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BITSAT Physics Test - 3 - JEE MCQ


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30 Questions MCQ Test BITSAT Mock Tests Series & Past Year Papers 2025 - BITSAT Physics Test - 3

BITSAT Physics Test - 3 for JEE 2024 is part of BITSAT Mock Tests Series & Past Year Papers 2025 preparation. The BITSAT Physics Test - 3 questions and answers have been prepared according to the JEE exam syllabus.The BITSAT Physics Test - 3 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for BITSAT Physics Test - 3 below.
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BITSAT Physics Test - 3 - Question 1

For measuring an unknown resistance, a meter bridge with a standard resistance of 90 Ώ is used. The null point indicated by the jockey is 40 cm on the standard resistance. If the scale used to measure the length of the standard resistance is accurate with a deviation of ±1 mm, the value of the unknown resistance and the deviation is

Detailed Solution for BITSAT Physics Test - 3 - Question 1

For a balance meter bridge,

Therefore, the value of the unknown resistance is 60 Ώ.
Deviation in the value of X:
As we know,

Therefore, the deviation is ±0.25.
So, the value of the unknown resistance is (60 ± 0.25).
Therefore, it is the correct option.

BITSAT Physics Test - 3 - Question 2

In the given circuit, the value of current flowing across the resistances A and B is

Detailed Solution for BITSAT Physics Test - 3 - Question 2

Apply KCL at node A.

Simplifying the above equations, we get

Current,

This is our required solution.

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BITSAT Physics Test - 3 - Question 3

A coil with 400 turns and 0.1 square metre area is placed perpendicularly in a magnetic field of strength 4 × 10-5 Wb/m2. If it rotates through 90 degrees in 0.1 sec, what will be the voltage of average emf that will be induced in the coil?

Detailed Solution for BITSAT Physics Test - 3 - Question 3

The induced emf will be given by:

Therefore, it is the correct option.

BITSAT Physics Test - 3 - Question 4

What will be the focal length of the eyepiece and the objective lens of a telescope, if the angular magnification of the telescope is given to be 5 for far object. The final image is formed at infinity and the distance between the eyepiece and the objective lens of the telescope is 54 cm. (Focal length of eyepiece = fE; Focal length of objective = fO)

Detailed Solution for BITSAT Physics Test - 3 - Question 4


The image formed by the objective lens will be at the focus of the objective as it is used to focus the far objects.
Let the point of image formation by objective be IO.
Therefore, POIO = fO
Image IO must lie on the first focus of the eyepiece, as the final image is formed at infinity.
Therefore,
PEIO = fE
Therefore, fO + fE = 54 cm
Also, magnification is given to be 5.
Therefore,

Therefore, it is the correct option.

BITSAT Physics Test - 3 - Question 5

For a hydrogen atom, when the electron comes in the ground state from an excited state, then

Detailed Solution for BITSAT Physics Test - 3 - Question 5

Correct answer is A. When an electron in a hydrogen atom makes a transition from an excited state to the ground state, its kinetic energy increases, its potential energy decreases and its total energy remains constant.

BITSAT Physics Test - 3 - Question 6

Three particles A, B and C carrying respectively one, two and three positive charges on them enter into a magnetic field which is perpendicular to their velocities. If the masses of the particles A, B and C are in the ratio 1 : 2 : 3, if all of them have same kineatic energy then choose the correct statement among the following:

Detailed Solution for BITSAT Physics Test - 3 - Question 6

We know, the radius of the particle deflecting in the magnetic field is given by:

Therefore, particle C will have the least radius and thus will deflect the most.
Therefore, it is the correct option.

BITSAT Physics Test - 3 - Question 7

A wire of length l is in the form of a circular loop having one turn only. The loop is carrying a current I and induces a magnetic field of strength B1 at the centre of the loop. If the same wire is used to make a coil having two turns, then the strength of the magnetic field at the centre is B2 providing the current flowing the loop is same. The correct option is

Detailed Solution for BITSAT Physics Test - 3 - Question 7

BITSAT Physics Test - 3 - Question 8

A circular wire of radius 50 mm with 500 turns is acting as an inductor in a circuit. The approximate value of self-inductance for the wire will be

Detailed Solution for BITSAT Physics Test - 3 - Question 8

We know,
NBA = LI
And

Therefore, it is the correct option.

BITSAT Physics Test - 3 - Question 9

In the graph below, the deviation of a ray of light is plotted against the angle of incident i when it passes through a prism.

