-1K-1; Heat of vaporisation of water = 2256 × 103 J kg-1.
Let the temperature of the flame be T1
T2 = boiling point of water = 100°C
Thermal conductivity of brass,
k = 109 J s-1m-1K-1
Base area of the boiler, A = 0.15 m2
Thickness of the boiler, L = 1 cm = 0.01 m
⇒ Heat of vaporisation of water
Lsteam = 2256 × 103 J kg-1
Rate of boiling water
= 6 kg min-1 = $$\frac{6}{60}$$ kg s-1
= 0.1 kg s-1
∴ Power consumed = 0.1 × Lsteam
= 0.1 × 2256 × 103W
= 2256 × 102W
∴ Powe transmitted through the boiler
= 2256 × 102 W
$$\Rightarrow \frac{\mathrm{KA}\left(\mathrm{T}_{1}-\mathrm{T}_{2}\right)}{\mathrm{L}}$$ = 2256 × 102
⇒ $$\frac{109 \times 0.15 \times\left(T_{1}-100\right)}{0.01}$$ = 2256 × 102
⇒ T1 – 100 = 138
⇒ T1 = 238°C
∴ Temperature of the flame = 238°C

Question 21.
Explain why:
1. a body with large reflectivity Is a poor emitter.

2. a brass tumbler feels much colder than a wooden tray on a chilly day.

3. an optical pyrometer (for measuring high temperatures) calibrated for an ideal black body radiation gives too low a value for the temperature of a red hot iron piece in the open but gives a correct value for the temperature when the same piece is in the furnace.

4. The earth without its atmosphere would be inhospitably cold.

5. heating systems based on circulation of steam are more efficient in warming a building than those based on the circulation of hot water.
1. We know that a + r + 1 = 1, where ‘a’ is absorbance, ‘r’ is reflectance and ‘t’ is transmittance or emittance. According to Kirchhoff’s law, emittance ∝ absorbance, i.e., good absorbers are also good emitters and so are poor reflectors. Therefore, if reflectivity is large (‘r’ is large) then ‘a’ is small and hence emittance is smaller (the object behaves as a poor emitter).

2. The coefficient of thermal conductivity of brass is higher than that of wood. When a brass tumbler is touched, heat quickly flows from human body to the tumbler and so the tumbler appears colder. But in the case of the wooden tray heat does not flow from the human body to the wooden tray, and so it feels relatively hotter.

3. Let the temperature of the hot iron in the furnace be‘T’ and that of the external environment be ‘T0’. According to Stefan’s Law, heat energy radiated per second E = σT4 in the open. Clearly E > E’. Hence the low readings of the optical pyrometer when the iron piece is in the open.

4. The earth’s atmosphere acts like an insulating blanket around it and does not allow heat to escape out but reflects it back to the earth. If it were absent, the heat would have escaped and the planet would be very cold.

5. Steam has a higher heat capacity (due to the latent heat) than boiling water. Therefore steam-based water heating systems are more effective.

Question 22.
A body cools from 80°C to 50°C in 5 minutes. Calculate the time it takes to cool from 60°C to 30°C. The temperature of the surroundings is 20°C.
The average temperature of 80° C and 50° C is 65° C, which is 45° C above the room temperature. Under these conditions, the body cools 30°C in 5 minutes. Using the equation.
$$\frac{\text { Change in temperature }}{\text { Time }}=\mathrm{KT}$$
we have $$\frac{30^{\circ} \mathrm{C}}{5 \mathrm{min}}$$ = K (45° C) ……(1)
The average of 60° C and 30° C is 45° C which is 25° C above the room temperature. K is the same for this situation as the original.
Hence $$\frac{30^{\circ} \mathrm{C}}{\mathrm{t}}$$ = K (25° C) ……(2)
Divide (1) by (2), we get,

∴ Time taken = 9 minutes.

### 1st PUC Physics Thermal Properties of Matter One Mark Questions and Answers

Question 1.
Define coefficient of thermal conductivity.
Coefficient of thermal conductivity of a material is defined as the quantity of heat conducted per second through a unit area of a slab of unit thickness when the temperature difference between its ends is 1K.

Question 2.
Heat radiation from a body due to its temperature is called thermal radiation.

Question 3.
What is a perfect black body?
A body, which absorbs all the radiation incident on it is called a perfect black body.

Question 4.
Define emissivity of a surface.
Emissivity of a surface is defined as the ratio of the quantity of thermal radiation emitted per sec per unit area of the surface to the quantity of thermal radiation emitted per sec per unit area of a perfect black body under identical conditions.

Question 5.
Define absorptivity of a surface.
Absorptivity of a surface is defined as the ratio of the quantity of thermal radiation absorbed per sec to the quantity of radiation incident on it per second.

Question 6.
What is the unit of coefficient of thermal conductivity?
Wm-1K-1

Question 7.
What is the dimensional formula of coefficient of thermal conductivity?
MLT-3 θ-1

Question 8.
What happens to the heat supplied to a good conductor when it is under steady-state?
Heat is entirely conducted from the hot end to the cold end.

Question 9.
Why is coal tar heated before use?
Coal tar is heated to reduce its viscosity.

Question 10.
What is the value of emissivity of a perfect black body?
Emissivity of a perfect black body is unity.

Question 11.
Give dimensions of solar constant and water equivalent.
Solar constant: [MT-3]
Water equivalent: [M]

### 1st PUC Physics Thermal Properties of Matter Two Marks Questions and Answers

Question 1.
Explain three methods of transfer of heat.

1. Conduction:
It is a process of transmission of heat in solids, without the actual movement of the particles.

2. Convection:
It is a process of transmission of heat in fluids (liquids and gases) with the actual movement of particles.

It is a process of transmission of heat without any medium.

Question 2.
When a solid is heated at one end, heat starts flowing from hot end to cold end. Heat supplied to any element of the solid is partly used in heating it, partly conducted through it and rest is radiated from its surface. After a certain stage, there is no variation of temperature with time and heat supplied is entirely conducted and radiated.

Such a state is called steady state. Under steady-state, the temperature at different parts of the solid remains constant and is independent of time. The ratio of the difference in temperature to the distance between two points in a heated body under steady state is called the temperature gradient.

Question 3.
Mention the properties of thermal radiation.

1. It travels through a vacuum with the velocity of light.
2. It travels in a straight line.
3. It undergoes reflection, refraction and total internal reflection.
4. It exhibits the phenomenon of interference, diffraction and polarisation.
5. Intensity of radiation decreases with increase in distance.

Question 4.
Explain Ferry’s black body.

Ferry’s black body consists of a hollow double-walled sphere with a fine hole and a pointed projection in front of the hole. The space between the two walls is evacuated to reduce any loss of radiation due to convection. The inner walls are coated with lamp black. Any radiation falling on the hole is completely absorbed due to multiple reflections, hence the hole acts as a black body.

Question 5.
State and explain Kirchhoff’s law.
At a given temperature, the ratio of the emissive power to the absorptive power for all bodies is a constant and is equal to the emissive power of a perfect black body.
If e is the emissive power and a is the absorptive power then, $$\frac{\mathrm{e}}{\mathrm{a}}$$ = constant  =1.
∴ e = a(∵ emissive power of a perfectly black body = 1)
Thus, the emissive power of a body is equal to its absorptive power, which means good emitter of thermal radiation is also a good absorber of heat.

Question 6.
State and explain Newton’s law of cooling.
The rate of cooling of a body is directly proportional to difference in temperature of the body and the surrounding. If the temperature of a body
decreases from θ1 to θ2 in t seconds, then rate of cooling is, $$\frac{\theta_{1}-\theta_{2}}{t}$$
Average temperature of the body θ = $$\frac{\theta_{1}+\theta_{2}}{2}$$
According to Newton’s law,
$$\frac{\theta_{1}-\theta_{2}}{t} \propto\left(\theta-\theta_{0}\right)$$
θ0 – temperature of the surrounding.

Question 7.
State and explain Stefan’s law.
The total radiation emitted by a perfect black body per second per unit area of the surface is directly proportional to the fourth power of its absolute temperature. If E is radiation emitted per second per unit area of a perfect black body at temperature T then, E ∝ T4 or E = σT4 where s is called Stefan’s constant.

