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BITSAT Practice Test - 15 - JEE MCQ


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30 Questions MCQ Test BITSAT Mock Tests Series & Past Year Papers 2025 - BITSAT Practice Test - 15

BITSAT Practice Test - 15 for JEE 2024 is part of BITSAT Mock Tests Series & Past Year Papers 2025 preparation. The BITSAT Practice Test - 15 questions and answers have been prepared according to the JEE exam syllabus.The BITSAT Practice Test - 15 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for BITSAT Practice Test - 15 below.
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BITSAT Practice Test - 15 - Question 1

A body starts from rest and moves with uniform acceleration. Which of the following graphs represents it motion?

Detailed Solution for BITSAT Practice Test - 15 - Question 1

BITSAT Practice Test - 15 - Question 2

One mole of a monatomic ideal gas is taken through the cycle shown in diagram
   A → B adiabatic expansion
   B → C cooling at constant volume
   C → D adiabatic compression
  D → A heating at constant volume
The pressure and temperature at A, B, etc., are denoted by PA, TA; PB, TB, etc., respectively.
Given TA = 1000 K, PB = (2/3) PA and PC = (1/3) PA.
Calculate the work done by the gas in the process A → B, the heat lost by the gas in the process B → C and temperature TD.
Given (2/3)2/5 = 0.85 and R = 8.31 J/mol K.

Detailed Solution for BITSAT Practice Test - 15 - Question 2

As for adiabatic change PVγ = constt.,

For AB in the cyclic process,

Hence, the work done in AB process,

i.e., WAB = (3/2) x 8.31 x 150 = 1869.75 J

For B → C, V = constt. so ΔW = 0.

So from first law of thermodynamics,

Now as along path BC, V = constt., P ∝ T,

Negative heat means heat is lost by the system.

As A and D are on the same isochor,

But C and D are on the same adiabatic,

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BITSAT Practice Test - 15 - Question 3

Four point positive charges of same magnitude(Q) are placed at the corners of a square. The electric field at the center of the square of side 'a' is :

Detailed Solution for BITSAT Practice Test - 15 - Question 3

BITSAT Practice Test - 15 - Question 4

Two particles of masses ma and mb and and same charge are projected in a perpendicular magnetic field. They travel along circular paths of radius ra and rand such that r> rb. Then which is true?

Detailed Solution for BITSAT Practice Test - 15 - Question 4

BITSAT Practice Test - 15 - Question 5

Two progressive waves having equation x1=3 sin ωt sin and x2=4 sin (ωt + 90o) are super imposed. The amplitude of the resultant wave is :

Detailed Solution for BITSAT Practice Test - 15 - Question 5

x1 = 3 sin ωt

x2=4 sin (ωt + 90o)

The phase difference between the two waves is 90o.

So,resultant amplitude

BITSAT Practice Test - 15 - Question 6

Spot the wrong statement:

The acceleration due to gravity g decreases if,

Detailed Solution for BITSAT Practice Test - 15 - Question 6

(i) Effect due to rotation: Acceleration due to gravity is maximum at poles because there is no effect of earth's rotation at poles which are situated on the axis of rotation. Also, the earth is flattened at the poles due to which the distance from centre of the earth decreases and cause an increase in gravity. 

Acceleration due to gravity is minimum at the equator because there is maximum possible effect of the earth's rotation.

So, as we move from the equator to the poles, the acceleration due to gravity g increases. 

(ii) Effect due to radial distance from centre: The acceleration due to gravity is maximum at earth's surface. It decreases on moving towards the centre and above the surface.

BITSAT Practice Test - 15 - Question 7

A machine gun fires six bullets per second into a target. The mass of each bullet is 3 g and the speed 500 m s−1. The power delivered to the bullets is

Detailed Solution for BITSAT Practice Test - 15 - Question 7

The power delivered is the work done per second which again is equal to the total kinetic energy of the six bullets fired per second.

Hence power P=6 (kinetic energy of each bullet)

BITSAT Practice Test - 15 - Question 8

A hollow cylinder of radius R and height h (<< R) is completely filled with a liquid having coefficient of viscosity η. Now a cylinder of radius R is placed over it and is rotated with constant angular velocity ω by an external agent as shown.

Detailed Solution for BITSAT Practice Test - 15 - Question 8

BITSAT Practice Test - 15 - Question 9

The wavelength of characteristic X-rays Kα line emitted by hydrogen like atom is 0.32 Å. The wavelength of Kβ line emitted by the same element is

Detailed Solution for BITSAT Practice Test - 15 - Question 9

From Rydberg's relation for Kα.

From Rydberg's relation for Kβ,

On dividing (1) by (2)

BITSAT Practice Test - 15 - Question 10

Find the rms value of the saw tooth wave form shown in figure -

Detailed Solution for BITSAT Practice Test - 15 - Question 10

Equation of straight line from t=0 to t=1 using straight line equation y=mx+c  where m is slope and c is the intercept
We get

BITSAT Practice Test - 15 - Question 11

Assertion: string can never remain horizontal, when loaded at the middle, howsoever large the tension may be.

Reason: For horizontal string, angle with vertical, 

Detailed Solution for BITSAT Practice Test - 15 - Question 11

If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.

