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Physics: Topic-wise Test- 2 - NEET MCQ


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30 Questions MCQ Test - Physics: Topic-wise Test- 2

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Physics: Topic-wise Test- 2 - Question 1

A uniform disc of radius 20 cm and mass 2 kg is fixed at its centre and can rotate about an axis through the centre and perpendicular to its plane. A massless cord is round along the rim of the disc. If a uniform force 2 newton is applied on the cord, tangential acceleration of a point on the rim of the disc will be

Detailed Solution for Physics: Topic-wise Test- 2 - Question 1

Physics: Topic-wise Test- 2 - Question 2

Fig shows a hemisphere of radius 4R A ball of radius R is released from position P. It rolls . A ball of without slipping along the inner surface of the hemisphere. Linear speed of its centre of mass when the ball is at position Q is 

Physics: Topic-wise Test- 2 - Question 3

Three point masses, each m, are placed at the vertices of an equilateral triangle of side ‘α" Moment of inertia of the system about the axis COD which passes through the mass at O and lies in the plane of triangle and perpendicular to OA is 

Detailed Solution for Physics: Topic-wise Test- 2 - Question 3

The two masses are at a distance of “a” and “a/2” so,

Physics: Topic-wise Test- 2 - Question 4

An object of mass m is projected with a velocity u at an angle 45° with the horizontal. When  the object is at maximum height, its angular momentum about the point of projection is 

Detailed Solution for Physics: Topic-wise Test- 2 - Question 4
max height ,H= u^2sin^2θ/2g

velocity at max height = ux= ucosθ

angular momentum= muxH

and putting θ =45degree

angular momentum = mu^3/g4√2
Physics: Topic-wise Test- 2 - Question 5

At what height above the earth’s surface does the acceleration due to gravity fall to 1% of its value at the earth’s surface?

Detailed Solution for Physics: Topic-wise Test- 2 - Question 5

Let the acceleration due to gravity at that height is g′.

The acceleration due to gravity at some height (g′ ) = 1% of the acceleration due to gravity(g ) at the surface of the earth.

According to question

g′ = 1% of g

g′ = 1 / 100 × g

g′ = g / 100

The acceleration due to gravity at a height h above the surface of the earth is;

g′ = g (R / R+h)2 , Where R is the radius of the earth.

Thus,

g / 100 = g(R / R+h)2

1 / 100 = (R / R+h)2

Taking square root on both sides we get,

1 / 10 = (R / R+h)

R+h = 10R

h = 9R

Physics: Topic-wise Test- 2 - Question 6

A ring has a total mass m but not uniformly distributed over its circumference. The radius of the ring is R. A point mass m is placed at the centre of the ring. Work done in taking away the point mass from centre to infinity is

Detailed Solution for Physics: Topic-wise Test- 2 - Question 6

W = increase in potential energy of system

=Uf - Ui

=m (Vf - Vi)

(V = gravitational potential)

Note: Even if mass is nonuniformly distributed potential at centre would be -GM/R

Physics: Topic-wise Test- 2 - Question 7

Imagine 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 R−5/2 , then T2 is proportional to

Detailed Solution for Physics: Topic-wise Test- 2 - Question 7

Physics: Topic-wise Test- 2 - Question 8

Two point masses of mass 4m and m respectively separated by d distance are revolving under mutual force of attraction. Ratio of their kinetic energies will be :

Detailed Solution for Physics: Topic-wise Test- 2 - Question 8

The center of mass (CM) for two point masses can be calculated using the formula:

Here, let m= 4m (mass at position 0) and m= m (mass at position d).

Now, we can find the distances of each mass from the center of mass:

For circular motion, the centripetal force acting on each mass is given by:

F= m⋅r⋅ω2

Where r is the distance from the center of mass and ω is the angular velocity.

For mass 4m:
Fc,4m = 4m⋅d / 5⋅ω2

For mass m:
Fc,m = m⋅4d / 5⋅ω2

 

Since both masses are in circular motion due to their mutual gravitational attraction, we can set the centripetal forces equal to each other:

4m⋅d / 5⋅ω= m⋅4d / 5⋅ω2

 

The kinetic energy (KE) for each mass in rotational motion is given by:

K.E. = 1 / 2Iω2

Where I is the moment of inertia. The moment of inertia for point masses is given by I = m⋅r2.

