# Circular Motion HC Verma Exercise Questions Solutions Ch-7

Circular Motion HC Verma Exercise Questions Solutions Ch-7 Concept of Physics Vol-1 for ISC Class-11. Step by Step Solution of Exercise Questions of Ch-7 Circular Motion HC Verma Concept of Physics . Visit official Website CISCE for detail information about ISC Board Class-11 Physics.

## Circular Motion Exercise Questions Solutions HC Verma Ch-7 Concept of Physics Vol-1 for ISC Class-11

 Board ISC and other board Publications Bharti Bhawan Publishers Ch-7 Circular Motion Class 11 Vol 1st writer H C Verma Book Name Concept of Physics Topics Solution of Exercise Questions Page-Number 114, 115, 116

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### Circular Motion Exercise Questions Solutions

(HC Verma Ch-7 Concept of Physics Vol-1 for ISC Class-11)

Page-114

#### Question-1 :-

Find the acceleration of the moon with respect to the earth from the following data : Distance between the earth and the moon = 3.85 × 105 km and the time taken by the moon to complete one revolution around the earth = 27.3 days.

#### Answer-1 :-

Distance between the Earth and the Moon :

r=3.85×105km =3.85×108m

Time taken by the Moon to revolve around the Earth :

T=27.3 days

= 24 × 3600 × 27.3s=2.36 × 106s

Velocity of the Moon :

#### Question-2 :-

Find the acceleration of a particle placed on the surface of the earth at the equator due to earth’s rotation. The diameter of earth = 12800 km and it takes 24 hours for the earth to complete one revolution about its axis.

#### Answer-2 :-

Diameter of the Earth = 12800 km
So, radius of the Earth, R = 6400 km = 6.4 × 106 m

Time period of revolution of the Earth about its axis :

#### Question-3 :-

A particle moves in a circle of radius 1.0 cm at a speed given by v = 2.0 t where v is cm/s and t in seconds.
(a) Find the radial acceleration of the particle at t = 1 s.
(b) Find the tangential acceleration at t = 1 s.
(c) Find the magnitude of the acceleration at t = 1 s.

#### Answer-3 :-

Speed is given as a function of time. Therefore, we have:
v = 2t
Radius of the circle = r = 1 cm
At time = 2 s, we get :
(a) Radial acceleration

#### Question-4 :-

A scooter weighing 150 kg together with its rider moving at 36 km/hr is to take a turn of a radius 30 m. What horizontal force on the scooter is needed to make the turn possible ?

#### Answer-4 :-

Given:
Mass = m = 150 kg
Speed = v = 36 km/hr = 10 m/s
Radius of turn = r = 30 m
Let the horizontal force needed to make the turn be F. We have :

#### Question-5 :-

If the horizontal force needed for the turn in the previous problem is to be supplied by the normal force by the road, what should be the proper angle of banking?

#### Answer-5 :-

Given:
Speed of the scooter = v = 36 km/hr = 10 m/s
Radius of turn = r = 30 m
Let the angle of banking be θ

We have :

#### Question-6 :-

A park has a radius of 10 m. If a vehicle goes round it at an average speed of 18 km/hr, what should be the proper angle of banking?

#### Answer-6 :-

Given:
Speed of the vehicle = v = 18 km/h = 5 m/s
Radius of the park = r = 10 m
Let the angle of banking be θ

#### Question-7 :-

If the road of the previous problem is horizontal (no banking), what should be the minimum friction coefficient so that scooter going at 18 km/hr does not skid?

#### Answer-7 :-

If the road is horizontal (no banking),

we have :

mv²/R = fs

N = mg

Here, fs is the force of friction and N is the normal reaction.

If μ is the friction coefficient, we have :

Friction force = fs = μ N

So, mv²/R = μ mg

Here,
Velocity = v = 5 m/s
Radius = R = 10 m

#### Question-8 :-

A circular road of radius 50 m has the angle of banking equal to 30°. At what speed should a vehicle go on this road so that the friction is not used?

#### Answer-8 :-

Given:
Angle of banking = θ = 30°
Radius = r = 50 m
Assume that the vehicle travels on this road at speed v so that the friction is not used.
We get :

#### Question-9 :-

In the Bohr model of hydrogen atom, the electron is treated as a particle going in a circle with the centre at the proton. The proton itself is assumed to be fixed in an inertial frame. The centripetal force is provided by the Coulomb attraction. In the ground state, the electron goes round the proton in a circle of radius 5.3 × 10−11 m. Find the speed of the electron in the ground state. Mass of the electron = 9.1 × 10−31 kg and charge of the electron = 1.6 × 10−19 C.

#### Answer-9 :-

Given :

Radius of the orbit of the ground state = r = 5.3 × 10−11 m

Mass of the electron = m = 9.1 × 10−31 kg

Charge of electron = q = 1.6 × 10 −19 c

We know:

Centripetal force = Coulomb attraction

Therefore, we have :

#### Question-10 :-

A stone is fastened to one end of a string and is whirled in a vertical circle of radius R. Find the minimum speed the stone can have at the highest point of the circle.