Question 8.
State and explain Wein’s displacement law.
The wavelength (λm) corresponding to maximum energy emitted by a black body is inversely proportional to its absolute temperature.
i. e., λm ∝ $$\frac{1}{T}$$ or λm T = constant.
The constant is called Wein’s Constant.

Question 9.
State and explain Planck’s law.
According to this law, radiation of a particular frequency (v) is emitted or absorbed in discrete amounts (E) given by the relation E = h v where h is a Planck’s constant. The value of h is 6.625 × 10-34 Js. These packets of energy are called quantum or photon.

Question 10.
Define solar constant. What is its value?
The solar constant is defined as the amount of heat radiation absorbed normally per second per unit area of a perfect black body placed on the earth at a mean distance from the sun in the absence of atmosphere.
The value of solar Constant is 1350 W/m2.

Question 11.
State

2. State Stefan’s law of radiation

1. Kirchhoff’s law:
At a given temperature, the ratio of the emissive power to the absorptive power is a constant for all bodies and is equal to the emissive power of a perfect black body.

2. Stefan’s law:
The total radiation emitted by a perfectly black body per second per unit area is directly proportional to the 4th power of its absolute temperature.

Question 12.
Calculate the heat of combustion of coal, when 10 gm of coal, on burning raises the temperature of 2 litres of water from 20° C to 55° C.
mass of 2l of water = 2kg
∆T = 55° C – 20° C = 35° C
s = 4.2 × 103 J kg-1 °C-1
Heat produced by burning coal = ms ∆T
= 2 × 4.2 × 103 × 35
= 2.94 × 105 J

Question 13.
2 kg of water at 80° C is mixed with 3 kg of water at 20°C. Assuming no heat losses, find the final temperature of the mixture.
m1 = 2 kg,
T1 = 80°C,
s = 4.2 × 103 J kg-1 °C-1,
m2 = 3kg,
T2 = 20° C
Let the final temperature be T3
Heat lost by 3 kg water = Heat gained by 2 kg water.
⇒ m1s(T1 – T3) = m2s(T3 – T2)
⇒ 2 × (80 – T3) = 3 × (T3 – 20)
⇒ T3 = 44° C
∴ Final temperature = 44° C

Question 14.
What are the basic requirements of a cooking utensil in respect of:

1. Specific heat and
2. thermal conductivity

For the temperature of the utensil to rise quickly, it is necessary for the utensil to have low specific heat. In order for the heat to spread quickly to the vegetables, the utensil must have high thermal conductivity.

### 1st PUC Physics Thermal Properties of Matter Three Marks Questions and Answers

Question 1.
State Wien’s displacement law. Draw the graph showing energy emitted versus wavelength for a black body at different temperatures.
Wein’s displacement law:
The wavelength (λm) Corresponding to maximum energy emitted by a black body is inversely proportional to its absolute temperature. i.e. λm ∝ $$\frac{1}{T}$$ or λm T = b. where b is a constant, called Wein’s constant. The radiation emitted by a black body is called black body radiation. The distribution of energy of a black body radiation at different temperatures is as shown in the figure.

The energy distribution is not uniform. There is a particular wavelength λm at which the energy emitted is maximum. The wavelength λm for which the intensity is maximum decreases with increase in temperature.

Question 2.
Draw energy distribution curves for a black body at two different temperatures T1 and T2 (T1 > T2). Write any two conclusions that can be drawn from these curves.
The radiation emitted by a black body is called black body radiation. The distribution of energy of a black body radiation at different temperatures with its wavelength is as shown in the figure.

Question 3.
Draw labelled curves which show how the energy in the black body radiation spectrum varies with wavelength at various temperatures. What result does this lead to?
The radiation emitted by a black body is called black body radiation. The distribution of energy of a black body radiation at different temperatures with its wavelength is as shown in the figure.

### 1st PUC Physics Thermal Properties of Matter Four Or Five Marks Questions and Answers

Question 1.
State and explain the laws of thermal conductivity.

Consider a metal slab of area of cross-section A and thickness d. Let the end faces (ABCD & EFGH) be maintained at temperatures θ1 and θ21 > θ2).
The amount of heat conducted (Q) from the hotter to colder face is,
1. directly proportional to the area of cross-section (A)
2. directly proportional to the time for which heat flows (t)
3. directly proportional to the temperature difference between the faces (θ1 – θ2)
4. inversely proportional to the distance between the faces (d)
This is called the law of thermal conductivity.
According to the law Q ∝ $$\frac{\mathrm{At}\left(\theta_{1}-\theta_{2}\right)}{\mathrm{d}}$$
$$\mathrm{Q}=\frac{\mathrm{KAt}\left(\theta_{1}-\theta_{2}\right)}{\mathrm{d}}$$
K is a constant of proportionality called coefficient of thermal conductivity.

Question 2.
What is black body radiation? Explain the characteristics of It.
The radiation emitted by a black body is called black body radiation. The distribution of energy of a black body radiation at different temperatures with its wavelength is as shown in the figure.

1. A black body emits radiation of all wavelengths lying in the region of ultraviolet, visible light and infrared.
2. As the temperature of the black body increases, the total amount of radiation emitted by it increases. (According to Stefan’s law)
3. The energy distribution is not uniform. There is a particular wavelength λm at which the energy emitted is maximum.
4. The wavelength λm for which the energy emitted is maximum decreases with increase in temperature.

Question 3.

It consists of a concave mirror M made of copper and plated with nickel. The mirror has a small hole at the centre at which an eyepiece E is fitted. A diaphragm D is placed at the focus of M; which has a hole to allow radiation to pass through it. A blackened metal strip S is placed behind D. To the other surface of S, one junction of a thermocouple TT is connected. The thermocouple is connected to a millivoltmeter. A screen R is used to protect the strip from direct radiation. The mirror can be adjusted for focusing.

Question 4.
Why is ice packed in sawdust? Explain.
Ice is packed in sawdust because it prevents ice from melting. Sawdust is a bad conductor of heat.

Question 5.
Explain the different modes of transmission of heat. Explain Wein’s displacement law with the help of graph of intensity versus wavelength.
1. Conduction:
It is a process of transmission of heat in solids, without the actual movement of the particles.
Eg: When metal rod is heated at one end, after some time another end also becomes hot.

2. Convection:
It is a process of transmission of heat in fluids (liquids and gases) with the actual movement of particles.
Eg: Smoke carries heat energy from the flame by convection.

It is a process of transmission of heat without any medium in the form an electromagnetic waves.
Eg: Heat from the sun reaches the earth by radiation.

Wein’s displacement law:
The wavelength (λm) Corresponding to maximum energy emitted by a black body is inversely proportional to its absolute temperature. i.e. λm ∝ $$\frac{1}{T}$$ or λm T = b. where b is a constant, called Wein’s constant. The radiation emitted by a black body is called black body radiation. The distribution of energy of a black body radiation at different temperatures is as shown in the figure.

The energy distribution is not uniform. There is a particular wavelength λm at which the energy emitted is maximum. The wavelength λm for which the intensity is maximum decreases with increase in temperature.

Question 6.
What is a black body? Draw the curves showing the energy distribution among black body radiations at different temperature. Hence define Wein’s displacement law. Give one application of Wein’s displacement law.
1. black body:
The radiation emitted by a black body is called black body radiation. The distribution of energy of a black body radiation at different temperatures with its wavelength.

Wein’s displacement law:
The wavelength (λm) Corresponding to maximum energy emitted by a black body is inversely proportional to its absolute temperature. i.e. λm ∝ $$\frac{1}{T}$$ 17 or λm T = b. where b is a constant, called Wein’s constant.

Question 7.
Discuss briefly energy distribution of black body radiation. Hence deduce Wien’s displacement law and Stefan’s law.
Wein’s displacement law:
The wavelength (λm) Corresponding to maximum energy emitted by a black body is inversely proportional to its absolute temperature. i.e. λm ∝ $$\frac{1}{T}$$ or λm T = b. where b is a constant, called Wein’s constant. The radiation emitted by a black body is called black body radiation. The distribution of energy of a black body radiation at different temperatures is as shown in the figure.

The energy distribution is not uniform. There is a particular wavelength λm at which the energy emitted is maximum. The wavelength λm for which the intensity is maximum decreases with increase in temperature.

