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# IIT JAM Physics MCQ Test 13

## 60 Questions MCQ Test Mock Test Series for IIT JAM Physics | IIT JAM Physics MCQ Test 13

Description
This mock test of IIT JAM Physics MCQ Test 13 for Physics helps you for every Physics entrance exam. This contains 60 Multiple Choice Questions for Physics IIT JAM Physics MCQ Test 13 (mcq) to study with solutions a complete question bank. The solved questions answers in this IIT JAM Physics MCQ Test 13 quiz give you a good mix of easy questions and tough questions. Physics students definitely take this IIT JAM Physics MCQ Test 13 exercise for a better result in the exam. You can find other IIT JAM Physics MCQ Test 13 extra questions, long questions & short questions for Physics on EduRev as well by searching above.
QUESTION: 1

### The trajectory of a particle of mass m is described in the cylindrical polar coordinate by  and  where ω and r0 are constant. The radial force is

Solution:

r = r0 cosh(ωt)
r = r0ωsinh(ωt)
r = r0ω2cosh(ωt) = rω2

So. the radial fo rce F = m

QUESTION: 2

### Two block of mass 5 kg and 2 kg are kept one above the other on the floor of lift. Which is accelerating in downward with an acceleration of 3 m/sec2. Find the normal reaction between two blocks.

Solution:

Let us draw the free body diagram of two blocks.

Both the blocks are accelerating with 3 m/sec2 in downward direction.
Applying Newton's second law
2 g - N1, = 2 * 3 = 6
5g + N1 - N2 = 15
⇒ 7g - N2 = 21
N= 49
N = 14, so normal reaction between two bodies is 14N.

QUESTION: 3

### A block of mass 5 kg is resting on a friction less plane. It is struck by a jet releasing water at 2 kg/sec and a speed 10 m/sec. Find initial acceleration of the block.

Solution:

QUESTION: 4

Two blocks of masses 3 kg and 4 kg are hung from the ceiling by three cables as shown.

The cable in the middle is exactly horizontal and the cable on the left makes an angle of 45° with respect to the vertical. The angle θ made by the cable on the right with respect to the vertical is

Solution:

At point A
T = 3g
T1 =3√2 g
At point B

QUESTION: 5

A man climbs down a hemispherical hill of radius 50 m from the topmost point. If the coefficient of friction between the shoes and the hill is μ = 0.05, then approximately how much distance does he have to walk before he slips?

Solution:

The free body diagram is given in figure

N - mgcosθ
⇒ N = mgcosθ
mgsinθ - f = 0
mgsinθ = f
θ is maximum if f is maximum

mgsinθ = fmax = μmgcosθ
tanθ = μ = 0.05
0 ≈ 0.05 rad
So, distance travelled before he slips
x = Rθ = 50 * 0.05 = 2.5 m

QUESTION: 6

Two satellites S1, and S2 revolve round a planet in coplanar circular orbit in the same sense. Their periods of revolutions are 1 hour and 8 hour respectively. The radius of the orbit of S1, = 104 km. When S2 is closest to S1 then the angular speed of S2 as obseived by an astronaut in S1.

Solution:

Let the radius of the orbit S2 be r2 km. We know that

r2 = 4 * 104 km
km per hr.
km per hr.
km per hr.
then relative angular velocity w21  =

QUESTION: 7

Assume a light planet revolving around a very massive star in a circular orbit of radius R with a period of revolution T. If the gravitational force of attraction between the planet and the star is proportional to R-3/2, then T is proportional to

Solution:

2R = Fgr

QUESTION: 8

Consider a spherical planet rotating about an axis passing through its centre. The velocity of a point on its equator is three-forth of the value of g measured at one of the poles, then the escape velocity for a particle shot upwards from that pole will be

Solution:

gat equator =
gat poles = GM/R2
Now. g at equator = 3/4 g at poles

QUESTION: 9

A particle with time-varying mass

where are positive constants, moves along the x-axis under the action of a constant positive force F for f the particle is at rest at time t = 0. then at time t. its velocity V will be —

Solution:

We know for action of constant positive force

P → constant momentum
P = mV = Ft

QUESTION: 10

The kinetic energy of body is increased by 44%. What isthe percentage increase in the magnitude of linear momentum of the body?

Solution:

QUESTION: 11

A point particle at rest is released from the top of a sphere of radius R and slides down frictionlessly under gravity. At what angle from the vertical does it leave the sphere?

