NCERT Solutions Class 11th Physics. FREE Flip-BOOK by Study Innovations. 461 Pages

No
No
No
Yes
Escape velocity of a body from the Earth is given by the relation:

g = Acceleration due to gravity
R = Radius of the Earth
It is clear from equation (i) that escape velocity vesc is independent of the mass of the body
and the direction of its projection. However, it depends on gravitational potential at the
point from where the body is launched. Since this potential marginally depends on the
height of the point, escape velocity also marginally depends on these factors.

Question 8.8:
A comet orbits the Sun in a highly elliptical orbit. Does the comet have a constant (a)
linear speed, (b) angular speed, (c) angular momentum, (d) kinetic energy, (e) potential
energy, (f) total energy throughout its orbit? Neglect any mass loss of the comet when it
comes very close to the Sun.
Answer

No
No
Yes
No
No

Yes
Angular momentum and total energy at all points of the orbit of a comet moving in a
highly elliptical orbit around the Sun are constant. Its linear speed, angular speed, kinetic,
and potential energy varies from point to point in the orbit.

Question 8.9:
Which of the following symptoms is likely to afflict an astronaut in space (a) swollen
feet, (b) swollen face, (c) headache, (d) orientational problem?
Answer

Answer: (b), (c), and (d)
Legs hold the entire mass of a body in standing position due to gravitational pull. In
space, an astronaut feels weightlessness because of the absence of gravity. Therefore,
swollen feet of an astronaut do not affect him/her in space.
A swollen face is caused generally because of apparent weightlessness in space. Sense
organs such as eyes, ears nose, and mouth constitute a person’s face. This symptom can
affect an astronaut in space.
Headaches are caused because of mental strain. It can affect the working of an astronaut
in space.
Space has different orientations. Therefore, orientational problem can affect an astronaut
in space.

Question 8.10:
Choose the correct answer from among the given ones:
The gravitational intensity at the centre of a hemispherical shell of uniform mass density
has the direction indicated by the arrow (see Fig 8.12) (i) a, (ii) b, (iii) c, (iv) O.

Answer

Answer: (iii)
Gravitational potential (V) is constant at all points in a spherical shell. Hence, the

gravitational potential gradient is zero everywhere inside the spherical shell. The

gravitational potential gradient is equal to the negative of gravitational intensity. Hence,

intensity is also zero at all points inside the spherical shell. This indicates that

gravitational forces acting at a point in a spherical shell are symmetric.

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

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

Question 8.11:
Choose the correct answer from among the given ones:

For the problem 8.10, the direction of the gravitational intensity at an arbitrary point P is
indicated by the arrow (i) d, (ii) e, (iii) f, (iv) g.

Answer

Answer: (ii)
Gravitational potential (V) is constant at all points in a spherical shell. Hence, the

gravitational potential gradient is zero everywhere inside the spherical shell. The

gravitational potential gradient is equal to the negative of gravitational intensity. Hence,

intensity is also zero at all points inside the spherical shell. This indicates that

gravitational forces acting at a point in a spherical shell are symmetric.

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

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

Question 8.12:
A rocket is fired from the earth towards the sun. At what distance from the earth’s centre
is the gravitational force on the rocket zero? Mass of the sun = 2 ×1030 kg, mass of the
earth = 6 × 1024 kg. Neglect the effect of other planets etc. (orbital radius = 1.5 × 1011 m).
Answer

Mass of the Sun, Ms = 2 × 1030 kg
Mass of the Earth, Me = 6 × 10 24 kg
Orbital radius, r = 1.5 × 1011 m
Mass of the rocket = m

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

Question 8.13:
How will you ‘weigh the sun’, that is estimate its mass? The mean orbital radius of the
earth around the sun is 1.5 × 108 km.
Answer

Orbital radius of the Earth around the Sun, r = 1.5 × 1011 m
Time taken by the Earth to complete one revolution around the Sun,
T = 1 year = 365.25 days
= 365.25 × 24 × 60 × 60 s
Universal gravitational constant, G = 6.67 × 10–11 Nm2 kg–2
Thus, mass of the Sun can be calculated using the relation,

Hence, the mass of the Sun is 2 × 1030 kg.