Stefan’s law:
The total radiation emitted by a perfect black body per second per unit area of the surface is directly proportional to the fourth power of its absolute temperature. If E is radiation emitted per second per unit area of a perfect black body at temperature T then, E ∝ T4 or E = σT4 where s is called Stefan’s constant.

1st PUC Physics Thermal Properties of Matter Numerical Problems Questions and Answers

Question 1.
A circular plate has a radius of 10 cm and thickness 1 cm. If one face of the plate is kept in contact with steam and the other face is in contact with ice, calculate the amount of heat flowing per hour across the plate. K as 85 W-1m-1K-1.
Solution:
The amount of heat conducted in t seconds is given by,
$$Q=K \frac{A\left(\theta_{1}-\theta_{2}\right) t}{d}$$
Here, K = 85 W m-1K-1
A = π r2 = 3.14 × (0.1)2
t = 1 hr = 60 × 60 s.
d = 10-2 m2
1 – θ2) = 373 -273 = 100 K
Q= $$\mathrm{K} \frac{\mathrm{A}\left(\theta_{1}-\theta_{2}\right) \mathrm{t}}{\mathrm{d}}$$
= $$\frac{85 \times 3.14 \times(0.1)^{2} \times 100 \times 60 \times 60}{10^{-2}}$$
= 96.08×106 J.

Question 2.
A metal rod 0.314 m long and 0.02 m in diameter has one end in boiling water at 373K and the other end in melting ice at 273K. Find the amount of ice that will melt per hour? K = 385 Wm-1K-1.
Latent heat of Ice = 3.36 × 105 Jkg-1.
Solution:
If ‘m’ is mass of ice that melts in one hour, then quantity of heat conducted is Q = mL
But Q = K $$\frac{A\left(\theta_{1}-\theta_{2}\right) t}{d}$$
∴ Q = mL
m = $$\frac{\mathrm{KA}\left(\theta_{1}-\theta_{2}\right) t}{\mathrm{d} \times L}$$
Here, K =385Wm-1K-1
A = p r2 =3.14 × (0.01 )2
L = 3.3 × 105 Jkg-1
θ1 – θ2 = 373 – 273 = 100 K
t = 1 hour = 3600 s
d = 0.314 m
∴ m= $$\frac{385 \times 3.142 \times(0.01)^{2} \times 100 \times 60 \times 60}{0.314 \times 3.36 \times 10^{5}}$$
= 41.28 x 10-2kg

Question 3.
Two metal rods of the same area of cross-section and with their lengths in the ratio 1:3 are joined together to form a compound rod. One end of this compound rod is kept at 100°C and the other at 0° C. Calculate the temperature at the Junction if the thermal conductivity is in the ratio 1:3.
Solution:

Let AB and BC be the two rods of length L1 and L2 and having the same area of cross-section A, joined together to form a compound rod as shown in the figure.
If ‘ θ ’ is the steady temperature at the junction B during the steady-state, then, rate of flow of heat through AB = rate of flow through BC

Question 4.
Calculate rate of heat radiated per unit area from the surface of a furnace at a temperature 2000K, assuming it to be a black body. If the emissivity is 0.37, then what Is the amount of heat radiated per second? G =5.67×10-8Wm-2K-4.
Solution:
From Stefan’s law of radiation, the amount of heat radiated per second is E = σT4.
E =5.67 × 10-8 × (2000)4
= 5.67 × 24 × 104
= 5.67 × 16 × 104
= 9.072 × 105Js-1.
If e is the emissivity of the surface, then the heat radiated per second R1 = eR
= 0.37 × 9.072 × 105
= 3.35 × 105 Js-1

Question 5.
A slab of stone of area 0.36m2 and thickness 0.1m is exposed on the lower surface to steam at 373 K. A block of ice at 273 k rests on the upper surface of the slab. If in one hour, 4.8kg of Ice is melted, calculate the conductivity of the stone. Latent heat of Ice = 3.35 x 105 JK-1.
Solution:
A = 0.36m2,
d = 0.1m
θ1 = 373 K
θ2 = 273 K
t = 1 hr = 3600s,
m = 4.8kg,
K = ?
L = 3.35 × 105 Jkg-1
Heat conducted through the rod = Heal used in melting ice
$$\frac{\mathrm{KA}\left(\theta_{1}-\theta_{2}\right) \mathrm{t}}{\mathrm{d}}$$ = mL
K = $$\frac{m L d}{A\left(\theta_{1}-\theta_{2}\right) t}=\frac{4.8 \times 3.35 \times 10^{5} \times 0.1}{0.36 \times 100 \times 3600}$$
= 1.241 Wm-1 K-1.

Question 6.
The filament of 40 W electric lamp of length 0.1m and diameter 0.1mm is kept at a temperature of 2773 K. Calculate the emissive power if all the heat is lost by radiation. σ = 5.67 × 10-8 Wm-2 K-4].
Solution :
P = 40 W,
l = 0.1 m,
r = 0.05 mm
T = 2773 K
e = ?
Area = 2π rl =3.142 x 10-5 m2
Energy emitted by a source in t sec
E = e A σ T4t
$$\frac{E}{t}$$ = e A σ T4
P = e A σ T4
e = $$\frac{P}{A \sigma T^{4}}$$
$$=\frac{40}{3.142 \times 10^{-5} \times 5.67 \times 10^{-8} \times(2773)^{4}}$$
= 0.3798

Question 7.
The thickness of ice in a lake at a certain moment is 0.03m. At what rate the thickness of the ice increasing and how long will it take for the thickness of ice layer to be doubled, if the temperature of air is -20° C. Given coefficient of thermal conductivity of ice=0.168 Wm-1K-1 degree. Latent heat of fusion of ice=3.35 × 10 Jkg-1 density of ice= 9.2 × 102 Kg/m3
Solution:
When the layer of ice covers the lake, the lower part of the ice is in contact with ice is at 0°c and outer surface with air at – 20°c
∴ θ1 =0°c and θ2 = – 20°c
Let the thickness of ice increase by small amount ∆x by an area A in a time ∆t.
Then,
mass of ice formed = A × ∆x × ρ
Quantity of heat, Q = A × ∆x × ρ × L    …..(1)
But, Q = $$\frac{\mathrm{k} \mathrm{A}\left(\theta_{1}-\theta_{2}\right) \Delta \mathrm{t}}{\mathrm{d}}$$ ….(2)
from equation (1) & equation (2)
A × ∆x × r × L = $$\frac{\mathrm{k} \mathrm{A}\left(\theta_{1}-\theta_{2}\right) \Delta \mathrm{t}}{\mathrm{d}}$$
$$\frac{\Delta x}{\Delta t}=\frac{k A\left(\theta_{1}-\theta_{2}\right)}{d x A x \rho x L}$$
$$=\frac{0.168 \times 20}{0.03 \times 9.2 \times 10^{2} \times 3.35 \times 10^{5}}$$
= 3.634 × 10-7ms-1
The time taken for the ice layer to double is
t = $$\frac{\Delta x / \Delta t}{d}$$
$$=\frac{3.634 \times 10^{-7}}{0.06}$$
= 60.57 × 10-7sec

Question 8.
Estimate the rate at which ice melts in a wooden box 0.04m thick, of inside measurement 0.6 m × 0.4m × 0.4m. Assume that the outside temperature is 300 K and the conductivity of wood is 0.08 wm-1K-1
Solution:
Area of the box
= A = 4 × (0.6 × 0.4) + 2 × (0.4 × 0.4)= 1.28m2
d = 0.04m
k= 0.08wm-1K-1
L= 3.36 × 105Jkg-1
Let Q be the heat transmitted per second.
Then, Q = $$\frac{\mathrm{k} \mathrm{A}\left(\theta_{1}-\theta_{2}\right) \mathrm{t}}{\mathrm{d}}$$
$$=\frac{0.08 \times 1.28 \times(300-273) \times 1}{0.04}=69.12 J$$
To melt 1 kg of ice, 3.36 × 105J of heat is required
∴ Rate at which ice melts
$$=\frac{69.12}{3.36 \times 10^{-5}}$$
= 20.57 × 10-5kgs-1.

