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QUESTION: 1

A body of mass m is suspended by two strings makings angle α and β with the horizontal. Find the tension T_{1} in the strings.

Solution:

Force acting on the system are shown in Fig.

For equilibrium in horizontal direction.

T_{1} cos∝ = T_{2} cosβ .....(1)

For equilibrium in vertical direction,

T_{1} sin∝ = T_{2} sinβ - mg = 0 .....(2)

From(1) ....(3)

Put in (2)

T_{1}(sinα cosβ+cosαsinβ) = mgcosβ

T_{1}sin (α+β) = mg cosβ

Put in (3)

QUESTION: 2

A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. The magnitude and possible direction of the acceleration of the body are

Solution:

Here, m = 5kg,

Resultant force,

θ = 36^{0}52'

This is the direction of resultant force and hence the direction of acceleration of the body. Fig.

Also,

QUESTION: 3

The relation between time t and distance x is t = ax^{2 }+ bx where a and b` are constants. The acceleration is

Solution:

QUESTION: 4

A particle moves from rest at P_{1} on the surface of a smooth circular cylinder of radius R as shown in Fig. At P_{2 }the particle leaves the cylinder. The relation between θ_{1} and θ_{2} is:

Solution:

As the normal force does no work on the particle, its energy is conserved.

So (L.E. + P.E.)_{P1}= (K.E. + P.E.)_{P2}

(K + U)_{P1} = (K + U)P2......(1)

Initially the particle in rest at P_{1} so kinetic energy (V = 0) will be zero at P_{1}.

So from equation(1)

mgR cosθ_{1} is potential energy at

P_{1 }= mgh

h → height = Rcosθ_{1}

At point P_{2} potential energy is mgh = mg Rsinθ_{2}

[using 2]

QUESTION: 5

The velocity of pendulum bob, which has been pulled aside from its equilibrium position through an angle θ and then released and it will pass through the equilibrium position, (where L is the length of the pendulum):

Solution:

Using the conservation of energy law in the form

Loss of PE = gain of KE

From Fig.

h = I - I cosθ

QUESTION: 6

A rocket motor consumes one quintal of fuel per second. The exhaust speed of gasses w.r.t. rocket is 5 km/s. The force exerted on the rocket and velocity acquired by the rocket, when its mass reduces to 1/100th of its initial mass is:

Solution:

u = -5km/s = -5 x 10^{3}m/s

F = -(-5 x 10^{3}) (100) = 5 x 10^{5}N

= 5 x 10^{3}log_{e}100 = (5 x10^{3})

= 5 x 10^{3} x 2.303 x 2

= 2.303 x 10^{4} m/s.

QUESTION: 7

A bullet is fired horizontally in the north direction with a velocity of 500m/sec. at 30^{0}N latitude. The horizontal component of Coriolis acceleration is:

Solution:

If X- axis is taken vertically, Z-axis towards north and Y-axis along east, then the velocity of the bulled is v = 500km/sec. and angular velocity w = w because the angular velocity vector w of the earth is directed parallel to its axis and is inclined at 30^{0} to the horizontal

Here,

Hence Coriolis acceleration

= 2w x v

m/sec^{2 }towards west.

QUESTION: 8

A canon launches a metal ball horizontal using a spring as the firing mechanism. The canon has mass m_{canon} and the m_{ball}. The spring constant is k. Initially, the spring is compressed a distance d from its equilibrium length. What is n_{canno}n/n_{ball}, the ratio of the cannon's recoil velocity to the ball's velocity?

Solution:

By conservation of momentum, P_{cannon} = -P_{ball}'

QUESTION: 9

A satellite is in elliptical orbit about the earth (radius = 6400km). At perigee it has an altitude of 1100 km and at the apogee its altitude is 4100km.The major axis of the orbit is at the distance.

Solution:

By the property of ellipse

Major axis 2a = maximum distance + minimum distance

Maximum distance = maximum altitude + radius of earth = (4100+6400)

Minimum distance from centre of earth = 6400 + 1100

So, 2a = (4100+6400)+(6400+1100) = 18000km

QUESTION: 10

A particle describes a circular orbit under the influence of an attractive central force directed towards a point on the circle,then the force is inversely proportional to,

Solution:

For attractive central force,

So, ...(1)

...(2)

Now substitute equation(2) in euqation of orbit

and from equation(1), seq q= 2au

QUESTION: 11

If f α -r^{n}, then for what value of n, the circular orbit described is stable?