Question 8.14:
A Saturn year is 29.5 times the earth year. How far is the Saturn from the sun if the earth
is 1.50 ×108 km away from the sun?
Answer

Distance of the Earth from the Sun, re = 1.5 × 108 km = 1.5 × 1011 m
Time period of the Earth = Te
Time period of Saturn, Ts = 29. 5 Te
Distance of Saturn from the Sun = rs

From Kepler’s third law of planetary motion, we have
For Saturn and Sun, we can write

Hence, the distance between Saturn and the Sun is .

Question 8.15:

A body weighs 63 N on the surface of the earth. What is the gravitational force on it due
to the earth at a height equal to half the radius of the earth?

Answer

Weight of the body, W = 63 N
Acceleration due to gravity at height h from the Earth’s surface is given by the relation:

Where,
g = Acceleration due to gravity on the Earth’s surface
Re = Radius of the Earth

Weight of a body of mass m at height h is given as:

Question 8.16:
Assuming the earth to be a sphere of uniform mass density, how much would a body
weigh half way down to the centre of the earth if it weighed 250 N on the surface?
Answer

Weight of a body of mass m at the Earth’s surface, W = mg = 250 N
Body of mass m is located at depth,
Where,

= Radius of the Earth

Acceleration due to gravity at depth g (d) is given by the relation:

Weight of the body at depth d,

Question 8.17:
A rocket is fired vertically with a speed of 5 km s–1 from the earth’s surface. How far
from the earth does the rocket go before returning to the earth? Mass of the earth = 6.0 ×
1024 kg; mean radius of the earth = 6.4 × 106 m; G= 6.67 × 10–11 N m2 kg–2.
Answer

Answer: 8 × 106 m from the centre of the Earth
Velocity of the rocket, v = 5 km/s = 5 × 103 m/s
Mass of the Earth,
Radius of the Earth,
Height reached by rocket mass, m = h
At the surface of the Earth,
Total energy of the rocket = Kinetic energy + Potential energy

At highest point h,
Total energy of the rocket
From the law of conservation of energy, we have
Total energy of the rocket at the Earth’s surface = Total energy at height h

Height achieved by the rocket with respect to the centre of the Earth

Question 8.18:
The escape speed of a projectile on the earth’s surface is 11.2 km s–1. A body is projected
out with thrice this speed. What is the speed of the body far away from the earth? Ignore
the presence of the sun and other planets.
Answer

Escape velocity of a projectile from the Earth, vesc = 11.2 km/s
Projection velocity of the projectile, vp = 3vesc
Mass of the projectile = m
Velocity of the projectile far away from the Earth = vf

Total energy of the projectile on the Earth
Gravitational potential energy of the projectile far away from the Earth is zero.

Total energy of the projectile far away from the Earth =
From the law of conservation of energy, we have

Question 8.19:
A satellite orbits the earth at a height of 400 km above the surface. How much energy
must be expended to rocket the satellite out of the earth’s gravitational influence? Mass of
the satellite = 200 kg; mass of the earth = 6.0 ×1024 kg; radius of the earth = 6.4 ×106 m;
G = 6.67 × 10–11 N m2 kg–2.
Answer

Mass of the Earth, M = 6.0 × 1024 kg
Mass of the satellite, m = 200 kg
Radius of the Earth, Re = 6.4 × 106 m
Universal gravitational constant, G = 6.67 × 10–11 Nm2kg–2
Height of the satellite, h = 400 km = 4 × 105 m = 0.4 ×106 m

Total energy of the satellite at height h

Orbital velocity of the satellite, v =

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

Question 8.20:
Two stars each of one solar mass (= 2× 1030 kg) are approaching each other for a head on
collision. When they are a distance 109 km, their speeds are negligible. What is the speed
with which they collide? The radius of each star is 104 km. Assume the stars to remain
undistorted until they collide. (Use the known value of G).
Answer