For mass 4m:

For mass m:

Now, we can find the ratio of the kinetic energies:

Thus, the ratio of the kinetic energies of the two masses is:

Ratio of Kinetic Energies = 1 : 4

 

 

Physics: Topic-wise Test- 2 - Question 9

If G is the universal gravitational constant and ρ is the uniform density of spherical planet.
Then shortest possible period of the planet can be

Detailed Solution for Physics: Topic-wise Test- 2 - Question 9

The fastest possible rate of rotation of a planet is that for which the gravitational force on material at the equator just barely provides the centripetal force needed for the rotation. Let M be the mass of the planet R its radius and m the mass of a particle on its surface. Then

Physics: Topic-wise Test- 2 - Question 10

A planet is moving in an elliptical path around the sun as shown in figure. Speed of planet in positional P and Q are υ1 and υ2 respectively with SP = r1 and SQ = r2 , then υ12 is equal to

Detailed Solution for Physics: Topic-wise Test- 2 - Question 10

Angular momentum of planet about the sun is constant.

i.e. mυr sin θ = constant

At position P and Q, θ = 900 and m = mass of planet = constant

∴ υr = constant

Physics: Topic-wise Test- 2 - Question 11

Consider two configurations of a system of three particles of masses m, 2m and 3m. The work done by external agent in changing the configuration of the system from figure (i) to figure (ii) is

 

Physics: Topic-wise Test- 2 - Question 12

A spherical ball of mass 2 kg is rolling without slipping with a speed 4 m/s on a rough floor. Its rotational kinetic energy is :

Detailed Solution for Physics: Topic-wise Test- 2 - Question 12

Physics: Topic-wise Test- 2 - Question 13

A thin wire of length L and uniform linear mass density r is bent into a circular loop with centre O as shown. The moment of inertia of the loop about the axis XX is

Physics: Topic-wise Test- 2 - Question 14

Refrigerators X and Y are removing 1000 J of heat from the freezer. Refrigerator X is working between -5° C and 25° C and refrigerator Y is working between -20° C and 20 °C. Find efficiency of refrigerator X and Y?

Detailed Solution for Physics: Topic-wise Test- 2 - Question 14

We know that the efficiency of refrigeration for a refrigerator is T2 / T1 + T2
Where T1 is source temperature and T2 is sink temperature
For refrigerator X we have T1 = 298K and T2 = 268K
Hence the efficiency of refrigeration = 268 / 298 - 268
= 268 / 30
= 8.93
For refrigerator Y we have T1 = 293K and T2 = 253K
Hence the efficiency of refrigeration = 253 / 293 - 253
= 253 / 40
= 6.35

Physics: Topic-wise Test- 2 - Question 15

Kelvin- Planck statement states that

Detailed Solution for Physics: Topic-wise Test- 2 - Question 15

It works on the principle that no machine is 100% efficient. For instance if taking an ideal machine into consideration, work can be completely converted to heat. Now if the conditions are reversed, all heat cannot be converted to equal amount of work until and unless backed up by an external force.

Physics: Topic-wise Test- 2 - Question 16

Which of the following is an example of heat pump?

Detailed Solution for Physics: Topic-wise Test- 2 - Question 16

A heat pump is an electrical device that heats a building by pumping heat in from the cold outside. In other words, it’s the same as a refrigerator, but its purpose is to warm the hot reservoir rather than to cool the cold reservoir (even though it does both).

Physics: Topic-wise Test- 2 - Question 17

For proper utilization of exergy, it is desirable to make first law efficiency ____ and the source and use temperatures should ____.

Detailed Solution for Physics: Topic-wise Test- 2 - Question 17

If first law efficiency is close to unity, the all the energy carried in by heat transfer is used and no heat is lost to the surroundings.

Physics: Topic-wise Test- 2 - Question 18

The second law of thermodynamics says

Detailed Solution for Physics: Topic-wise Test- 2 - Question 18

The second law of thermodynamics gives a fundamental limitation to the efficiency of a heat engine and the coefficient of performance of a refrigerator. It says that the efficiency of a heat engine can never be unity or 100%, this implies that the heat released to the cold reservoir can never be made zero.
For a refrigerator the second law says that the coefficient through performance can never be infinite, this implies that the external work can never be zero.

Physics: Topic-wise Test- 2 - Question 19

If the door of refrigerator is left open inside a closed room, what would happen to the temperature of the room?

Detailed Solution for Physics: Topic-wise Test- 2 - Question 19

If you leave the door open, heat is merely recycled from the room into the refrigerator, then back into the room. A net room temperature increase would result from the heat of the motor that would be constantly running to move energy around in a circle.

Physics: Topic-wise Test- 2 - Question 20

Suppose we have a box filled with gas and a piston is also attached at the top of the box.What are the ways of changing the state of gas (and hence its internal energy)?

Detailed Solution for Physics: Topic-wise Test- 2 - Question 20

As change of state depends on both temperature and pressure, so, either method A or C can be used. We can understand this by ideal gas equation PV = nRT

Physics: Topic-wise Test- 2 - Question 21

Hot coffee in a thermos flask is shaken vigorously, considering it as a system which of the statement is not true?

Detailed Solution for Physics: Topic-wise Test- 2 - Question 21

No, heat is not transferred as the flask is insulated from the surroundings ∴dQ=0

Physics: Topic-wise Test- 2 - Question 22

Which wall would allow the flow of thermal energy between systems A and B to achieve thermal equilibrium?