Question 9.
A man wraps himself in flannel 4 mm thick. How much heat will be lost per square metre of his body per hour if the atmosphere temperature is 20° and the body temperature is 37°c(The co-efficient of thermal conductivity of flannel is 0.05 Wm-1k-1)
Solution:
k=0.05Wm-1k-1
d = 4mm = 4 × 103
m = 0.004m
θ1 =37°c,
θ2 =20°c
t =1 hr=3600 seconds
A =1m2
The quantity of heat conducted is
$$Q=\frac{k A\left(\theta_{1}-\theta_{2}\right) t}{d}$$
$$=\frac{0.05 \times 1(37-20) 3600}{0.004}$$
= 7.65 × 105J

Question 10.
Calculate the rate at which heat is lost through a glass window of area 0.1m2 thickness 4mm when the temperature inside is at 40° c and outside is -5°c. Co-efficient of thermal conductivity of a glass is 1.1wm-1k-1.
Solution:
A=0.1m2
d = 4mm = 4 × 10-3
m=0.004m
θ1 = 40°C
θ2 = -5°c
k = 1.1Wm-1k-1
Heat lost through the glass window is
$$Q=\frac{K A\left(\theta_{1}-\theta_{2}\right) t}{d}$$
$$=\frac{1.1 \times 0.1(40+5) \times 1}{0.004}$$
= 1237.5J

Question 11.
The two faces of metal plate of thickness 40mm and area 102m2 maintained at 273K and 373K. The coefficient of thermal conductivity of the metal is 380 Wm-1k-1 Find the quantity of heat flows, through the plate in one hour.
Solution:
d = 40mm = 40 × 10-3
m = 0.04m
A = 102m2
θ2 = 273k
θ1 =37K
K = 380Wm-1k-1
t = 1hr = 3600 seconds
The quantity of heat that flows in one hour is
$$Q=\frac{k A\left(\theta_{1}-\theta_{2}\right) t}{d}$$
$$=\frac{380 \times 10^{2}(373-273) \times 3600}{0.04}$$
Q = 3.42 × 1011J

Question 12.
A spherical body of radius 0.01m is at 400K. Calculate the energy radiated by the body per second per unit area. Assume the body to be perfect black body.
Solution:
From stefan’s law of radiation, amount of heat radiated per second per unit area is
E = σ T4
= 5.67 × 10-8 × (400)4 = 1451.5s-1
The total energy radiated per second is
E = A σ T4
E = P r2 σ T4
= 3.142 × (0.01 )2 × 5.67 × 10-8 × (400)4
= 0.4561s-1

Question 13.
A metal rod 0.3m long & 0.02m in diameter has one end at 100°C & other end at 0°C. Calculate the total amount of heat conducted in 2 minutes. Given k = 385 JS-1m-1K-1.
Solution:
l = 0.3m,
d = 0.02m
∴ r = 0.01m
θ1 = 100°C
θ2 = O°C
t = 2 minutes = 120 S
k = 385 JS-1m-1K-1.
Q = ?
t= π r2 = 3.142 × 0.012 = 3.142 × 104m2
$$\therefore Q=\frac{k A t\left(\theta_{1}-\theta_{2}\right)}{\ell}$$
$$=\frac{385 \times 3.142 \times 10^{4} \times 120(109)}{0.3}$$
= 4.8 × 103 J

Question 14.
A metallic rod 1.5m long and 2×104 m2 in area is heated at one end. The coefficient of thermal conductivity of the material of the metallic rod is 210 Js-1m-1K-1. In the steady-state the temperature of the ends of the rod are 373K and 263 K. Calculate.

1. Temperature gradient In the rod.
2. Temperature at a point 0.8 m from the hot end.
3. Rate of heat transmission.

d=1.5m,
A = 2 × 104m2,
K = 210 W/m/k, θ1 = 373K and θ2 = 263K.
$$\frac{\theta_{1}-\theta_{2}}{d}=\frac{373-263}{1.5}=\frac{110}{1.5}=73.33 \mathrm{K} / \mathrm{m}$$

2. Let θ be the temperature at a distance d1 = 0.8m from the hot end then temperature gradient = $$\frac{\theta_{1}-\theta}{d^{1}}$$
i.e., 73.33 = $$\frac{373-\theta}{0.8}$$ or θ =314.336 K.

3. Rate of heat transmission
$$\frac{\mathrm{q}}{\mathrm{t}}=\mathrm{KA}\left(\frac{\theta_{1}-\theta_{2}}{\mathrm{d}}\right)$$
= 210 × 2 × 10-4 × 73.33
= 3.08 J/s.

Question 15.
The two faces of a metal plate of thickness 40 mm and area 10-2m2 are maintained at 273 K and 373 K. The coefficient of thermal conductivity of metal is 380 w/m/K. Find the quantity of heat that flows through the plate in one hour.
d = 40 × 10-3m,
A = 10-2m2,
t = 1 hr = 3600s,
θ1 = 373K,
θ1 = 2.73K and K= 380 W/m/K
$$\mathrm{q}=\frac{\kappa \mathrm{At}\left(\mathrm{q}-\theta_{2}\right)}{\mathrm{d}}$$
= 34.6×106J.

Question 16.
The time taken for a substance to cool form 100° C to 50° C is 5 minutes. If the room temperature is 30°C, what time required for same substance to cool from 50°C to 40°C?
Case (i):
Rate of cooling
$$\frac{\mathrm{d} \theta}{\mathrm{dt}}=\frac{100-50}{5}$$ = 10° C/min,
Average temperature
$$\theta=\frac{100+50}{2}=75^{\circ} \mathrm{C}$$
Room temperate θ0 = 30°C.
According to Newton’s law of cooling,
$$\frac{\mathrm{d} \theta}{\mathrm{dt}}$$ = -K(θ – θ0).
10 = -K(75 – 30).
K= $$-\frac{10}{45} / \mathrm{min}$$ . ……(1).

Case(ii):
Rate of cooling
$$\frac{\mathrm{d} \theta}{\mathrm{dt}}$$ = $$\frac{50-40}{\mathrm{dt}}=\frac{10}{\mathrm{dt}}$$,
Average temperature
$$\theta=\frac{50+40}{2}=45^{\circ} \mathrm{C}$$
Room temperate e θ0 = 30°C.
According to Newton’s law of cooling,
$$\frac{\mathrm{d} \theta}{\mathrm{dt}}$$ = K(θ – θ0)
$$\frac{10}{\mathrm{dt}}=-\left(\frac{-10}{45}\right)(45-30)$$ (from)
dt =3 min.

1st PUC Physics Thermal Properties of Matter Hard Questions and Answers

Question 1.
A hot body placed in air is cooled down. According to Newton’s law of cooling, the rate of decrease of temperature being ‘k’ times the temperature of the surrounding. Starting from t = 0, find the time in which the body will lose 90% of the maximum heat it can lose.
We have,
$$\frac{\mathrm{d} \mathrm{T}}{\mathrm{dt}}$$ = – K(T – T0)
where T0 is the temperature of the surrounding and T is the temperature of the body at time t. Let T = T1 at t = 0.
Then

⇒ T – T0 = (T1 – T0) e-kt ……(1)
The body continues to lose heat until its temperature becomes equal to that of the surrounding. The loss of heat in this entire period is
∆H = ms(T -T0).
This is the maximum heat it can lose. If the body loses 90% of this heat, the decrease in to temperature will be
$$\frac{9 \Delta \mathrm{H}}{10 \mathrm{ms}}$$ = $$\frac{9}{10}$$ (T1 – T0)
If the body loses this heat in time t, the temperature at t1 will be
T1 – $$\frac{9}{10}$$ (T1 – T0)
$$=\frac{T_{1}}{10}+\frac{9 T_{0}}{10}$$
Putting these values of time and temperature in (1),
$$\frac{T_{1}+9 T_{0}}{10}$$ – T0 = (T1 – T2) e-kt1
⇒ $$\frac{T_{1}-T_{0}}{10}$$ = (T1 – T2) e-kt1
⇒ e-kt1 = 1/10
⇒ t1 = $$\frac{\ln 10}{k}$$
∴ Required time = $$\frac{\ln 10}{k}$$

Question 2.
Two thin metallic spherical shells of radii r1 and r2 (r1 < r2 are placed with their centres coinciding. A material of thermal conductivity k is filled in the space between the shells. The inner shell is maintained at temperature T1 and the outer shell at temperature T2 (T1 < T2). Calculate the rate at which heat flows radially through the material.