Mass of each star, M = 2 × 1030 kg
Radius of each star, R = 104 km = 107 m
Distance between the stars, r = 109 km = 1012m
For negligible speeds, v = 0 total energy of two stars separated at distance r

Now, consider the case when the stars are about to collide:
Velocity of the stars = v
Distance between the centers of the stars = 2R

Total kinetic energy of both stars

Total potential energy of both stars
Total energy of the two stars =
Using the law of conservation of energy, we can write:

Question 8.21:
Two heavy spheres each of mass 100 kg and radius 0.10 m are placed 1.0 m apart on a
horizontal table. What is the gravitational force and potential at the mid point of the line
joining the centers of the spheres? Is an object placed at that point in equilibrium? If so, is
the equilibrium stable or unstable?
Answer

Answer:
0;
–2.7 × 10–8 J /kg;
Yes;

Unstable
Explanation:
The situation is represented in the given figure:

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

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

Question 8.22:
As you have learnt in the text, a geostationary satellite orbits the earth at a height of
nearly 36,000 km from the surface of the earth. What is the potential due to earth’s
gravity at the site of this satellite? (Take the potential energy at infinity to be zero). Mass
of the earth = 6.0 × 1024 kg, radius = 6400 km.
Answer

Mass of the Earth, M = 6.0 × 1024 kg
Radius of the Earth, R = 6400 km = 6.4 × 106 m
Height of a geostationary satellite from the surface of the Earth,
h = 36000 km = 3.6 × 107 m
Gravitational potential energy due to Earth’s gravity at height h,

Question 8.23:
A star 2.5 times the mass of the sun and collapsed to a size of 12 km rotates with a speed
of 1.2 rev. per second. (Extremely compact stars of this kind are known as neutron stars.
Certain stellar objects called pulsars belong to this category). Will an object placed on its
equator remain stuck to its surface due to gravity? (Mass of the sun = 2 × 1030 kg).
Answer

Answer: Yes
A body gets stuck to the surface of a star if the inward gravitational force is greater than
the outward centrifugal force caused by the rotation of the star.
Gravitational force, fg
Where,
M = Mass of the star = 2.5 × 2 × 1030 = 5 × 1030 kg

m = Mass of the body
R = Radius of the star = 12 km = 1.2 ×104 m

Centrifugal force, fc = mrω2
ω = Angular speed = 2πν
ν = Angular frequency = 1.2 rev s–1
fc = mR (2πν)2
= m × (1.2 ×104) × 4 × (3.14)2 × (1.2)2 = 1.7 ×105m N
Since fg > fc, the body will remain stuck to the surface of the star.

Question 8.24:
A spaceship is stationed on Mars. How much energy must be expended on the spaceship
to launch it out of the solar system? Mass of the space ship = 1000 kg; mass of the Sun =
2 × 1030 kg; mass of mars = 6.4 × 1023 kg; radius of mars = 3395 km; radius of the orbit
of mars = 2.28 × 108kg; G= 6.67 × 10–11 m2kg–2.
Answer

Mass of the spaceship, ms = 1000 kg
Mass of the Sun, M = 2 × 1030 kg
Mass of Mars, mm = 6.4 × 10 23 kg
Orbital radius of Mars, R = 2.28 × 108 kg =2.28 × 1011m
Radius of Mars, r = 3395 km = 3.395 × 106 m
Universal gravitational constant, G = 6.67 × 10–11 m2kg–2

Potential energy of the spaceship due to the gravitational attraction of the Sun
Potential energy of the spaceship due to the gravitational attraction of Mars
Since the spaceship is stationed on Mars, its velocity and hence, its kinetic energy will be
zero.
Total energy of the spaceship

The negative sign indicates that the system is in bound state.
Energy required for launching the spaceship out of the solar system
= – (Total energy of the spaceship)

Question 8.25:
A rocket is fired ‘vertically’ from the surface of mars with a speed of 2 km s–1. If 20% of
its initial energy is lost due to Martian atmospheric resistance, how far will the rocket go
from the surface of mars before returning to it? Mass of mars = 6.4× 1023 kg; radius of
mars = 3395 km; G = 6.67× 10-11 N m2 kg–2.
Answer