The points X and Y in the plot respectively represent

Detailed Solution for BITSAT Physics Test - 3 - Question 9

The deviation of the prism is given by:

At minimum deviation,

The point X represents  and the point Y represents δm or minimum deviation.
Therefore, it is the correct option.

BITSAT Physics Test - 3 - Question 10

Choose the correct match from the options.

Detailed Solution for BITSAT Physics Test - 3 - Question 10

Moseley's law for X-ray spectrum is expressed aswhere b = constant and depends upon the series.

Rutherford and Soddy formulated the radioactive decay law which is expressed as 
De Broglie wavelength is the wavelength associated with a particle of mass m and having velocity v. The de Broglie wavelength is given by:

Therefore, it is the correct option.

BITSAT Physics Test - 3 - Question 11

Two bodies are projected from the same point with equal speeds in such directions that they both strike the same point on a plane whose inclination is β. If α be the angle of projection of the first body with the horizontal the ratio of their times of flight is

Detailed Solution for BITSAT Physics Test - 3 - Question 11


Let β be the inclination of plane
α angle of projection of 1st body 
angle of projection of 2nd body
The component of g perpendicular to the plane = −gcosβ
For the first component,
x = vt cos α, y = vt sin(α − β) − g cos βt2 
Finding Range,

The range of both the bodies is same.

BITSAT Physics Test - 3 - Question 12

Two stones are projected with same velocity v at an angle θ¸ and (90° − θ). If H1 and H2 are greatest heights in the two paths, what is the relation between R, H1 and H2?

Detailed Solution for BITSAT Physics Test - 3 - Question 12


  (Range remains same, if θ= 90 − θ1

BITSAT Physics Test - 3 - Question 13

A particle is projected with speed v at an angle θ(0 < θ < π/2) above the horizontal from a height H above the ground. If v = speed with which particle hits the ground and t = time taken by particle to reach ground, then

Detailed Solution for BITSAT Physics Test - 3 - Question 13

From figure,


It means speed is independent of the angle of projection

From this, where θ increases, t increases.

BITSAT Physics Test - 3 - Question 14

The semivertex angle of the conical pendulum through which a body can perform uniform circular motion, if the string support 2/√3 times its weight, is

Detailed Solution for BITSAT Physics Test - 3 - Question 14

Resolving the tension in horizontal and vertical direction we get, 
The component along the vertical direction is T cosθ
The component along the horizontal direction is T sinθ
The weight of the string acting in downward direction W = mg.

BITSAT Physics Test - 3 - Question 15

Two buses A and B are moving around concentric circular paths of radii rA​ and rB. If the two buses complete the circular paths in the same time. The ratio of their linear speeds is 

Detailed Solution for BITSAT Physics Test - 3 - Question 15

Let the time period of the buses are TA, TB,  angular velocity is ωA, ωB and linear velocity is vA, vB. We know that for circular motion

According to the question time period is same for both the buses

By the relation of v = ωr 

BITSAT Physics Test - 3 - Question 16

A particle moves in a straight line with variable acceleration = at ,SI units (a is +ve constant and t is time). Find the speed of the particle at t = 2s, if the velocity is +1 m/s at t = 1s

Detailed Solution for BITSAT Physics Test - 3 - Question 16

acceleration = at

BITSAT Physics Test - 3 - Question 17

Assertion: In case of projectile motion, the magnitude of rate of change of velocity is variable.
Reason: In projectile motion, magnitude of velocity first decreases and then increases during the motion.

Detailed Solution for BITSAT Physics Test - 3 - Question 17

Assertion: Magnitude of change in velocity is acceleration due to gravity which is constant throughout the motion.
Reason: Initially acceleration is opposite to the vertical component of velocity which decreases its magnitude till highest point. After crossing highest point acceleration is in the direction vertical component of velocity which decrease magnitude of velocity.
Hence, Assertion is false and reason is true.

BITSAT Physics Test - 3 - Question 18

In the given figure, a = 15 m s−2 represents the total acceleration of a particle moving in the clockwise direction in a circle of radius R = 2.5 m at a given instant of time. The speed of the particle is

Detailed Solution for BITSAT Physics Test - 3 - Question 18

BITSAT Physics Test - 3 - Question 19

A particle of mass m is projected with a velocity v making an angle of 30° with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height h is

Detailed Solution for BITSAT Physics Test - 3 - Question 19

Angular momentum of the projectile
L = mvhr⊥
= m(vcosθ)h (where h is the maximum height)

BITSAT Physics Test - 3 - Question 20

Assertion: Magnitude of instantaneous velocity is equal to instantaneous speed.
Reason: Distance is nearly equal to displacement if displacement is very small.