Consider two spherical shells of radii x and x + dx concentric with the given system. Let the temperatures at these shells be T and T + dt respectively. The amount of heat flowing radially inward through the material between x and x + dx is

## Karnataka 1st PUC Geography Question Bank Chapter 6 Hydrosphere

### 1st PUC Geography Hydrosphere One Mark Questions and Answers

Question 1.
What is Hydrological cycle? (T. B. Qn)
The cyclic (Evaporation, condensation and precipitation) movement of water between the atmosphere and the Earth’s surface is called ‘Hydrological cycle’.

Question 2.
Define Oceanography. (T. B. Qn)
The scientific study of water bodies (Sea, Oceans) is called ‘Oceanography’.

Question 3.
Which instrument is used to measure depth of the Ocean? (T. B. Qn)
Fathometer is the instrument used to measure the depth.

Question 4.
Name the region which covers largest area in the ocean floor. (T. B. Qn)
Deep sea plains account for about 82.7% of the total area of the ocean floor.

Question 5.
Which hemisphere is called water hemisphere?
Southern hemisphere is called water hemisphere.

Question 6.
Define hydrosphere
The total water mass on the earth’s surface is called hydrosphere.

Question 7.
Name the four oceans of the world.
Pacific Ocean, Atlantic ocean Indian Ocean, Antarctic Ocean and arctic oceans.

Question 8.
What is a nautical mile?
The length, width and areas of oceans is measured by means of nautical miles. One nautical mile measures a length of 6080 feet.

Question 9.
Which is the largest and deepest ocean?
Pacific is the largest and deepest ocean in the world

Question 10.
What is isobaths?
Lines drawn the map to show the places having some depth in the sea or ocean floor.

Question 11.
What are Bathy orographical maps?
The maps depicting the surface configuration of the ocean floor are called Bathhyorographical maps.

Question 12.
What is Hypsographic curve?
It is the curve representing the extent of the area, height and depths of the land as well as oceans.

Question 13.
What is the average depth of the oceans?
The average depth of the oceans 3600 meters.

Question 14.
Which is the deepest part of the oceans?
Ocean deep is the deepest part of the oceans.

Question 15.
What is sea level?
Sea level is considered as a standard for measuring heights of landforms.

Question 16.
Which is deepest part of the Pacific Ocean?
The challenger deep in the Marina trench near the Philippine Islands in the Pacific Ocean is the deepest part of the Ocean floor (10,898 meters).

Question 17.
What is continental Slope?
The portion of the ocean floor from the edge of the continental shelf to the deep sea plains is called the continental slope.

Question 18.
What is an ocean deep?
The ocean deeps are the depressions on the ocean floor with very steep sides.
The long, narrow and deep troughs on the ocean floor are known as ocean deeps or troughs.

Question 19.
What is Salinity? (T. B. Qn)
It refers to the amount of dissolved solids in the ocean water. In other words. It means the amount of salt in the ocean water.

Question 20.
Which lines are drawn on the map to show places having same salinity? (T.B.Qn)
Isobaths drawn on the map to show the places having same depth in the sea or ocean floor.

Question 21.
Define Iceberg.
The huge masses of floating ice are known as ‘Icebergs’. They are found near the polar areas. Usually many of them drift towards the equator during summer and lower the temperature of the ocean water.

Question 22.
Where does Sargasso sea locate?
In the North Atlantic Ocean.

Question 23.
Where does Great Barrier Reef locate?
East coast of Australia.

Question 24.
Where does the Australian current flow?
Indian Ocean.

Question 25.
What instrument is used to measure salinity?
Electric salinity meter

Question 26.
What are Isohalines?
Isohalines are the liens drawn on the map to show the places having the same amount of salinity.

Question 27.
What is the average salinity of ocean water?
The average salinity of the ocean water is 35 ppt (parts per thousand) means 35 gms of salt per 1000 grams of sea or ocean water.

Question 28.
Mention any two determining factors of salinity of ocean water?
Evaporation and precipitation are the major determining factors of salinity of ocean water.

Question 29.
What are Sea Waves? (T. B. Qn)
Series Waves are the undulations found on the surface of the sea or ocean.

Question 30.
Where do we find the ‘Agulhas current’? (T. B. Qn)
Agulhas currents are found in East of South Africa.

Question 31.
What is meant by Ocean current?
Large scale movements of water in the oceans in a definite direction are called ocean currents.

Question 32.
What is wave time?
The time of movement from one crest to another or from one trough to another is called wave time.

Question 33.
What is the height of the wave?
The vertical distance between crest and trough is called the height of the wave.

Question 34.
What is Trough?
The lowest part of the wave is between two waves is called trough.

Question 35.
What is crest?
The highest point of the wave is called crest.

Question 36.
What is wave length?
The distance between two successive crests or troughs is called the wave length

Question 37.
Where do you find the Gulf Stream?
It is a Warm current, flows east wards from the newfound land, in the east coast of U.S.A

Question 38.
What are the two cold currents of the North Atlantic Ocean?
Canary current, East Greenland current and Labrador Current.

Question 39.
Where are the Falkland current and Beguile currents flow?
These are two cold currents in the South Atlantic circuit.

Question 40.
Where do you find Kurosiwo current?
Kurosiwo current found in East coast of Japan. It is north pacific current.

Question 41.
Name the two cold currents of the North Pacific Ocean.
Oyash.io and Okhotsk current are the cold currents of North Pacific Ocean.

Question 42.
Where do you find Agulhas current?
Agulhas current found in South Africa.

Question 43.
What is ‘drift’?
The movement of sea surface water dragged with the prevailing winds is known as ‘drift’.

Question 44.
Mention the types of current?
Warm and cold currents.

Question 45.
What are Warm currents?
The currents originating in the tropical region have warm waters called warm currents’.

Question 46.
What are cold currents?
The currents which flow from the polar areas carried cold water. These are known ass cold currents.

Question 47.
What is the direction of ocean currents?
The direction of the currents is clockwise in the northern hemisphere and anti-clockwise n the southern hemisphere.

Question 48.
Mention the divisions of Atlantic Ocean currents?
There are three divisions of Atlantic Ocean currents are a. Equatorial currents b. Current of North Atlantic Ocean and current of South Atlantic Ocean.

Question 49.
Which is the important occupation of the people living in the costal areas?
Fishing. Is the important occupation of the people living in the coastal areas.

Question 50.
Name the most important food item derived from the oceans.
Fish Molluscans, crustaceans and many other forms for the preparation of food items.

Question 51.
Name the important marine areas of crude oil production:
The Persian Gulf region of West Asia and the North sea bed off Europe is well known for petroleum production. Bombay high in India are the major areas of crude oil.

Question 52.
How does a tide occur? (T. B. Qn)
The regular rise and fall of water level in the world’s sea and oceans is called ‘tides’. Gravitation, attraction of the Moon, Gravitational pull by the Sun and Centrifugal force of the Earth. Causes for the occurance of tides.

Question 53.
What is Tidal period? (T. B. Qn)
The time between successive-high tide and low tide is known as ‘Tidal period’.

Question 54.
What is Tide?
The regular rise and fall of water level in the world’s sea and oceans is called ‘tides’.

Question 55.
What is tidal current?
The different types of inward and outward movement of water are known as Tidal current.

Question 56.
What are neap tides?
Neap tides are occur when the earth, sun and moon are not in straight line. The sun and moon are not exerting, sun and moon are more or less in a straight line.

Question 57.
What are spring tides?
The tides which occur when the earth, sun and moon are more or less in a straight line.

Question 58.
What is High tide?
The part of the earth facing the moon experience highest rise of sea water called high tide.

Question 59.
What is low tide?
The opposite part of the earth experience lowest water lends called low tide or Ebb tide.

Question 60.
Mention the place petroleum production in India?
Bombay high

Question 61.
Where is minimita?
It is in Japan. It is the example of marine pollution due to the effects of industries located on the bank of Minamata bay in Japan.

Question 62.
Expand UNCLOS?
United nation conservation on Law of the Sea.

Question 63.
When did sea conservation act passed by UNO?
The UNO has passed the sea convention Law in 1982.