Initial velocity of the rocket, v = 2 km/s = 2 × 103 m/s
Mass of Mars, M = 6.4 × 1023 kg
Radius of Mars, R = 3395 km = 3.395 × 106 m
Universal gravitational constant, G = 6.67× 10–11 N m2 kg–2
Mass of the rocket = m

Initial kinetic energy of the rocket =

Initial potential energy of the rocket

Total initial energy
If 20 % of initial kinetic energy is lost due to Martian atmospheric resistance, then only
80 % of its kinetic energy helps in reaching a height.

Total initial energy available
Maximum height reached by the rocket = h
At this height, the velocity and hence, the kinetic energy of the rocket will become zero.

Total energy of the rocket at height h
Applying the law of conservation of energy for the rocket, we can write:



Question 9.1:
A steel wire of length 4.7 m and cross-sectional area 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?
Answer

Length of the steel wire, L1 = 4.7 m
Area of cross-section of the steel wire, A1 = 3.0 × 10–5 m2
Length of the copper wire, L2 = 3.5 m
Area of cross-section of the copper wire, A2 = 4.0 × 10–5 m2
Change in length = ΔL1 = ΔL2 = ΔL
Force applied in both the cases = F
Young’s modulus of the steel wire:

… (i)
Young’s modulus of the copper wire:

Dividing (i) by (ii), we get:

The ratio of Young’s modulus of steel to that of copper is 1.79 : 1.

Question 9.2:
Figure 9.11 shows the strain-stress curve for a given material. What are (a) Young’s
modulus and (b) approximate yield strength for this material?

Answer

It is clear from the given graph that for stress 150 × 106 N/m2, strain is 0.002.
∴Young’s modulus, Y

Hence, Young’s modulus for the given material is 7.5 ×1010 N/m2.
The yield strength of a material is the maximum stress that the material can sustain
without crossing the elastic limit.
It is clear from the given graph that the approximate yield strength of this material is 300
× 106 Nm/2 or 3 × 108 N/m2.

Question 9.3:

The stress-strain graphs for materials A and B are shown in Fig. 9.12.

The graphs are drawn to the same scale.
Which of the materials has the greater Young’s modulus?
Which of the two is the stronger material?
Answer

Answer: (a) A (b) A
For a given strain, the stress for material A is more than it is for material B, as shown in
the two graphs.

Young’s modulus
For a given strain, if the stress for a material is more, then Young’s modulus is also
greater for that material. Therefore, Young’s modulus for material A is greater than it is
for material B.
The amount of stress required for fracturing a material, corresponding to its fracture
point, gives the strength of that material. Fracture point is the extreme point in a stress-
strain curve. It can be observed that material A can withstand more strain than material B.
Hence, material A is stronger than material B.

Question 9.4:
Read the following two statements below carefully and state, with reasons, if it is true or
false.

The Young’s modulus of rubber is greater than that of steel;
The stretching of a coil is determined by its shear modulus.
Answer

Answer: (a) False (b) True
For a given stress, the strain in rubber is more than it is in steel.
Young’s modulus,
For a constant stress:
Hence, Young’s modulus for rubber is less than it is for steel.
Shear modulus is the ratio of the applied stress to the change in the shape of a body. The
stretching of a coil changes its shape. Hence, shear modulus of elasticity is involved in
this process.

Question 9.5:
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 the steel and the brass wires.