Solution:

At the top of sphere, the particle is at rest its

PE = mgR
At point B PE = mgRcosθ
By the conseivation of energy

= mg(3cosθ- 2)
A particle will leave the sphere if N = 0

QUESTION: 12

Three masses m 2m and 3m are moving in x - y plane with speeds 3u, 2u ad u respectively, as shown in the figure. The velocity of the resulting mass would be

Solution:

Let velocity of resulting mass is
Px; 3mu - 2m.2ucos30 - 3m.ucos30 = 6mVx

Py = -2m.2usin30 + 2musin30 = 6 mVy = -2mu + mu = 6mVy = -mu = 6mVy

So,

QUESTION: 13

A mass of 1 kg moving with a speed 4 m/sec, hits another mass 2 kg moving in the same direction with a speed of 2 m/sec, then Kinetic energy of the centre of mass is

Solution:

QUESTION: 14

A block of mass 1 kg is initially held at rest on the frictionless surface of a wedge of mass 10 kg. The wedge itself lies on horizontal frictionless surface. The block is now released so that it is allowed to slide down the slope. How far down the slope does the block move when the wedge has moved a distance of 20 cm?

Solution:

Vcm = Const.

Distance travelled by block along the slope = 20 sec 60° = 40cm

QUESTION: 15

A ball of mass m moving at a speed V makes a head on collision with an identical ball at rest. The kinetic energy of the balls after the collision is 1/2 th of the original. Find the coefficient of restitution.

Solution:

2 + 2e2 = 2
e = 0

QUESTION: 16

Four masses each of mass m are at the corners of a square of side a as shown in the figure. The product of inertia Ixy of the system about O is

Solution:

QUESTION: 17

A hollow sphere of moment of inertia 2/3 MRand a thin loop of moment of inertia MR2 roll without slipping down an inclined plane. The ratio of their times of arrival   at the bottom of the incline is given by

Solution:

From the free body diagram
Mgsinθ - f = Ma
f R = I∝
For rolling

For hollow sphere

For thin loop

QUESTION: 18

Consider the uniform solid right cone depicted in the figure on the right. This cone has mass m and a circular base of radius r. If the moment of inertia of the cone about an axis parallel to the x axis passing through the centre of mass Ocm (see figure) is given by

Then the moment of inertia about another axis parallel to the x axis, but passing through the point 0r ( (see figure), is

Solution:

The moment of inertia about the point Ocm is ___

Point O is at h/4 distance from Ocm point. So. the moment of inertia using parallel axis theorem is

But this axis is parallel to x-axis and the radius of circular cross-section is “r”. Now the resultant moment of ineitia about axis parallel to x-axis and passing through Ois

QUESTION: 19

Three identical rings of radius a are placed on the xy-plane so that each ring touches tine other two tangentially. Each ring, which may be considered to be infinitesimally thin, has a mass M. The centres of three equidistant from the z-axis this common distance being 2a/√3 . What is the moment of inertia of this  system about the z-axis.

Solution:

Using parallel axis theorem
Ml of each ring about z-axis

So. Ml of whole system about z-axis

QUESTION: 20

I f pressure at half the depth of a lake is equal to  2/3 pressure at the bottom of the  lake then what is the depth of the lake.

Solution:

Pressure at bottom of the lake = P0 + pgh and pressure at half the depth of a lake  According to given condition

QUESTION: 21

The directed beam from a small but powerful searchlight placed on the ground tracks a small plane flying horizontally at a fixed height h above the ground with a uniform velocity V as shown in the figure below.

If the searchlight starts rotating with an instantaneous angular velocity ω0 at time t = 0 when the plane was directly overhead, then at a later time t its instantaneous angular velocity ω(t) is given by

Solution:

QUESTION: 22

A ball is dropped vertically from a height H on to a plane surface and permitted to bounce repeatedly along a vertical line. After every bounce, its kinetic energy becomes a quarter of its kinetic energy before the bounce. The ball will come to rest after time

Solution:

⇒

For

⇒ total time is required to come in rest —

QUESTION: 23

A horizontally placed hollow tube has a cross-sectional Area A at the beginning of the tube that gradually tapers off to A/2 at the end. An incompressible, ideal fluid of density p enters the tube with a velocity V at the beginning of the tube. What is the difference in pressure at the two ends of the tube?

Solution:

Let velocity at the other end A/2 is V
By equation of continuity AV = A'V
V' = 2V
Using Bernoulli equation

QUESTION: 24

A solid sphere of mass m and radius r is released from rest at point A on a track in vertical plane. The track is rough enough to support rolling between A and B and from B onwards it is smooth. The maximum height attained by sphere from ground on its journey from B onwards is

Solution:

At point B

At highest point

QUESTION: 25

The pulley shown in figure has moment of ineitia I about its axis and radius R. Find the acceleration of two blocks.

