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This mock test of Physics Test 1 - Unit And Dimensions, Kinematics, Friction, Laws Of Motion, Work Energy Power for JEE helps you for every JEE entrance exam.
This contains 25 Multiple Choice Questions for JEE Physics Test 1 - Unit And Dimensions, Kinematics, Friction, Laws Of Motion, Work Energy Power (mcq) to study with solutions a complete question bank.
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QUESTION: 1

A national roadway bridge over a canal is in the form of an arc of a circle of radius 49 m. What is the maximum speed with which a car can move without leaving the ground at the highest point? (Take g = 9.8 m/second square)

Solution:

QUESTION: 2

An athlete completes one round of a circular track of radius R in 40 sec. What will be his displacement at the end of 2 min. 20 sec

Solution:
2 minutes 20 seconds=140 seconds or 40+40+40+40/2 seconds One round takes 40 second so it is clear that 3 and1/2 rounds are done. after three round final and initial status are same after 1/2 more round athlete completes half of the circle So final and inial status are on the edges of diameter. Displacement is diatance between final and initial status So it is equal to diameter or 2R.

QUESTION: 3

A body slides down a smooth curved track which is the quadrant of a circle having radius 10 m. The speed of the body at the bottom of the track is

Solution:

QUESTION: 4

A microscope has its main scale with 20 divisions in 1 cm and vernier scale with 25 divisions, the length of which is equal to the length of 24 divisions of main scale. The least count of microscope is:

Solution:

25* vernier scale reading (vsd) = 24* main scale reading (msd)

1vsd = 0.96 msd

Least count = 1msd - 1vsd=1msd-0.96 msd = 0.04 msd

1 msd = 1/20 cm

So, least count

QUESTION: 5

A gardener pushes a lawn roller through a distance of 20m. If he applies a force of 20 kg weight in a direction inclined at 60° to the ground, find the work done by him. (g=9.8m/s2)

Solution:

F = 20kg wt = 20×9.8N

s = 20m

θ = 60°

W = Fscosθ = 20×9.8×20×cos60°

W = 1960J

*Multiple options can be correct

QUESTION: 6

A particle collides with another particle such that the second particle starts moving and the first particle stops just after the collision. Then which of the following conditions must always be valid (m/M= ratio of masses)

Solution:

** Answer :- a,d**

**Solution :- If nothing is given about the nature of collision, we can not presume it as perfectly elastic. Let e be the coefficient of restitution of collision. For inelastic collision e<1. The collision cannot be perfectly inelastic because the particles do not stick together after collision.**

**Just after collision**

**Conservation of linear momentum yields**

**mv=Mv'**

**⇒v'/v=m/M**

**Since e=v'v (numerically)**

**m/M=e**

**Since e≤1, m/M≤1**

**Hence a and d are correct.**

QUESTION: 7

One kilometer is equal to how many miles?

Solution:

QUESTION: 8

If two numerically equal forces P & P acting at a point produce a resultant force of magnitude P itself, then the angle between the two original forces is

Solution:

QUESTION: 9

The angular velocity (in rad/s) of a body rotating at N r.p.m. is

Solution:

Angular velocity is defined as the rate of change of angular displacement with respect to time. It is usually expressed by a Greek letter ω (omega).

Mathematically, angular velocity,

ω =dθ/dt

If a body is rotating at the rate of N r.p.m. (revolutions per minute), then its angular velocity,

ω = 2πΝ / 60 rad/s

QUESTION: 10

An object with a mass of 10kg moves at a constant velocity of 10m/s. A constant force then acts for 4 sec on the object and gives it a speed of 2m/sec in opposite direction. The acceleration produced in it is

Solution:

Change in velocity

-2-10=-12

Time required is 4 seconds

a=Δv/t

=-12/4

=-3 m/s²

QUESTION: 11

Three blocks of masses M_{1},M_{2} & M_{3} are connected by mass-less strings as shown on a friction-less table. They are pulled with a forceT_{3}=40 N. If M_{1}=10kg, M_{2} = 6kg, M_{3} = 4kg then TensionT_{2} will be:

Solution:

QUESTION: 12

Collision between two carom coins is an example for

Solution:
In ideal case,we can take each collision as elastic collision with no loss in energy to the surrounding.
In real case, all collision are inelastic as their is some loss of energy from system to the surrounding.
In case of carrom coins,some part of the kinetic energy of the coin on collision gets dissipated in tha form of heat,sound, friction etc. This leads to decrease in kinetic energy.