Answer

Elongation of the steel wire = 1.49 × 10–4 m
Elongation of the brass wire = 1.3 × 10–4 m
Diameter of the wires, d = 0.25 m

Hence, the radius of the wires, = 0.125 cm

Length of the steel wire, L1 = 1.5 m
Length of the brass wire, L2 = 1.0 m

Total force exerted on the steel wire:

F1 = (4 + 6) g = 10 × 9.8 = 98 N
Young’s modulus for steel:

Where,
ΔL1 = Change in the length of the steel wire

A1 = Area of cross-section of the steel wire
Young’s modulus of steel, Y1 = 2.0 × 1011 Pa

Total force on the brass wire:
F2 = 6 × 9.8 = 58.8 N

Young’s modulus for brass:

Elongation of the steel wire = 1.49 × 10–4 m
Elongation of the brass wire = 1.3 × 10–4 m

Question 9.6:
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?
Answer

Edge of the aluminium cube, L = 10 cm = 0.1 m
The mass attached to the cube, m = 100 kg
Shear modulus (η) of aluminium = 25 GPa = 25 × 109 Pa

Shear modulus, η
Where,
F = Applied force = mg = 100 × 9.8 = 980 N
A = Area of one of the faces of the cube = 0.1 × 0.1 = 0.01 m2
ΔL = Vertical deflection of the cube

= 3.92 × 10–7 m
The vertical deflection of this face of the cube is 3.92 ×10–7 m.

Question 9.7:
Four identical hollow cylindrical columns of mild steel support a big structure of mass
50,000 kg. The inner and outer radii of each column are 30 cm and 60 cm respectively.
Assuming the load distribution to be uniform, calculate the compressional strain of each
column.
Answer

Mass of the big structure, M = 50,000 kg
Inner radius of the column, r = 30 cm = 0.3 m
Outer radius of the column, R = 60 cm = 0.6 m
Young’s modulus of steel, Y = 2 × 1011 Pa

Total force exerted, F = Mg = 50000 × 9.8 N = 122500 N
Stress = Force exerted on a single column
Young’s modulus, Y

Where,
Area, A = π (R2 – r2) = π ((0.6)2 – (0.3)2)

= 7.22 × 10–7
Hence, the compressional strain of each column is 7.22 × 10–7.

Question 9.8:
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?
Answer

Length of the piece of copper, l = 19.1 mm = 19.1 × 10–3 m
Breadth of the piece of copper, b = 15.2 mm = 15.2 × 10–3 m
Area of the copper piece:
A=l×b
= 19.1 × 10–3 × 15.2 × 10–3

= 2.9 × 10–4 m2
Tension force applied on the piece of copper, F = 44500 N
Modulus of elasticity of copper, η = 42 × 109 N/m2

Modulus of elasticity, η

= 3.65 × 10–3

Question 9.9:
A steel cable with a radius of 1.5 cm supports a chairlift at a ski area. If the maximum
stress is not to exceed 108 N m–2, what is the maximum load the cable can support?
Answer

Radius of the steel cable, r = 1.5 cm = 0.015 m
Maximum allowable stress = 108 N m–2
Maximum stress =
∴Maximum force = Maximum stress × Area of cross-section
= 108 × π (0.015)2
= 7.065 × 104 N
Hence, the cable can support the maximum load of 7.065 × 104 N.

Question 9.10:
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 ratio of their
diameters if each is to have the same tension.
Answer

The tension force acting on each wire is the same. Thus, the extension in each case is the
same. Since the wires are of the same length, the strain will also be the same.
The relation for Young’s modulus is given as:

Where,
F = Tension force
A = Area of cross-section
d = Diameter of the wire

It can be inferred from equation (i) that
Young’s modulus for iron, Y1 = 190 × 109 Pa
Diameter of the iron wire = d1
Young’s modulus for copper, Y2 = 110 × 109 Pa
Diameter of the copper wire = d2
Therefore, the ratio of their diameters is given as:

Question 9.11:
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.
Answer

Mass, m = 14.5 kg
Length of the steel wire, l = 1.0 m
Angular velocity, ω = 2 rev/s
Cross-sectional area of the wire, a = 0.065 cm2
Let δl be the elongation of the wire when the mass is at the lowest point of its path.
When the mass is placed at the position of the vertical circle, the total force on the mass
is:
F = mg + mlω2
= 14.5 × 9.8 + 14.5 × 1 × (2)2
= 200.1 N

Young’s modulus for steel = 2 × 1011 Pa

Hence, the elongation of the wire is 1.539 × 10–4 m.