Solution:

Mg - T1 = Ma
T2 - mg = ma

QUESTION: 26

A force F acts tangentially at the highest point of a solid sphere of mass m kept on a rough horizontal plane. If the sphere rolls without slipping. Find the acceleration of the centre of sphere.

Solution:

Suppose that the static friction F on the sphere acts towards right. Let r be the radius of sphere and a linear acceleration of centre of sphere. The angular acceleration about the centre is  as there is no slipping.
F + f = ma
Fr - fr = l∝

QUESTION: 27

A particle of mass m is thrown upward with velocity V and there is retarding air resistance proportional to the square of the velocity with proportionality constant K. If the particle attains a maximum height after time t and g is the gravitational acceleration, what is the velocity?

Solution:

Equation of motion

QUESTION: 28

A small spherical ball is released from a point at a height h on a rough block shown in figure. Assuming that it does not slip anywhere, find its linear speed when it rolls on the horizontal part of the track?

Solution:

QUESTION: 29

An xyz coordinate system is rotating with respect to an xyz coordinate system. The angular velocity of the xyz system relative to the xyz system is given by  where t is time. The position vector   then Coriolis acceleration at time t = 1 sec is

Solution:

we know that

QUESTION: 30

On the opposite sides of a wide vertical vessel filled with water two identical holes are opened each having the cross sectional area a=0.50 cm2. The height difference between them is equal to Δh = 50 cm. Find the resultant force of reaction of the water flowing out of the vessel.

Solution:

Let the depth of hole 1 and 2 from the surface of water is h1 and h2
Thrust force F1 = 2apgh1

Net thrust force F= F- F1

*Multiple options can be correct
QUESTION: 31

A particle moving in the yz plane undergoes a displacement   while a constant  acts on the particle then which of the following are correct.

Solution:

work W = F.ds =
W= 16 J

*Multiple options can be correct
QUESTION: 32

A particle of mass m , moving with a velocity  collides elastically with another particle of mass 2m which is at rest initially. Here, V0 is constant. Which of the following are correct ?

Solution:

*Multiple options can be correct
QUESTION: 33

A particle is released at x=1 in a force  which of the following are true?

Solution:

Velocity of center of mass

speed of particle of mass m before collision in center of mass frame

speed of particle of mass 2m before collision in center of mass frame

After collision, magnitude of velocity of particle before and after collision is same in cm frame. So. after collision speed of particle is

*Multiple options can be correct
QUESTION: 34

For a completely inelastic collision of two particles in one dimension, which of the following statements are correct.

Solution:

•  is exact differential work done by this force is independent of path. Force is conservative.
• Torque about origin is zero . So angular momentum of particle is constant about origin.
• Equilibrium position of particle is given by

F(x)= 0

So Potential energy is minimum at X = √ 2
So, x = √2 is point of stable equilibrium
Hence, the particle will move towards x = 1

*Multiple options can be correct
QUESTION: 35

A particle of mass 2/3 kg is subjected to a potential energy V ( x ) = (3x2 - 2x3)J. where x ≥ 0 and expressed in meters, then

Solution:

V { x ) = X2 ( 3 - 2x)
The position of maxima and minima are given by

x = 0.1

So. x=0 is a point of minima

x = 1 is a point of maxima

*Multiple options can be correct
QUESTION: 36

A planet moving in an elliptical orbital has its smallest and greatest speeds Vmin and Vmax. If the eccentricity of the planet’s orbit is e. Then which of the following are correct.

Solution:

We know that
By the conservation of angular momentum

*Multiple options can be correct
QUESTION: 37

A solid cylinder of mass m, radius r, rolls down an inclined plane of inclination θ without slipping

Solution:

For motion of the centre of mass

Now the maximum friction that can act is μN,

Though friction is present, it is not dissipative. So, the principle of conservation of mechanical energy can be used friction present here only aids in rolling We have get

*Multiple options can be correct
QUESTION: 38

Consider a uniform square plate of mass m and side a as shown.

Solution:

*Multiple options can be correct
QUESTION: 39

A bullet is fired horizontally in the north direction with a speed u at θ latitude. It hits with a target l meters away then

Solution:

Coriolis acceleration =

The deflection is given by
towards the east.

*Multiple options can be correct
QUESTION: 40

For a particle moving in a central field

Solution:

Since for a particle moving in a central force field, angular momentum remain conserved. Hence, motion is confined in a plane.