QUESTION: 13

The density of material in CGS system of units is 4g/ cm^{2}.In a system of units in which unit of lengths is 10 cm and unit of mass is 100 g, the value of density of material will be

Solution:

QUESTION: 14

A person is holding a bucket by applying a force of 10N. He moves a horizontal distance of 5m and then climbs up a vertical distance of 10m. Find the total work done by him

Solution:

F = 10N, s = 5m, θ = 90°

Work done, W_{1}=Fscosθ = 10×5×cos90° = 0

For vertical motion, the angle between force and displacement is 0°.

Here, F = 10N, s = 10m, θ=0°

Work done, W_{2}=10×10×cos0 = 100J

Total work done = W_{1}+W_{2} = 100J

QUESTION: 15

On the basis of dimensional equation, the maximum number of unknown that can be found, is

Solution:

Since we define the dimension of the physical quantity into three parameters- mass

Thus we can find maximum of three unknown on the basis of dimensional equation.

QUESTION: 16

Two blocks of masses M = 5 kg and m = 3 kg are placed on a horizontal surface as shown in this Fig. The coefficient of friction between the blocks is 0.5 and that between the block M and the horizontal surface is 0.7 . What is the maximum horizontal force F that can be applied to block M so that the two blocks move without slipping ?Take g = 10 ms

Solution:

Let F be the maximum horizontal force that can be applied to block M so that the two blocks move without slipping , i.e. block m doesn't move to backward position.

Lets say, the coefficient of frostion between surface and M is µ_{1} & that between M and m is µ_{2}

When force F acts on the block combination, friction f acts in the opposite direction as shown in the fig above.

So, the acceleration of the combination

Because of this accelaration, a force F' = ma acts in the opposite direction of accelaration. For this the friction between m and M acts in the same direction of accelaration as shown below.

QUESTION: 17

A heavy uniform chain lies on a horizontal table top. If the coefficient of friction between the chain and the table surface is 0.25, then the maximum fraction of the length of the chain, that can hang over one edge of the table is

Solution:

QUESTION: 18

A particle is projected up a 45° rough incline with a velocity 'v'. The coefficient of friction is 0.5 the speed with which it returns back to the straight point is v^{1} then *v ^{1}/v* is

Solution:

QUESTION: 19

A stone is projected upwards from the top of a building with some initial speed and reaches the ground in 5 sec. Now it is allowed to fall with the same initial speed downwards and reaches the ground in 3 sec. If the stone is allowed to fall freely under gravity from the same point, it will reach the ground in

Solution:

QUESTION: 20

If radian correction is not considered in specific heat measurement. The measured value of specific heat will be

Solution:

*Answer can only contain numeric values

QUESTION: 21

Let the resultant of three forces of magnitude 5N, 12N, & 13N acting on a body be zero. If sin23° = 5/13, then angle between 5N & 13 N force (in degree) will be:

Solution:

*Answer can only contain numeric values

QUESTION: 22

A very large number of balls are thrown vertically upwards in quick succession in such a way that the next ball is thrown when the previous one is at maximum height 5 m. Then number of balls thrown per minute will be :-

Solution:

For max. height h = 5 m

initial velocity

So time taken for max-height

So balls are thrown at interval of 1 sec.

So per min. thrown balls are = 60

*Answer can only contain numeric values

QUESTION: 23

A smooth wire of length 20π m is bent in to a circle and kept in a vertical plane. A bead can slide smoothly on the wire. When the circle is rotating with angular speed ω about the vertical diameter AB, as shown in figure, the bead is at rest with respect to the circular ring at position p as shown. Then the value of √3(ω^{2}) will be:-

Solution:

When wire is bend to form a circle,

L = 2πr

20π = 2πr

r = 10m.

F.B.D of bead :

As bead is at rest,

*Answer can only contain numeric values

QUESTION: 24

A block of mass m is placed at the distance of 1m from the centre of a round turn table (A table whose surface can rotate about central axis). Table starts from rest and rotates with constant angular acceleration of 3 rad/s^{2}. The coefficient of friction between block and table is µ = 0.5. At time t = x/3 sec from the start of motion the block is just about to slip. Find value of x.

Solution:

There will be net force due to both the radial and tangential acceleration,

*Answer can only contain numeric values

QUESTION: 25

A block of mass 2 kg is suspended from a string, is released from a height of 5m as shown in figure then what will be the impulse when string. Just becomes tight

Solution:

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