#### Answer-10 :-

Let m be the mass of the stone.
Let v be the velocity of the stone at the highest point.
R is the radius of the circle.
Thus, in a vertical circle and at the highest point,

we have :

mv²/R = mg

⇒ v² = R g

⇒ v = √Rg

#### Question-11 :-

A ceiling fan has a diameter (of the circle through the outer edges of the three blades) of 120 cm and rpm 1500 at full speed. Consider a particle of mass 1 g sticking at the outer end of a blade. How much force does it experience when the fan runs at full speed? Who exerts this force on the particle? How much force does the particle exert on the blade along its surface?

#### Answer-11 :-

Diameter of the fan = 120 cm
∴ Radius of the fan = r = 60 cm = 0.6 m
Mass of the particle = M = 1 g = 0.001 kg
Frequency of revolutions = n = 1500 rev/min = 25 rev/s
Angular velocity = ω = 2πn = 2π × 25 = 57.14 rev/s

Force of the blade on the particle:
F = Mrw2
= (0.001) × 0.6 × (157.14)2
=14.8 N
The moving fan exerts this force on the particle.
The particle also exerts a force of 14.8 N on the blade along its surface.

#### Question-12 :-

A mosquito is sitting on an L.P. record disc rotating on a turn table at revolutions per minute. The distance of the mosquito from the centre of the turn table is 10 cm. Show that the friction coefficient between the record and the mosquito is greater than π2/81. Take g =10 m/s2.

#### Answer-12 :-

r = 10 cm = 0 . 1 m

g = 10 m/ s²

It is given that the mosquito is sitting on the L.P. record disc.Therefore,we have : Friction force ≥ Centrifugal force on the mosquito

#### Question-13 :-

A simple pendulum is suspended from the ceiling of a car taking a turn of radius 10 m at a speed of 36 km/h. Find the angle made by he string of the pendulum with the vertical if this angle does not change during the turn. Take g = 10 m/s2.

#### Answer-13 :-

Speed of the car = v = 36 km/hr = 10 m/s
Acceleration due to gravity = g = 10 m/s

Let T be the tension in the string when the pendulum makes an angle θ with the vertical.
From the figure, we get :

(Page-115)

#### Question-14 :-

The bob of a simple pendulum of length 1 m has mass 100 g and a speed of 1.4 m/s at the lowest point in its path. Find the tension in the string at this instant.

#### Answer-14 :-

Given:
Mass of the bob = m = 100 gm = 0.1 kg
Length of the string = r = 1 m
Speed of bob at the lowest point in its path = 1.4 m/s
Let T be the tension in the string.
From the free body diagram,

we get :

#### Question-15 :-

Suppose the bob of the previous problem has a speed of 1.4 m/s when the string makes an angle of 0.20 radian with the vertical. Find the tension at this instant. You can use cos θ ≈ 1 − θ2/2 and SINθ ≈ θ for small θ.

#### Answer-15 :-

Given:
Mass of the bob = m = 0.1 kg
Length of the circle = R = 1 m
Velocity of the bob = v = 1.4 m/s
Let T be the tension in the string when it makes an angle of 0.20 radian with the vertical.

From the free body diagram, we get :

#### Question-16 :-

Suppose the amplitude of a simple pendulum having a bob of mass m is θ0. Find the tension in the string when the bob is at its extreme position.

#### Answer-16 :-

Let T be the tension in the string at the extreme position.
Velocity of the pendulum is zero at the extreme position.
So, there is no centripetal force on the bob.
∴ T = mgcosθ

#### Question-17 :-

A person stands on a spring balance at the equator.

(a) By what fraction is the balance reading less than his true weight?

(b) If the speed of earth’s rotation is increased by such an amount that the balance reading is half the true weight, what will be the length of the day in this case?

#### Answer-17 :-

(a)

Balance reading = Normal force on the balance by the Earth.
At equator, the normal force (N) on the spring balance :
N = mg − mω2r

True weight = mg
Therefore, we have :

(b)

When the balance reading is half, we have :

#### Question-18 :-

A turn of radius 20 m is banked for the vehicles going at a speed of 36 km/h. If the coefficient of static friction between the road and the tyre is 0.4, what are the possible speeds of a vehicle so that it neither slips down nor skids up?

#### Answer-18 :-

Given:
Speed of vehicles = v = 36 km/hr = 10 m/s
Radius = r = 20 m
Coefficient of static friction = μ = 0.4
Let the road be banked with an angle θ

We have :

if R == N

When the car travels at the maximum speed, it slips upward and μN1 acts downward.
Therefore we have :

Similarly, for the other case, it can be proved that :

Thus, the possible speeds are between 14.7 km/hr and 54 km/hr so that the car neither slips down nor skids up.

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