*Answer can only contain numeric values
QUESTION: 41

A particle of mass 2 kg is moving along the x-axis. There is no force on the particle except during the time interval between t = 0 sec and t = 5 sec, when a force of the form F(t) = 6t2N/S2 acts on it. If the velocity of the particle at t = 0 is 1000m/sec then calculate its velocity at t=10 sec.

Solution:

Force acting on particle

*Answer can only contain numeric values
QUESTION: 42

An Aircraft, which weights 12000 kg when unloaded, is on a relief mission, carrying 4000 food packets weighing 1 kg each. The plane is gliding horizontally with its engine off at a uniform speed of 540 kmph when the first food packet is dropped. Assume that the horizontal air drag can be neglected and the aircraft keeps moving horizontally. If one food packet is dropped every second, then the distance between the last two packet drops will be _____ m.

Solution:

The mass of an aircraft = 12000 kg
Uniform speed of aircraft is
The distance between the last two packet drops will be equal to the covered distance in one second = 150 m

*Answer can only contain numeric values
QUESTION: 43

A particle of mass 1 kg is moving in a central force field given by

Assume that the particle is moving in a circular orbit with angular momentum 2 J-sec. Find out the radius of the orbit.

Solution:

Central force acting on 1 kg mass is given by

Where V is central force potential

Effective potential

Where L= angular momentum of particle

Let radius of circular orbit is rfor circular orbit . Ve is minimum at r = r0

r = 1 m

*Answer can only contain numeric values
QUESTION: 44

A bullet of mass 0.1 kg travelling at a speed of 200 m/sec hits one end of a stick of length 1.0 m and 0.5 kg lying on a horizontal table and pivoted at the other end, as shown in figure. After hitting the stick, the bullet continues to move along its original path but with a reduced speed of 175 m/sec. Assume there is no friction, find the angular speed of the stick after it is hit (in rad/sec).

Solution:

Angular momentum of system before collision
Li = mvl ∵ ( speed of bullet before collision )
= 0.1x200x1 = 20
Angular momentum of system after collision

(ω = angular velocity of stick , v' = velocity of bullet after collision)

Applying law of conservation of angular momentum
Li= lf

*Answer can only contain numeric values
QUESTION: 45

A constant tangential force facts at the top of a solid sphere of mass M and radius R as shown in figure so that the sphere rolls on a horizontal surface without slipping. Find magnitude of the friction force exerted by the sphere on the surface.

Solution:

Net force acting on the sphere Fcos60 + f
Where f is a friction force
F cos60 + f = Ma

Magnitude of friction Force

*Answer can only contain numeric values
QUESTION: 46

The mass ratio m2/m1 = 0.67. The coefficient to friction between the body and the inclined plane is equal to μ =0.1. The masses of the pulley and the threads are negligible. Find the magnitude of acceleration of body mwhen the uniformly stationary system of masses starts moving.

Solution:

Substituting all the values; a = 0.048 g
≈ a = 0 . 48 m/sec2

*Answer can only contain numeric values
QUESTION: 47

A ball suspended by a thread swings in a vertical plane so that its acceleration values in the extreme and the lowest position are equal. Find the thread deflection angle at the extreme position.

Solution:

Let V be the speed of the ball at its lowest position and I be length of the thread. Then according to the problem
...(i)
where α is the maximum deflection angle from newton's law in
F = ma
By the energy conservation
v2 = 2gl(1- cos α)...(ii)
By the equation 1=2
sin α = 2 ( 1 - cos α)

*Answer can only contain numeric values
QUESTION: 48

The heat required to convert 1 gm of water into steam under standard atmospheric pressure if the entropy of water at the boiling point is 0.31 and for 100% steam at the steam temperature is 1.74 _______ (in cal).

Solution:

Q = 533.4 cals
ds = dQ/T

Q = T(S1 - S2)
Q = 373(1.74 - 0.31) = 533.4 cals
Q = 533.4 cals

*Answer can only contain numeric values
QUESTION: 49

A planet of mass m moves along a circle around the sun with velocity v=34.9km/sec (relative to the heliocentric reference frame). Find the period of revolution of this planet around the sun (in days)

Solution:

we have
and
and we know that

*Answer can only contain numeric values
QUESTION: 50

A wide cylindrical vessel 50 cm in height is filled with water and rests on a table Assuming the viscosity to be negligible. Find at what height (in cm) from the bottom of the vessel a small hole should be perforated for the water jet coming out of it to hit the surface of the table at the maximum distance lmax Cmaxfrorn the vessel.