Detailed Solution for BITSAT Physics Test - 3 - Question 20

The instantaneous speed is always equal to the magnitude of instantaneous velocity.
This is because, while calculating instantaneous velocity, the time interval taken into consideration is infinitesimally small.

When the time interval is extremely small, distance is approximately equal to displacement. 
Instantaneous speed = Instantaneous velocity

BITSAT Physics Test - 3 - Question 21

A closed cylinder of length ' l ' containing a liquid of variable density ρ(x) = ρ0(1 + αx) is rotating about a vertical axis with an angular speed ω. Find the net force exerted by the liquid on the axis of rotation. (Take the cylinder to be massless and A = cross sectional area of cylinder, ignore the gravity) -

Detailed Solution for BITSAT Physics Test - 3 - Question 21

The position of COM is given by

Net mass is given by

Net force on Axis

BITSAT Physics Test - 3 - Question 22

A uniformly tang vessel is filled with a liquid of density 900 kg m−3. The force that acts on the base of the vessel due to the liquid is (g = 10 m s−2)

Detailed Solution for BITSAT Physics Test - 3 - Question 22

Force acting on the base = P × A
= (dhg)A
= 0.4 × 900 × 10 × 2 × 10-3
= 7.2N

BITSAT Physics Test - 3 - Question 23

If two soap bubbles of different radii are connected by a tube

Detailed Solution for BITSAT Physics Test - 3 - Question 23

Let radius of curvature of the common internal film surface of the double bubble formed be r'. Then, excess of pressure as
compared to atmosphere inside 

The pressure difference is

BITSAT Physics Test - 3 - Question 24

Two capillary tubes of radii 0.2 cm and 0.4 cm are dipped in the same liquid. The ratio of heights through which liquid will rise in the tubes is

Detailed Solution for BITSAT Physics Test - 3 - Question 24

The expression of capillarity is 

BITSAT Physics Test - 3 - Question 25

A layer of glycerine of thickness 1 mm is present between a large surface area and a surface area of 0.1 m2. With what force the small surface is to be pulled, so that it can move with a velocity of 1 ms−1 1? (Given that coefficient of viscosity = 0.07 kg m−1 s−1)

Detailed Solution for BITSAT Physics Test - 3 - Question 25

Given,
ɳ = 0.07 kg m−1s−1, dv = 1 ms−1,
dx = 1 mm = 1 × 10−3 m,   A = 0.1m2   

BITSAT Physics Test - 3 - Question 26

To what height should a cylindrical vessel be filled with a homogeneous liquid to make the force with which the liquid presses on the sides of the vessel equal to the force exerted by the liquid on the bottom of the vessel. If should be

Detailed Solution for BITSAT Physics Test - 3 - Question 26

Let h be the desired height of liquid in cylinder for which the force on the bottom and sides of the vessel is equal
Force on bottom = ρ gh × πR2
Force on the walls of vessel
= ρ g πh2R = ρ g h π Rh or R = h

BITSAT Physics Test - 3 - Question 27

A spherical drop of water has radius 1 mm if surface tension of water is 70 × 10−3 Nm−1, difference of pressure between inside and outside of the spherical drop is

Detailed Solution for BITSAT Physics Test - 3 - Question 27

Excess pressure  in case of liquid drop,

BITSAT Physics Test - 3 - Question 28

A mercury drop of radius 1 cm is broken into 106 droplets of equal size. The work done is (S = 35 × 10−2 Nm−1)

Detailed Solution for BITSAT Physics Test - 3 - Question 28

If r is the radius of small droplet and R is the radius of big drop, then according to question,

Work done = surface tension × increase in area

BITSAT Physics Test - 3 - Question 29

A spherical solid ball of volume V is made of a material of density ρ2< ρ1). [Assume that the liquid applies a viscous force on the ball that is proportional to the square of its speedv,ie,Fviscous = −kv2(k > 0)]. The terminal speed of the ball is

Detailed Solution for BITSAT Physics Test - 3 - Question 29

The force acting on the ball are gravity force, buoyancy force and viscous force. When ball acquires terminal speed, it is in dynamic equilibrium, let terminal speed of ball is vT.

BITSAT Physics Test - 3 - Question 30

Eight equal drops of water are falling through air with a steady velocity of 10 cms−1. if the drops combine to form a single drop big size, then the terminal velocity of this big drop is

Detailed Solution for BITSAT Physics Test - 3 - Question 30

Terminal velocity is proportional to square of radius of the sphere.
v ∝ r2 

and using volume conservation,

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