Question 9.12:
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.
Answer

Initial volume, V1 = 100.0l = 100.0 × 10 –3 m3
Final volume, V2 = 100.5 l = 100.5 ×10 –3 m3
Increase in volume, ΔV = V2 – V1 = 0.5 × 10–3 m3
Increase in pressure, Δp = 100.0 atm = 100 × 1.013 × 105 Pa

This ratio is very high because air is more compressible than water.

Question 9.13:
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?
Answer

Let the given depth be h.
Pressure at the given depth, p = 80.0 atm = 80 × 1.01 × 105 Pa
Density of water at the surface, ρ1 = 1.03 × 103 kg m–3
Let ρ2 be the density of water at the depth h.
Let V1 be the volume of water of mass m at the surface.
Let V2 be the volume of water of mass m at the depth h.
Let ΔV be the change in volume.

For equations (i) and (ii), we get:

Therefore, the density of water at the given depth (h) is 1.034 × 103 kg m–3.
Question 9.14:
Compute the fractional change in volume of a glass slab, when subjected to a hydraulic
pressure of 10 atm.

Answer

Hydraulic pressure exerted on the glass slab, p = 10 atm = 10 × 1.013 × 105 Pa
Bulk modulus of glass, B = 37 × 109 Nm–2

Where,
= Fractional change in volume

Hence, the fractional change in the volume of the glass slab is 2.73 × 10–5.

Question 9.15:
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.
Answer

Length of an edge of the solid copper cube, l = 10 cm = 0.1 m
Hydraulic pressure, p = 7.0 ×106 Pa
Bulk modulus of copper, B = 140 × 109 Pa

Where,
= Volumetric strain

ΔV = Change in volume
V = Original volume.

Original volume of the cube, V = l3

Therefore, the volume contraction of the solid copper cube is 5 × 10–2 cm–3.

Question 9.16:
How much should the pressure on a litre of water be changed to compress it by 0.10%?
Answer

Volume of water, V = 1 L
It is given that water is to be compressed by 0.10%.

Therefore, the pressure on water should be 2.2 ×106 Nm–2.

Question 9.17:
Anvils made of single crystals of diamond, with the shape as shown in Fig. 9.14, 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?

Answer

Diameter of the cones at the narrow ends, d = 0.50 mm = 0.5 × 10–3 m

Compressional force, F = 50000 N
Pressure at the tip of the anvil:

Therefore, the pressure at the tip of the anvil is 2.55 × 1011 Pa.
Question 9.18:
A rod of length 1.05 m having negligible mass is supported at its ends by two wires of
steel (wire A) and aluminium (wire B) of equal lengths as shown in 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 (a) equal stresses and (b)
equal strains in both steel and aluminium wires.

Answer

Answer: (a) 0.7 m from the steel-wire end

0.432 m from the steel-wire end
Cross-sectional area of wire A, a1 = 1.0 mm2 = 1.0 × 10–6 m2
Cross-sectional area of wire B, a2 = 2.0 mm2 = 2.0 × 10–6 m2
Young’s modulus for steel, Y1 = 2 × 1011 Nm–2
Young’s modulus for aluminium, Y2 = 7.0 ×1010 Nm–2
Let a small mass m be suspended to the rod at a distance y from the end where wire A is
attached.

If the two wires have equal stresses, then:

Where,
F1 = Force exerted on the steel wire
F2 = Force exerted on the aluminum wire

The situation is shown in the following figure.

Taking torque about the point of suspension, we have:

Using equations (i) and (ii), we can write:

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

If the strain in the two wires is equal, then:

Taking torque about the point where mass m, is suspended at a distance y1 from the side
where wire A attached, we get:
F1y1 = F2 (1.05 – y1)

… (iii)

Using equations (iii) and (iv), we get:

In order to produce an equal strain in the two wires, the mass should be suspended at a
distance of 0.432 m from the end where wire A is attached.