Detailed Solution for Physics: Topic-wise Test- 2 - Question 22

Wall that permits *heat" to flow through them,such as engine block is called diathermic wall.
wall Perfectly insulating ball that doesn't allow the flow heat to them are called adiabatic walls.

Physics: Topic-wise Test- 2 - Question 23

Two identical rectangular strips, one of copper and the other of steel, are riveted as shown to form a bi-metal strip. On heating, the bi-metal strip will

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Detailed Solution for Physics: Topic-wise Test- 2 - Question 23

On heating, the copper strip will suffer greater elongation and hence the bimetal strip will bend with the steel strip on the concave side. 

[Bimetal strips are widely used in thermal switching applications such as automatic electric iron].

Physics: Topic-wise Test- 2 - Question 24

Four cylindrical rods of different radii and lengths are used to connect two heat reservoirs at fixed temperatures t1 and t2 respectively. From the following pick out the rod which will conduct the maximum quantity of heat:

Detailed Solution for Physics: Topic-wise Test- 2 - Question 24

The rate of heat transfer is directly proportional to the bisectional surface area of the solid and inversely proportional to its parallel length. 
That is heat conduction rate say H ∝ r2
∝ 1/L
I.e ∝ r2/L
Hence the conduction would be
maximum in case in which  r2/L ratio is largest.

Physics: Topic-wise Test- 2 - Question 25

The temperature for which the reading on Celsius and Fahrenheit scales are identical is

Detailed Solution for Physics: Topic-wise Test- 2 - Question 25

The Celsius and Fahrenheit are two important temperature scales. The Fahrenheit scale is used primarily in the United States, while Celsius is used throughout the world. The two scales have different zero points and the Celsius degree is bigger than the Fahrenheit one. There is one point on the Fahrenheit and Celsius scales where the temperatures in degrees are equal. This is -40 degree C and -40 degree F.

Physics: Topic-wise Test- 2 - Question 26

A steel tape is calibrated at 20° C. A piece of wood is being measured by steel tape at 10°C and reading is 30 cm on the tape. The real length of the wood is:

Detailed Solution for Physics: Topic-wise Test- 2 - Question 26

When heated the length between adjacent markings increases. Hence the actual length will be less than
30 cm.

Physics: Topic-wise Test- 2 - Question 27

The coefficient of liner expansion of a cubical crystal along three mutually perpendicular direction is showimage and showimage (1). What is the coefficient of cubical expansion of crystal?

Detailed Solution for Physics: Topic-wise Test- 2 - Question 27

Cubical expansion means there is expansion is length (a1), breadth (a2) and height (a3). Therefore the net coefficient of cubical expansion is α1 + α2 + α3.

Physics: Topic-wise Test- 2 - Question 28

An Indian rubber cube of side 10 cm has one side fixed while a tangential force of 1800 N is applied to the opposite side. Find the shear strain produced. Take η = 2 x 106 N/ m2.

Detailed Solution for Physics: Topic-wise Test- 2 - Question 28

To find the shear strain produced in the rubber cube, we use the following steps:

  • The shear modulus (η) is given as 2 x 106 N/m2.
  • The side of the cube is 10 cm, which is 0.1 m.
  • The tangential force applied is 1800 N.
  • The area of the side where the force is applied is:
    Area = Side x Side = 0.1 m x 0.1 m = 0.01 m2.
  • Calculate the shear stress (force per area):
    Shear Stress = Force / Area = 1800 N / 0.01 m2 = 180,000 N/m2.
  • The shear strain is the ratio of shear stress to shear modulus:
    Shear Strain = Shear Stress / η = 180,000 N/m2 / 2 x 106 N/m2 = 0.09.

Thus, the shear strain produced is 0.09.

Physics: Topic-wise Test- 2 - Question 29

The length of the wire is increased by 1 mm on the application of a given load. In a wire of the same material but of length and radius twice that of the first, on application of the same force, extension produced is

Detailed Solution for Physics: Topic-wise Test- 2 - Question 29

Young's Modulus of elasticity =stress/strain
Y= [F/a/△l/l] ​ or Y= [Fl​/a△l]
or △l= [Fl/aY] ​= Fl​/ πr2Y
In the given problem, △l∝ l​/ r2
When both l and r are doubled, △l is halved.

Physics: Topic-wise Test- 2 - Question 30

The volume of a spherical body is decreased by 10-3% when it is subjected to pressure of 40 atmospheres. Find the bulk modulus of body.
(1 atm = 1.01 x 105 N/m2).

Detailed Solution for Physics: Topic-wise Test- 2 - Question 30

We know that magnitude of bulk modulus
K=[P/(dV/V)]
Now, percentage change in volume is. 10-3 %
Therefore, (dV/V)x100=10-3
So, dV/V=10-5
Hence, k is,
K=40atm/10-5=40x1.01x105 /10-5 =4.04x1011 N/m

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