Solution:

Let H be the total height of the water column and the hole is made at a height h from the bottom

for the horizontal range

Now for maximum l

*Answer can only contain numeric values
QUESTION: 51

A bent tube is lowered into a water stream as shown in fig. The velocity of the stream relative to the tube is equal to v=2.5m/sec. The closed upper end of the tube located at the height ho=12 cm has a small orifice. To what height h will the water jet support ?(in cm)

Solution:

Let the velocity of the water jet near the orifice be u' then applying Bernoulli's theorem

here the pressure term on the both sides is the same and is equal to the atmospheric pressure. Now, If water rises up to a height whole of its kinetic energy will be converted into potential energy.

*Answer can only contain numeric values
QUESTION: 52

A wide vessel with a small hole in the bottom is filled with water and kerosene, neglecting the viscosity. Find the velocity of the water flow, if the thickness of the water layer is equal to h1 = 30 cm and that of the kerosene layer to h2 = 20 cm.

Solution:

The density of water is greater than that of kerosene oil.
pressure clue to water level = h1p1
pressure clue to kerosene level = h2p2
so net pressure become = h1p1g  +  h2p2g
From Bernoulli Theorem, at point A

*Answer can only contain numeric values
QUESTION: 53

A uniform disc of radius R=20 cm has a round cut as shown in figure, the mass of the remaining portion of the disc is equal to m = 7.3kg. find the moment of inertia of such a disc relative to the axis passing through its centre of inertia and perpendicular to the plane of the plane of the disc.

Solution:

let us take point O as origin and point x-axis towards right. Obviously the C.M of the shade position of given shape lies on the x-axis. Hence the c.m of the shade portion is given by

thus the sought moment of inertia

*Answer can only contain numeric values
QUESTION: 54

Find the value of speed (km/hr) with which a projectile should be launched from the surface of the earth so as to reach a height h equal to 1/4th of the radius r of the earth.

Solution:

Let m = mass of the earth
m = mass of the projectile

Increase in potential energy =

form law of conservation of energy

it is given h = R/4

5060m /sec or ≈ 5 km/sec

*Answer can only contain numeric values
QUESTION: 55

A particle is moving in a plane with a constant radial velocity 5 m/sec and constant angular velocity of 3 rad/sec. When the particle is at a distance r = 4 m from the origin, the magnitude of the instantaneous velocity of the particle in m/sec is

Solution:

*Answer can only contain numeric values
QUESTION: 56

Two point objects A and B have masses 1000 kg and 3000 kg respectively. They are initially at rest with a separation equal to 1 m. Their mutual gravitational attraction then draws them together. How far from A's original position (m) will they collide___.

Solution:

Since gravitational force is conservative, therefore they collide at their centre of mass —

*Answer can only contain numeric values
QUESTION: 57

After being hit. a golf ball reaches a maximum height of 60 m with a speed of 20 m /sec. Right after being hit the speed of the ball is (take g as 10 m /s2) ______ m/sec.

Solution:

*Answer can only contain numeric values
QUESTION: 58

A cylindrical tank of height 0.4m is open at the top and has a diameter 0.16 m. Water is filled in it upto a height of 0.16m. Calculate the time (in second) that will take to empty the tank through a hole of radius 5 x 10-3 m is its bottom.

Solution:

Let rate of drop of water is

t = 46 sec

*Answer can only contain numeric values
QUESTION: 59

A cylinder of 1 kg mass and 0.02 m diameter left at the top of an inclined plane of height 1 m rolls down without slipping. Find the velocity (m/sec) of centre of mass of cylinder on reaching the bottom of inclined plane ___ .

Solution:

3.6 to 3.7
The kinetic energy of cylinder when it reaches the bottom = potential energy of the cylinder at the to p = mgh = 1 * 9.8 * 1 = 9.8 J
Kinetic energy KE = KE of translation + KE of rotation

Kinetic energy at bottom = 9.8 J

*Answer can only contain numeric values
QUESTION: 60

A stone is dropped vertically from the top of a tower of height 40 m. At the same time a gun is aimed directly at the stone from the ground at horizontal distance 30 m from the base of the tower and fired. If the bullet from the gun is to hit the stone before it reaches the ground, the minimum velocity of the bullet must be, _________(m/sec) approximately.

Solution:

The initial y-coordinate of stone is 40 m and that it falls through a distance gun  in a time t.

∴ y-coordinate of stone at any time after release is y =
Now y-coordinate for bullet at any time

when bullet hit the stone, y-coordinate for both one same

a collision will take place when