Physics Test

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Physics Test - 1

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Weekly Quiz Competition

Question 1
3 / -1

What will be the maximum percentage error in the measurement of the density of an iron rod whose radius and length are measured as (4 ± 0.004) cm and (8 ± 0.008) cm respectively? The mass of the wire is measured to be (30 ± 0.03) g.

A
0.04

B
0.4

C
4

D
40

Solution
Question 2
3 / -1

A block of mass m is attached to two inextensible massless strings passes over two ideal massless pullies as shown in figure. The other end of the strings is attached to two masses of 5 kg each. The angle between the strings is 90o in the equilibrium position. The value of mass m (in kg) will be

A
5

B
5√2

C
10

D
5/√2

Solution

From the free body diagram of the central block:
Question 3
3 / -1

In the figure given, two masses of 5 kg and 10 kg are attached to a massless string and the string passes over an ideal frictionless pulley. The system is released from the rest at t = 0 sec. Find the work done by the string on the mass of 5 kg during the time t = 1 sec, just after releasing the system from the rest.

(use acceleration due to gravity, g = 10 m/s2)

A
-50 J

B
-111 J

C
50 J

D
111 J

Solution

From the FBD below

Acceleration of the blocks in the given problem
Question 4
3 / -1

The total kinetic energy of a body rolling without slipping is 65.6 J. The mass of the body is 20 kg and it is moving with a linear velocity of 2 m/s. If the radius of the body is 1 m, find the radius of gyration of the body.

A
0.4 m

B
0.8 m

C
4 m

D
8 m

Solution

Kinetic energy of a rolling body is given as:

This is our required solution.
Question 5
3 / -1

For a two-particle system with total mass M and separated by some distance d, if one of the particles has mass m, what would be the ratio of the masses m/M so that the gravitational force between two particles is maximum?

A
1 : 1

B
1 : 2

C
1 : 3

D
2 : 3

Solution

By the law of gravitation,

This is our required solution.
Question 6
3 / -1

An iron chain of uniform material is lying on a smooth frictionless table. The chain is pulled from one end with a constant force of 10 N. If mass of the chain is given to be 2 kg and the length of the chain is 1 metre, what will be the tension in the rope at a distance of 600 mm from the point of application of force?

A
0.4 N

B
4 N

C
0.6 N

D
6 N

Solution

Acceleration of the chain,

We have to find the tension at point P.

Both the parts OP and PQ will move with the same acceleration.

Equation of motion for OP

T = ma

This is our required solution.
Question 7
3 / -1

A force F = 20 N is acted upon a stationary body of mass 40 kg. If the applied force is in the line of motion of the object, work done by this force in the first second will be

A
5 J

B
0.5 J

C
0.25 J

D
1 J

Solution

W = F.s

F = 20 N

s = ?

By the equation of motion
Question 8
3 / -1

The ratio of moments of inertia of two rings of radii R and nR made of same material about an axis passing through their respective centres is 1 : 8. The value of n is

A
2.828

B
2

C
4

D
0.5

Solution

We know that moment of inertia of a thin circular ring, I = MR²

Where ρ is the linear mass density , is same as materials of both rings is same
So, we have
Question 9
3 / -1

Two bodies each of mass M are kept at a distance of 2a metres from each other. If another body of mass m is projected vertically upward from the midpoint of the line joining the centre of the two bodies with mass M, what would be the minimum initial velocity of the body with mass m to escape the gravitational field of the two-body system?

A

B

C

D

Solution

Let the minimum velocity be v.

By the conservation of energy law,

This is our required solution.
Question 10
3 / -1

N number of water droplets, each of radius r, are accumulated to make a bigger droplet of radius R. The ratio of the final surface energy of the droplets to the initial surface energy will be

A
R/r

B
r/R

C
Nr/R

D
R/Nr

Solution

Let R be the radius of the bigger droplet and 'r' be the radius of the smaller droplet.

Equating the volume of the droplets:

Energy of droplet with radius R = 4πR²T

Energy of N droplets with radius r = N × 4πr²T

Ratio of final surface energy to the initial surface energy

This is our required solution.