Solution:

f(r) a -r^{n}

f = -kr^{n}

Where k is a constant

= ar^{n+1 }where (new constant)

Condition for stable orbit

If the potential energy function for the central force is of the form and centrifugal energy

The condition for the stability in the radial motion is

So, differentiate equation (1) w.r.to r

...(2)

Thus any circular orbit with r = r_{0 }under a central force is stable

QUESTION: 12

For a rigid body, linear velocity and angular velocity 'w' are related as;

Solution:

hence

QUESTION: 13

The linear mass density l of a rod, of uniform area of cross-section, 1 m long, varies with the distance x from its left end as given by

The centre of mass of the rod is located, from the left end, at a distance of:

Solution:

The centre of mass Q(dm = λdx)

QUESTION: 14

Figure shows a system of three initially resting particles of masses m_{1} = 4.1kg, m_{2 }= 8.2kg./ and m_{3} = 4.1kg.The particles are each acted on by different net external forces, which have magnitudes F_{1} = 6N, F_{2 }= 12N and F_{3} = 14N. The directions of the forces are shown in the figure. Where is the center of mass of this system?

Solution:

The position of the center of mass is marked by a dot in the figure. As figure suggests, we treat this point as if it held a real particle, assigning to it a mass M equal to the system mass of m_{1} + m_{2} m_{3} = 16.4kg and assuming that all external forces are applied at that point. We find the center of mass from equation

The x component of the net external force acting on the center of mass is

QUESTION: 15

Find the centre of mass of a uniform disc of radius a from whhich a circular section of radius b has been removed. The centre of hole is at a distance c from the centre of the disc.

Solution:

Suppose the circular disc of radius a with centre O is made up of

(i) Circular section of radius b with centre O_{1} and

(ii) Remaining portion of disc with cm at O_{2}

Taking O as origin, and O_{1}, O_{2} on X-axis, (y = 0, z = 0), figure the position of cm of disc is given by

...(i)

If s is surface density of material of the disc.

*Multiple options can be correct

QUESTION: 16

A particle is moving in space with O as the origin. Some possible expressions for its position velocity and acceleration in cylindrical coordinate (p,φ,z) are given below, which of the following are correct?

Solution:

In this coordinate system, we have the three coordinates as

These coordinate are related to the cartesian coordinates by the relations

Position vector r is given by

Hence, the velocity is given by

Differentiating above equation with respect to time, we get acceleration

*Multiple options can be correct

QUESTION: 17

For a particle moving in a central potential, which one of the following statements is correct?

Solution:

Two types of central force always produce closed orbits: F(r) = αr (a linear force) and F(r) = α / r^{2} (an inverse-square law).

*Multiple options can be correct

QUESTION: 18

A ball of mass m is thrown upward with velocity v and return back with v'. If air exerts an average resisting force F. Then

Solution:

Retarding force = mg +F.

Retardation a =

Distance s =

Downward motion: Force = mg -F

Acceleration a' =

Distance's =

Sinces = S, we have

*Multiple options can be correct

QUESTION: 19

A bullet of mass 10gm is fired horizontally in the north direction with a velocity of 500m/sec at 30^{0}N latitude and it hits a target 250 meters away. Then

Solution:

If X-axis is taken vertically, Z-axis towards north and Y-axis along east, then the velocity of the bullet is v = 500 km/sec. and angular velocity - UTP2 because the angular velocity vector ww of the earth is directed parallel to its axis and is inclined at 30^{0} to the horizibtal.

Hence coriolis acceleration

Time of journey, t

Deflection of the bullet due to the coriolis acceleration

Vertical displacement of the bulet due to the gravity

Coriolis force = -2m w x v

=3.6 x 10^{-4} newton towwards east.

*Multiple options can be correct

QUESTION: 20

A body of mass 2 kg moves under the influence of a central force, with a potential energy function Joule, where r is the radius circular orbit and r = 5m

The corresponding angular momentum and total energy are L and E. Then

Solution:

The body moves in a circular orbit of radius r = 5ms

So, energy of the system will always be equal to v_{e} at r = 5m

Also, ve will be maximum at r = 5m and dve/dr will be zero at r = 5m

The body moves in a circular orbit of radius r = 5m

So, energy of the system will always be equal to v_{e} at r = 5m

Also, v_{e} will be maximum at r = 5m and dv_{e}/dr will be zero at r = 5m

*Answer can only contain numeric values

QUESTION: 21

A man travelling in a car with a maximum constant speed of 20m/s watches his friend start off at a distance of 100m on a motor cycle with constant accelerationa. Then man in the car will reach his friend when a is .......m/sec^{2}.

Solution:

0 to 2

Let the man was initially at A and moves towards B with uniform velocity u = 20m/s.