Question 63.
What is Maritime?
The climate which is influenced much by the oceans is known as Maritime.

Question 63.
Why marine conservation is important?
Oceans are the most important parts of the Earth’s natural environment. It is the home for a great variety of life which is threatened by marine pollution. This subchapter introduces causes for marine pollution and needs of its conservation.

### 1st PUC Geography Hydrosphere Two Marks Questions and Answers

Question 1.
Name the four submarine relief features of the Ocean floor. (T. B. Qn)
The major submarine relief features are

1. The Continental slope
2. The Continental Slope
3. The Deep Sea Plains d. The Ocean Deep.

Question 2.
Mention the important features of Continental Shelf. (T. B. Qn)
The shallow submerged extension of continent is called the continental shelf. It is the shallowest part of the ocean with depth varying between 20 to 600m. Its width vary from one ocean to another ranging between 65 to 75km. The width of the shelf is often related to slope of the adjacent land. It is wide along the low lying land and narrow along the mountainous coasts. Nearly 70% of continental shelves are covered with thick deposits of silt, sand, mud and sediments derived from the land.

Question 3.
Why is ocean deep or trench important in the Ocean bottom? (T. B. Qn)
An arc-shaped depression in the deep ocean floor is called ocean deep or trench. These are the deepest parts of the ocean floor. Ocean trenches are produced by the subduction of oceanic crust under continental crust. They resemble step sided valleys or canyons on the land. Trenches or deeps are generally parallel of the coasts facing mountains and along the islands. Great earthquakes and tsunamis generally take birth in this region.

Question 4.
What is sea? Give an example?
A part of the ocean either completely or partially enclosed by land is known as a sea. Sea is an enclosed or semi-enclosed body of salt water. In size it is smaller than an ocean. There are partly enclosed seas like the Caribbean Sea, Mediterranean sea, Arabian Sea, Sea of Japan, The coral sea, the Red sea, etc.

Question 5.
What is Gulf? Give an example?
A gulf is an enlarged bay. It is extensive and penetrated far into the land .A gulf is an inlet of the sea, more enclosed than a bay, penetrating further inland. The Gulf of Mexico, the Persian Gulf, the Gulf of Aden.

Question 6.
What is bay? Given an example?
A bay is a wide curving indentation in a coastline lying between two headlands. Examples of bays are the Bay of Bengal, the Bay of Biscay, the Hudson Bay, etc.

Question 7.
What is Strait? Give an example?
A narrow passage of water connecting two larger water bodies is called a strait, gulf is an enlarged bay. It is a narrow sea passage joining two larger areas of water bodies. The strait of Gibraltar, the Palk’Strait, the Sunda strait, etc. are examples of straits.

Question 8.
What is Echo sounding?
It is a technique of using sound pulses to find the depth of water. Eco sounder is a device for measuring depth of water by sending pressure waves down from the surface and recording the time until the echo returns from the bottom.

Question 9.
What are the three sources of power generation from ocean water?
Three sources of power generation from ocean water are:

1. Tidal energy
2. geothermal energy
3. Energy due to ocean temperature differences.

Question 10.
What is Geo thermal energy?
Geo thermal energy in the oceans is associated with fracture zones and volcanoes. It is used as power resource by many countries like U.S.A, New Zealand.

Question 11.
State any four factors that influence on the distribution of temperature in the ocean.
Factors affecting the distribution of temperature are latitudes, prevailing winds, ocean currents, unequal distribution of land and water, salinity.

Question 12.
Mention any four dissolved minerals in the Ocean water. (T. B. Qn )
The total amount of dissolved solids in the ocean water is called salinity. The ocean’s salinity consists of various elements. They are Sodium chloride, Magnesium chloride, Magnesium sulphate, Calcium sulphate, potassium sulphate, Calcium carbonate, Magnesium bromide etc.

Question 13.
What factors affect the salinity in the Ocean water? (T. B. Qn )
Factors affecting the salinity in the ocean water are: Evaporation, Precipitation, and Fresh water mixing with ocean water, ground water and glaciers increase the percentage of salt content in the ocean. The trade winds drive away saline water to less saline areas resulting the variation of salt content.

Question 14.
Mention the types of distribution of temperature of ocean water?
The distribution of temperature of ocean water is of two types. They are:

1. Horizontal distribution of temperature of ocean water.
2. Vertical distribution of temperature of ocean water.

Question 15.
What is Horizontal distribution of temperature?
The horizontal distribution of temperature of ocean water is influenced by many factors, such as latitudes, depth, salinity prevailing winds, ocean currents, inland water bodies and coastal areas, seasons, rate of evaporation, ice bergs, surrounding lands, etc.

Question 16.
What is vertical distribution of temperature?
The decrease in temperature from the ocean surface with increasing depth is known as vertical distribution of temperature. The temperature of the ocean water decreases steeply with depth except in polar areas. This is called vertical distribution of temperature of ocean water.

Question 17.
Name the two types of Ocean currents. Where are they found? (T. B. Qn)
On the basis latitude and temperature two types of ocean currents are identified. They are:
a. Warm Currents: These are the high temperature ocean currents flowing from the equatorial regions to Polar Regions.
b. Cold currents: These are the cool ocean currents flowing from Polar Regions to equatorial regions.

Question 18.
Mention any four South Indian Ocean currents.

• Mozambique current
• West Australian cold current
• West wind drift.

Question 19.
Name three types of movements in ocean water.
Three types of movements in ocean water

1. area Waves
2. Tides
3. Ocean currents.

Question 20.
Distinguish between crest and trough?
The highest point of the wave is called crest. Each wave has two parts the crest and the trough. The upper part of the wave is called ‘Crest’ and lower part is called “trough’. The horizontal distance between two successive crest or trough’s is called wavelength.

Question 21.
Name the currents of the North Atlantic Ocean.
There are two types current of the North Atlantic Ocean.

1. Warm current: Antilles current, Florida current Gulf stream Norwegian current.
2. Cold currents: There are three cold currents, canary current, East Greenland Current and Labrador Current.

Question 22.
What are the reasons for ocean currents?

• Density of water
• Winds
• Earth’s rotation
• Relief of the ocean floor
• Salinity
• Shape of the coast
• Temperature

These are the major reasons for ocean currents.

Question 23.
Distinguish between High tide and Low tide. (T. B. Qn)
Tides are universal and are an every day activity on the water bodies. Generally 4 tides are observed in a day. Whenever the sea level rises it is called ‘High tide’or ‘flood tide’and when the sea level falls it is known as Tow tide’ or ‘ebb tide’.

Question 24.
Mention any four uses of tides. (T. B. Qn)
Tides are useful to man and a society in various ways. They are:

• Tides increase the depth of water in shallow harbours and help navigation during high ties.
• Tides clean the entrance of ports, harbours and river mouths.
• Tides help fishing and other aquaculture activities.
• Tides promote salt and foam production in the coastal areas.
• Tides promote the generation of tidal energy

Question 25.
Why marine conservation is important?
Oceans are the most important parts of the Earth’s natural environment. It is the home for a great variety of life which is threatened by marine pollution. This subchapter introduces causes for marine pollution and needs of its conservation

Question 26.
Why oceans are the storehouse of natural resources?
As a major component of the natural environment, the oceans are highly useful to mankind. Man obtains resources from the water bodies both in organic resources from the water bodies both organic and inorganic form. At the same time the oceans serve mankind directly as well as indirectly. They are the natural highways are carrying large volume of goods. In addition, they modify the climatic condition. A large number of people are engaged in the extraction of resources from the oceans. Thus it provides a livelihood for millions of people. So the oceans are described as the storehouse of natural resources.

### 1st PUC Geography Hydrosphere Five Marks Questions and Answers

Question 1.
Explain the topography of the Ocean floor with a diagram. (T. B. Qn)
On the basis of the depth, the ocean floor can be divided into four zones, parts or relief features. They are.

1. The continental Shelf: The gently sloping portion of the continent or land that lies submerged below other sea is called the continental shelf. The continental shelf has a very gentle slope. It extends form the shore line to depths between 180 and 200 meters. Average width of the continental shelves is about 48km. The extent of the continental shelf depends on the relief of the broadening land masses.