Question 9.19:
A mild steel wire of length 1.0 m and cross-sectional area 0.50 × 10–2 cm2 is stretched,
well within its elastic limit, horizontally between two pillars. A mass of 100 g is
suspended from the mid-point of the wire. Calculate the depression at the midpoint.
Answer

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

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

Let T be the tension in the wire.
∴mg = 2T cosθ
Using the figure, it can be written as:

Expanding the expression and eliminating the higher terms:
Young’s modulus of steel, Y =

Hence, the depression at the midpoint is 0.0106 m.

Question 9.20:
Two strips of metal are riveted together at their ends by four rivets, each of diameter 6.0
mm. What is the maximum tension that can be exerted by the riveted strip if the shearing
stress on the rivet is not to exceed 6.9 × 107 Pa? Assume that each rivet is to carry one
quarter of the load.
Answer

Diameter of the metal strip, d = 6.0 mm = 6.0 × 10–3 m

Maximum shearing stress = 6.9 × 107 Pa

Maximum force = Maximum stress × Area
= 6.9 × 107 × π × (r) 2
= 6.9 × 107 × π × (3 ×10–3)2
= 1949.94 N
Each rivet carries one quarter of the load.
∴ Maximum tension on each rivet = 4 × 1949.94 = 7799.76 N

Question 9.21:
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?
Answer

Water pressure at the bottom, p = 1.1 × 108 Pa
Initial volume of the steel ball, V = 0.32 m3
Bulk modulus of steel, B = 1.6 × 1011 Nm–2
The ball falls at the bottom of the Pacific Ocean, which is 11 km beneath the surface.
Let the change in the volume of the ball on reaching the bottom of the trench be ΔV.

Therefore, the change in volume of the ball on reaching the bottom of the trench is 2.2 ×
10–4 m3.

Question 10.1:
Explain why
The blood pressure in humans is greater at the feet than at the brain
Atmospheric pressure at a height of about 6 km decreases to nearly half of its value at the
sea level, though the height of the atmosphere is more than 100 km
Hydrostatic pressure is a scalar quantity even though pressure is force divided by area.

Answer

The pressure of a liquid is given by the relation:
P = hρg
Where,
P = Pressure
h = Height of the liquid column
ρ = Density of the liquid
g = Acceleration due to the gravity
It can be inferred that pressure is directly proportional to height. Hence, the blood
pressure in human vessels depends on the height of the blood column in the body. The
height of the blood column is more at the feet than it is at the brain. Hence, the blood
pressure at the feet is more than it is at the brain.
Density of air is the maximum near the sea level. Density of air decreases with increase in
height from the surface. At a height of about 6 km, density decreases to nearly half of its
value at the sea level. Atmospheric pressure is proportional to density. Hence, at a height
of 6 km from the surface, it decreases to nearly half of its value at the sea level.
When force is applied on a liquid, the pressure in the liquid is transmitted in all directions.
Hence, hydrostatic pressure does not have a fixed direction and it is a scalar physical
quantity.

Question 10.2:

Explain why
The angle of contact of mercury with glass is obtuse, while that of water with glass is
acute.
Water on a clean glass surface tends to spread out while mercury on the same surface
tends to form drops. (Put differently, water wets glass while mercury does not.)
Surface tension of a liquid is independent of the area of the surface
Water with detergent dissolved in it should have small angles of contact.
A drop of liquid under no external forces is always spherical in shape
Answer

The angle between the tangent to the liquid surface at the point of contact and the surface
inside the liquid is called the angle of contact (θ), as shown in the given figure.

Sla, Ssa, and Ssl are the respective interfacial tensions between the liquid-air, solid-air, and
solid-liquid interfaces. At the line of contact, the surface forces between the three media
must be in equilibrium, i.e.,

The angle of contact θ , is obtuse if Ssa < Sla (as in the case of mercury on glass). This
angle is acute if Ssl < Sla (as in the case of water on glass).
Mercury molecules (which make an obtuse angle with glass) have a strong force of

attraction between themselves and a weak force of attraction toward solids. Hence, they
tend to form drops.

On the other hand, water molecules make acute angles with glass. They have a weak
force of attraction between themselves and a strong force of attraction toward solids.
Hence, they tend to spread out.