Let the man in car meets the man B at point Q, then

AQ= 100 + BQ

⇒

⇒

⇒

The value of t will be real, if

*Answer can only contain numeric values

QUESTION: 22

A thin uniform rod XY of mass 0.1 kg and length 0.3m is hinged at its lowe end at X to the horizontal floor. It originally stands vertical and is allowed to fall to the floor. It strikes the floor with an angular speed of (rad/sec).......

Solution:

By the law of conservation of energy

Total energy at A = total energy at B

*Answer can only contain numeric values

QUESTION: 23

A bullet of mass 0.01kg and travelling at a speed of 500m/s strikes a block of mass 2 kg which is suspended by a string of length 5m. The centre of gravity of the block is found to rise a vertical distance of 0.1m. What is the speed(m/s) of the bullet after it emerges from the block? (g=9.8m/s^{2}).

Solution:

Let V be the velocity acquire by the block.

If n' is speed of bullet on emerging out of block, then by law of conservation of momentum, we get

mn +M x 0 = MV +Mn'

*Answer can only contain numeric values

QUESTION: 24

A toy rocket of mass 0.1 kg has a small fuel of mass 0.02kg which it burns out in 3s. Starting from rest on a horizontal smooth track, it gets a speed of 20 ms^{-1} after the fuel is burnt out. What is the approximate thrust of the rocket? What is the energy (joules/kg) content per unit mass of the fuel? (Ignore the small mass variation of the rocket during fuel burning)

Solution:

Herem mass of rocket,

m_{1} = 0.1 kg,

mass of fuel, m_{2} = 0.02 kg, t = 3s

Initial velocity u = 0

final velocity n = 20m/s

from n = u+at

Thrust on the rocket,

If s is the distance moved by the rocket in 3 sec, then from

Energy content of the fuel = work done on

rocket = Thrust x distance =

Energy/kg = 20/0.02 = 1000 J/kg

*Answer can only contain numeric values

QUESTION: 25

A wooden block having a mass of 1 kg is placed on a table. The block just starts to move, when a force of 10N is applied at 45^{0} to the vertical to push the block. The coefficient of friction between the table and the block, (taking g = 10m/s^{2}) is approximately________

Solution:

Resolving the forces and finding their horizontal and vertical component

mg = R + Fcosθ

Frictional force = F sinθ

By Eq.(ii) we get

= 0.8

*Answer can only contain numeric values

QUESTION: 26

A particle moves in a plane with velocity such that the time dependence of the magnitude of the velocity is It is given that at At what time (sec) will θ become 4 radian?

Solution:

*Answer can only contain numeric values

QUESTION: 27

A boy stands at 78.4m from a building and throws a ball which just enters a window 39.2m above the ground Calculate the velocity of projection (ms^{-1}) of the ball

Solution:

Let a boy standing at A throw a ball with a velocity u at an angle q with the horizontal, which just enters window W.

As the boy is at 78.4 m from the building and the ball just enters the window 39.2m above the ground, therefore

And horizontal range,

Dividing (i) by (ii), we get

Substituting in (ii), we get

*Answer can only contain numeric values

QUESTION: 28

A satellite of mass 100 kg is placed initially in a temporary orbit 800km above the surface of earth. The satellite is to be placed now in a permanent orbit at 2000km above the surface of earth. Find the amount of work done (mega joule) to move the satellite from temporary to permaanent orbit. The radius of the earth is 6400km. Given g = 10m/s^{2}.

Solution:

Consider a satellite of mass m is orbiting around the earth with orbit velocity n. Let h be the height of satellite from the surface of earth and r be the orbital radius of the satellite. Then

Total energy the orbiting satellite (E)

Work done in shifting the satellite from height h to h_{1} is

W = difference in energy of satellite in two orbits

*Answer can only contain numeric values

QUESTION: 29

A rocket of initial 6000kg ejects mass at a constant rate of 16kg/s with constant relative speed of 11km/s. What is the acceleration (in m/s^{2}) of the rocket on minute after blast?

Solution:

= 34.92m/s^{2}

*Answer can only contain numeric values

QUESTION: 30

A soccer ball has a mass of 0.40kg. Initially, it is moving to the left at 20m/s but then it is kicked and given a velocity at 45^{0} upward and to the right with a magnitude of 30m/s figure. Find the average net force (inN), asuming a collision time dt= 0.010s.

Solution:

With our choice of axes, we find the following velocity components:

The x- component of impulse is equal to the x-component of momentum change, and the same is true for they y-components.

The components of the average net force on the ball are

The magnitude and directionn of the average force are

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