If the coastal area is a plateau area, the continental shelf will be very broad. On the other hand, if the coastal region is hilly or mountainous, the continental shelf will be very narrow or even absent for example the Atlantic Ocean has 2.3%, the Pacific Ocean has 5.7% and the India Ocean has 4.2%.

2. The continental slope: The zone of steep slope that descends from the edge of the continental shelf to the deep sea plains is called “continental slope”. It is the transitional zone lying between the continental shelf and the deep sea plains. The continental slope is very steep. It extends from 182 meters to 3.600 meters. The angle of the continental slope is 2 to 5 degre3es or even more. It occupies only 8.5% of the total area of the ocean floor. But it varies from ocean to ocean. The Atlantic Ocean has broader continental slopes and accounts for 12.4%. But it is 7% of the Pacific Ocean and 6.5% of the Indian Ocean.

3. The deep sea plains: The level and rolling areas of the ocean floor are generally called deep sea plains or abyssal plains or the ocean plains. They lie between 3,000 and 6,000 meters below other surface of the ocean. They occupy vast area of the ocean floor and account for about 82.7% of the total sea floor. They cover about 90% in the Indian Ocean. Their depth ranges from 5,000 to 6,000 meters. They are covered by oozes, which are the remains of deep sea creatures and plants, and of red volcanic dust.

4. The Ocean Deeps: The long narrow and deep troughs on the ocean floor are known as ‘ocean deep’ or ‘trough’. They cover only 1% of the ocean floor. They are most common neat the coasts where young fold mountains, volcanoes and earthquakes abound. Some they are tectonic in origin. They are the deepest portions of the ocean. Deeps may be caused due to tectonic forces, i.e. faulting earthquakes etc. There are 57known deeps. Of these 32 are found in Pacific Ocean, 19in the Atlantic Ocean and 6 in the Indian Ocean. The deepest trench in the world is Challenger deep located in Mariana Trench to the west of Philippines in the North Pacific Ocean.

Question 2.
Explain different parts of the oceans? Give examples.
Sea: A part of the ocean either completely or partially enclosed by land is known as a sea. Sea is an enclosed or semi-enclosed body of salt water. In size it is smaller than an ocean. There are partly enclosed seas like the Caribbean Sea, Mediterranean sea, Arabian Sea, Sea of Japan, The coral sea, the Red sea, etc.

Bay: Abay is a wide curving indentation in a coastline lying between two headlands. Examples of bays are the Bay of Bengal, the Bay of Biscay, the Hudson Bay

Gulf: A gulf is an enlarged bay. It is extensive and penetrated far into the land .A gulf is an inlet of the sea, more enclosed than a bay, penetrating further inland. The Gulf of Mexico, the Persian Gulf, the Gulf of Aden.

Strait: A narrow passage of water connecting two larger water bodies is called a strait. Gulf is an enlarged bay. It is a narrow sea passage Joining two larger areas of water bodies. The strait of Gibraltar, the Palk Strait, the Sunda strait, etc. are examples of straits.

Question 3.
Describe the salient features of Salinity of the Ocean water. (T. B. Qn)
Salinity of ocean water refers to the amount of dissolved solids in the ocean water. Ocean water consists of various kinds of chemical elements and minerals. Of these constituents, sodium chloride is the most important constituent of ocean water. The other constituents or salts like magnesium chloride, magnesium sulphate, calcium sulphate, potassium sulphate, etc.

Are also present in the ocean or sea water. The ocean water is saline by the rivers which bring huge amount of mineral salt dissolved in water. Another reason for salinity is the evaporation of sea water, by which mineral salt in the sea or ocean water increase every year. Salinity is measured in gram per kg of sea water and it is expressed as part per thousands for examples: 35% it means 35gram of salt in 1000 gram of sea water.

Factors affecting the salinity in the ocean water are: Evaporation, Precipitation, and Fresh water mixing with ocean water, ground water and glaciers increase the percentage of salt content in the ocean. The trade winds drive away saline water to less saline areas resulting the variation of salt content.

Latitudes, precipitation, mixing of fresh water etc. influence on the horizontal and vertical distribution of salinity. The regions near Tropic of Cancer and Capricorn record high salinity due to high temperature, more evaporation, low rainfall and extensive arid and semi-arid areas. While the equatorial region record low sanity because of high temperature and high rainfall. The polar region record least salinity due to very low temperature, evaporation and less rainfall.

Question 4.
Explain the types of distribution of temperature of ocean water.
Temperature of ocean water has much importance. If affects the climate of coastal areas and it is important for marine plans and animals.The average range of the temperature of ocean water is 12’C. But it is varies from place to place and from time to time. This is due to the regional variation of solar energy, prevailing winds, location of their extent etc.

The distribution of temperature in the oceans varies much owing to the latitude, depth and salinity. It is also much influenced by the prevailing winds, currents and nearness to land masses. The main source of heat for ocean water is the sun. The mechanism of heating and cooling ocean water is quite different from that of land. Ocean water warms up and cools down more slowly the diurnal range and annual range of temperature of ocean water is not significant. The temperature of the ocean water varies from place to place.
Distribution of temperature of Ocean water:

The distribution of temperature of ocean water is of two types. They are:

1. Horizontal distribution of temperature of ocean water.
2. Vertical distribution of temperature of ocean water.

a. Horizontal distribution of temperature of Ocean water: The temperature of ocean water decreases from the equator towards the poles. This is known as horizontal distribution of temperature.

b. Vertical Distribution of Temperature: The decrease in temperature from the ocean surface with increasing depth is known as vertical distribution of temperature.

Question 5.
Explain the factors influencing on the salinity of the ocean water.
Salinity of ocean water refers to the amount of dissolved solids in the ocean water. Ocean. water consists of various kinds of chemical elements and minerals.

The distribution of salinity is described both horizontally as well as vertically. Salinity in the ocean water varies from one place to another as well as with the variation of the depth. The amount of salinity to a large extent is determined by the difference between precipitation and evaporation.

a. Evaporation: The amount of salinity and evaporation are directly related. The process of evaporation of water results in the precipitation of salt. Hence higher the amount of evaporation more will be the amount of salinity. Salinity is highest in the belt of tropics, as there is the is excessive evaporation. In the Atlantic Ocean salinity is about 37% near the tropics.

b. Precipitation: In the region of heavy rain fall and cloudiness, more freshwater is added to the ocean water and also the amount of evaporation is reduced. So the amount of salinity is much lower. In the equatorial region the amount of salinity is 35% equivalent to the average of the world.

c. Flow of river water: In the coastal regions where freshwater of rivers enters the sea salinity is greatly reduced.

d. Movement of ocean water: There is a vertical movement of water due to the variation of salinity. On the surface water. Salinity is higher and it sinks due to density and the water rises up from below. Similarly light water from the oceans enters the enclosed seas and dense water flows towards the open seas.

e. Winds: Prevailing strong winds drag surface ocean water and cause the rise of water from below. Thus it reduces the amount of salinity.

Question 6.
Explain the Indian Ocean currents with the help of Maps. (T. B. Qn)
Indian Ocean is different from other ocean in the pattern of its current. This is due to the size of the Indian Ocean, the position of land masses and the seasonal change in the direction of monsoon winds. This is true of the North Indian Ocean. But in the south Indian Ocean, the currents flow as in the Atlantic Ocean and Pacific Ocean.

a. North Indian Ocean currents: The north Indian Ocean currents change their direction. twice a year as result of the alternating monsoon winds greatly influence the ocean currents in summer. The current flows off from the coast of Somali as the Somali current and drifts across the Arabian Sea as the south-west monsoon drift. Finally it joins the North Equatorial current and completes a clockwise circulation.

In winter, the North east winds influence the currents. Hence the currents starting form the stair of Malacca flows along the eastern and western coasts of India as “North East Monsoon Drift” and west ward. Then it turns near the Gulf of Aden to the west flows towards the south and turns east wards near the equator, finally it joins the North Equatorial Current and completes an antilock wise circulation.

b. South Indian Ocean currents: The currents of the south Indian Ocean are not affected by the monsoon winds. So the current of the south Indian Ocean are similar to those in the southern part of the Atlantic Ocean and Pacific Ocean. In the south Indian Ocean, the South-East trade winds blow towards the west and give rise to the equatorial current, i.e. the south equatorial current.