Surface tension is the force acting per unit length at the interface between the plane of a
liquid and any other surface. This force is independent of the area of the liquid surface.
Hence, surface tension is also independent of the area of the liquid surface.

Water with detergent dissolved in it has small angles of contact (θ). This is because for a
small θ, there is a fast capillary rise of the detergent in the cloth. The capillary rise of a
liquid is directly proportional to the cosine of the angle of contact (θ). If θ is small, then
cosθ will be large and the rise of the detergent water in the cloth will be fast.

A liquid tends to acquire the minimum surface area because of the presence of surface
tension. The surface area of a sphere is the minimum for a given volume. Hence, under no
external forces, liquid drops always take spherical shape.

Question 10.3:

Fill in the blanks using the word(s) from the list appended with each statement:

Surface tension of liquids generally . . . with temperatures (increases / decreases)

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

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)

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

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

Answer

decreases

The surface tension of a liquid is inversely proportional to temperature.

increases; decreases

Most fluids offer resistance to their motion. This is like internal mechanical friction,
known as viscosity. Viscosity of gases increases with temperature, while viscosity of
liquids decreases with temperature.

Shear strain; Rate of shear strain

With reference to the elastic modulus of rigidity for solids, the shearing force is
proportional to the shear strain. With reference to the elastic modulus of rigidity for
fluids, the shearing force is proportional to the rate of shear strain.

Conservation of mass/Bernoulli’s principle

For a steady-flowing fluid, an increase in its flow speed at a constriction follows the
conservation of mass/Bernoulli’s principle.

Greater

For the model of a plane in a wind tunnel, turbulence occurs at a greater speed than it
does for an actual plane. This follows from Bernoulli’s principle and different Reynolds’
numbers are associated with the motions of the two planes.

Question 10.4:

Explain why

To keep a piece of paper horizontal, you should blow over, not under, it

When we try to close a water tap with our fingers, fast jets of water gush through the
openings between our fingers

The size of the needle of a syringe controls flow rate better than the thumb pressure
exerted by a doctor while administering an injection

A fluid flowing out of a small hole in a vessel results in a backward thrust on the vessel

A spinning cricket ball in air does not follow a parabolic trajectory

Answer

When air is blown under a paper, the velocity of air is greater under the paper than it is
above it. As per Bernoulli’s principle, atmospheric pressure reduces under the paper. This
makes the paper fall. To keep a piece of paper horizontal, one should blow over it. This
increases the velocity of air above the paper. As per Bernoulli’s principle, atmospheric
pressure reduces above the paper and the paper remains horizontal.

According to the equation of continuity:

Area × Velocity = Constant

For a smaller opening, the velocity of flow of a fluid is greater than it is when the opening
is bigger. When we try to close a tap of water with our fingers, fast jets of water gush
through the openings between our fingers. This is because very small openings are left for
the water to flow out of the pipe. Hence, area and velocity are inversely proportional to
each other.

The small opening of a syringe needle controls the velocity of the blood flowing out. This
is because of the equation of continuity. At the constriction point of the syringe system,
the flow rate suddenly increases to a high value for a constant thumb pressure applied.

When a fluid flows out from a small hole in a vessel, the vessel receives a backward
thrust. A fluid flowing out from a small hole has a large velocity according to the
equation of continuity:

Area × Velocity = Constant

According to the law of conservation of momentum, the vessel attains a backward
velocity because there are no external forces acting on the system.

A spinning cricket ball has two simultaneous motions – rotatory and linear. These two
types of motion oppose the effect of each other. This decreases the velocity of air flowing
below the ball. Hence, the pressure on the upper side of the ball becomes lesser than that
on the lower side. An upward force acts upon the ball. Therefore, the ball takes a curved
path. It does not follow a parabolic path.

Question 10.5:

A 50 kg girl wearing high heel shoes balances on a single heel. The heel is circular with a
diameter 1.0 cm. What is the pressure exerted by the heel on the horizontal floor?

Answer