On reaching the eastern coast of Africa, it flow southwards and is divided in two to branches, as it is obstructed by Madagascar Island. One branch enters the Mozambique current. The other branch flows eastwards of Madagascar Island and is known as ‘Madagascar current’. These two currents join again. And flow southwards as the Agulhas current. At 40 S latitude under the influence of the westerlies, it flows eastwards as West wind drift. It is also known as Indian Ocean current.

On reaching Australia, it divides into two branches, One branch turns northwards along the west coast of Australia as the west Australian current. Again turns west and North West, and finally joins the south equatorial current. The other branch moves to the southern coast of Australia and enters the Pacific Ocean.

Question 7.
Briefly explain the types of tides. (T. B. Qn)
The regular and periodic fall and rise of the ocean water is called tides. The rising of water level is known as high tide or flood tide. The falling of water level is known as low tide or ebb tide. The difference between the high tide and low tide is called tidal range or range of tide.

Types of Tides:

There are different types of tides. The important types of tides are as follows:

1. Spring tides: Spring tides occur when the earth, sun and moon are more or less in a straight line. This happens on a full moon day. When the earth is between the sun and the moon or on a new moon day, when the moon is between the sun and the earth. The combined pull of the sun and moon on the earth is very great, and it produces very high tides and very low tides. Such tides are called spring tides.

2. Neap Tides: When the earth, sun and moon are not in a straight line, the sun and the moon are not exerting combined force.

3. So, the gravitational pull is much less. At half moon, the force exerted is the minimum. On the ocean water. Consequently the high tides are not very high and the low tides are not very low. They are called “Neap tides”.

Question 8.
Explain the uses of tides.
The tides are very useful to man in many respects. The main uses are as follows:

• Tides help large ships to enter ports and encourage trades.
• The tides help to keep the harbors clean. The high tides take away the sewage and refuse from the port as well as from mud and silt from river mouths.
• The tides are used to generate electricity and thermal units. They provide inexhaustible and pollution free energy. Example: in Japan and France.
• Tides also help fishermen. They can go out for fishing with the low tide and come back with the high tide. Further tides play an important role in the distribution of plankton and helps in fishing.
• Tides prevent the freezing of seawater along the coast and mouths of rivers.
• Tides helping in producing salt in the coastal area, as water are collected by high tide in the low lying areas.
• Tides prevent the freezing of seawater along the coast and mouths of rivers.
• The ebb tides prevent the harbors to be silted.
• The energy of tides can be utilized for generating hydroelectricity.
• Tides are responsible for eroding the coastline and removing the debris to the sea.

Question 9.
What is Conservation of Ocean? Mention the important measures. (T. B. Qn)
Conservation of ocean means rational uses of ocean resources. So that a harmony between man’s ocean resource requirements and their availability could be maintained. The rational uses of ocean resources by the present generation and the preservation of ocean resources for the future generations, is known as conservation of ocean. It also means the protection of oceans and ocean resources against pollution caused by dumping of oceans and of industrial, agricultural and municipal wastes into oceans by man, oil spill from oil tankers, and nuclear explosions in sea and oceans.

Need for conservation of Oceans: There is a need for conservation of oceans due to the following reasons:

• If oceans are not conserved, all the living organism in the oceans either die or become unsuitable for human consumption.
• Oceans are the storehouse of pearls, corals and sponges. These resources have to be conserved.
• Oceans have oil and natural gas reserves. These reserves have to be conserved.
• Oceans are rich in minerals. These mineral resources have to be rationally exploited.
• Oceans allow the growth of innumerable species of plants. These plants have to be conserved.
• Oceans are the breeding centers of marine fisheries. Thousands of fishes are found in oceans. The marine fishery resources should be exploited rationally.
• Literacy and education programmes on marine features must be initiated and promoted.
• Proper law to be enacted to save sea and ocean.

World wise awareness programme must be arranged to show the pro and cons of the marine pollution etc.

Question 10.
What is Ocean? Write the uses of Ocean.
Ocean refers to the large mass of salt water. About 71% the earth’s surface is covered by water and the remaining 29% is converged by land. There are five important oceans in the world. The five oceans in the order of their sixe are the Pacific Ocean, the Atlantic Ocean, the Indian Ocean, the Antarctica ocean and the Arctic Ocean. The Pacific Ocean is the largest ocean with an area of 166.3 million swq.kms and an average depth of 4kms.

The Atlantic Ocean is the second largest ocean lying between America and Europe and Africa. Its average depth is 3kms. The Indian Ocean is smaller than the Atlantic Ocean. The Antarctic Ocean surrounds the continent of Antarctica.

The Arctic Ocean lies in the north pole region. Uses of Oceans:

The major uses of oceans are:

• Oceans are a very high source of energy. Offshore wells provide petroleum and natural gas, ocean tides can be used to produce hydro-electricity.
• Ocean water can be used for domestic purposes after desalination, which is possible through modern technology
• The marine life provides valuable medicines.
• Oceans encourage commercial fishing.
• Oceans promote ports and harbors and encourage ship-building.
• The provide corals, pearls and shells and sponges
• Oceans serve as natural frontiers and protect a country from foreign invasion.
• Ocean serve as place for dumping wastes.
• Ocean are the source of water vapor and precipitation in general. The winds blowing form the oceans bring rainfall.
• Oceans have a direct control over temperature. This is clear form the fact that the coastal areas are cooled in summer and warmer in winter.
• Oceans are a rich source of food in the form of protein from fish, mammals, sea weds etc.
• Oceans have an abundance of salt and other chemicals, such as sulphate potassium chloride, magnesium chloride etc.

## Karnataka 1st PUC Accountancy Question Bank Practical Oriented Questions

### 1st PUC Practical Oriented Questions

Question 1.
Mention the accounting equation and also give examples with balance sheet.
Accounting equation => Capital = Assets + Liability
Example:
(i) Anil started business with cash Rs. 50000 Accounting equation => Capital = Assets + Liabilities
∴ Capital = 50,000, Assets (cash) = 50,000
∴ 50,000 =50,000

(ii) Purchase goods for Rs. 20,000
Capital = 50,000, Cash = 50,000-20,000 = 30,000 Stock = 2,000
∴ Capital = Cash + Stock
50,000 = 30,000 + 20,000

Question 2.
Write the specimen of accounting cycle.2009, 2011(S), 2008, 2009, 2010(N), 2012(N)

Question 3.
Pass journal entries from the ledger given. 2009(S)

Question 4.
Prepare the format of a bills of exchange. 2008(N), 2009, 2010. 2011(S), 2012(N) and (S)

Question 5.
Give the speciman of an Invoice. 2011(S), 20011(N)

Question 6.
Write the speciman of debit note. 2009, 2010(N) 2009(S)

Question 7.
Write the speciman of credit note.
Specimen of Credit note

Question 8.
Write the accounting equation find out missing figures. 2009(N)

Accounting equation
Asset = Capital + Liabilities
(i) 40,000 = Capital-20,000
Capital = 40,000 -20,000
Capital = 20,000

(ii) Assets = 15,000 + Liabilities + 20,000
∴ Assets = 45,000

(iii) 50,000 = Liabilities + 2,000
Liabilities = 50,000 – 20,000
∴ Liabilities = 30,000

Question 10.
Prepare a trading a/c with five imaginary figures. 2008(N), 2010(S), 2012(N)

Question 11.
Write the specimen of profit and loss account with imaginary figures.

Question 12.
Prepare Balance sheet speciman with imaginary figures. 2010(S)

Question 13.
Prepare a Trial balance with ten imaginary figures. 2008(S)

Question 14.
Prepare a capital ledger a/c for the following transaction.
01- 07 – 2008 Savith commenced business with cash Rs. 20000 machinery Rs. 10000 and building Rs. 15000.
18 – 07 – 2008 She withdraw Rs. 5000
20 – 07 – 2008 the netprofit earned rs. 1,000

Question 15.
Write the specimen of journal and ledger

Question 16.
Write the specimen of simple cash book with five imaginary items. 2010(N), 2008, 2010(S)

Question 17.
Prepare a machinery account for two years with imaginary figures under
(a) Straight line method
(b) Diminishing balance method.

Question 18.
Prepare a statement of affairs with five imaginary figures.

Question 19.
Draw a block diagram of main components of computer.