BITSAT Practice Test - 15 - Question 12

Two circular coils mounted parallel to each other on the same axis carry steady currents. If an observer between the coils reports that one coil is carrying a clockwise current i, while the other is carrying a counter clockwise current i, between the two coils, then there is

Detailed Solution for BITSAT Practice Test - 15 - Question 12

Since the current-carrying loop acts as an electromagnet, with application of Right-hand thumb rule, it can be stated that the face of the loop where the current flows in a clockwise direction behaves as a South pole and the other face where the current flows in an anti-clockwise direction behaves as a North pole.

Since, there is a force of attraction between the opposite poles, these coils will have a steady force of attraction between them.

BITSAT Practice Test - 15 - Question 13

A body of mass M is dropped from a height h on a sand floor. If the body penetrates x cm into the sand, the average resistance offered by the sand to the body is

Detailed Solution for BITSAT Practice Test - 15 - Question 13

If the body strikes the sand floor with a velocity v, then 

With this velocity v, when body passes through the sand floor it comes to rest after travelling a distance x.
 Let F be the resisting force acting on the body. Net force in downwards direction

Work done by all the forces is equal to change in KE

BITSAT Practice Test - 15 - Question 14

Which of the following device works on the principle that the charge between the two flow if they are connected by a conducting wire:

Detailed Solution for BITSAT Practice Test - 15 - Question 14

When both spheres are connected by a conducting wire then the charge q on smaller sphere will flow onto the larger sphere having charge Q.

Van de Graff generator work on this fact that charge given to a hollow conductor is transferred to outer surface and is distributed uniformly over it.

BITSAT Practice Test - 15 - Question 15

In a certain double slit experimental arrangement, interference fringes of width 1.0 mm each are observed when light of wavelength 5000 Å is used. Keeping the set-up unaltered if the source is replaced by another wavelength of 6000 Å, the fringe width will be -

Detailed Solution for BITSAT Practice Test - 15 - Question 15

BITSAT Practice Test - 15 - Question 16

The displacement x of a particle varies with time according to the relation  Then,

Detailed Solution for BITSAT Practice Test - 15 - Question 16

BITSAT Practice Test - 15 - Question 17

A horizontal force of 10 N is necessary to hold a stationary block against a wall. The coefficient of friction between the block and the wall is 0.2. The weight of the block is

Detailed Solution for BITSAT Practice Test - 15 - Question 17

N = F = 10N
mg = f = μN = 0.2 x 10
Weight = mg = 2 N

BITSAT Practice Test - 15 - Question 18

Two blocks of masses m and M are placed on a horizontal frictionless table connected by a spring as shown in the figure. The block of mass M is pulled to the right with a force F. If the acceleration of the block of mass m is 'a', then the acceleration of the block of mass M will be

Detailed Solution for BITSAT Practice Test - 15 - Question 18

For the block with mass m, the force experienced is the spring force.
∴ Spring force = Mass × acceleration = ma
For block with mass M, forces are spring force and the external F

BITSAT Practice Test - 15 - Question 19

A smooth sphere 'A' is moving on a frictionless horizontal surface with angular speed ω and centre of mass velocity v. It collides elastically and head-on with an identical sphere B at rest. Neglect friction everywhere. After the collision, their angular speeds are ωA and ωB, respectively. Then,

Detailed Solution for BITSAT Practice Test - 15 - Question 19

Since there is no friction between the sphere and the horizontal surface and also between the spheres themselves, there will be no transfer of angular momentum from sphere A to sphere B due to the collision. Since the collision is elastic and the spheres have the same mass, the sphere A only transfers its linear velocity v to sphere B. Sphere A will continue to rotate with the same angular speed ω at a fixed location. Hence the correct choice is (3).

BITSAT Practice Test - 15 - Question 20

The angular momentum of the Earth revolving around the Sun is proportional to Rn, where R is the distance between the Earth and the Sun. The value of n is

Detailed Solution for BITSAT Practice Test - 15 - Question 20

BITSAT Practice Test - 15 - Question 21

Directions: The question below is based on the following passage:

Two light springs of force constants k1 = 1.8 Nm-1 and k2 = 3.2 Nm-1 and a block of mass m = 200 g are arranged on a horizontal frictionless table as shown in the figure. One end of each spring is fixed on rigid supports and the other end is free. The distance CD between the free ends of the springs is 60 cm and the block moves with a velocity 'v' = 120 cms-1 between the springs.

When the block moves towards the spring k2, the time taken by it to move from D up to the maximum compression of spring k2 is (in seconds)

Detailed Solution for BITSAT Practice Test - 15 - Question 21

Time taken by block to move from D upto the maximum compression of spring k2 = half of the time period of oscillation of the block if it were attached to the free end of k2, i.e.,

The correct choice is (4).

BITSAT Practice Test - 15 - Question 22

A particle executes SHM on a linear path with an amplitude of 2 cm. When the displacement is 1 cm, the magnitude of its velocity is equal to that of its acceleration. The frequency of the oscillations is

Detailed Solution for BITSAT Practice Test - 15 - Question 22

Velocity of the particle executing SHM at position y is

Magnitude of the acceleration of the particle is

Hence, the frequency of the oscillation is

BITSAT Practice Test - 15 - Question 23

Transverse wave of amplitude 10 cm is generated at one end (x = 0) of a long string by a tuning fork of frequency 500 Hz. At a certain instant of time, the displacement of a particle A at x = 100 cm is -5 cm and of particle B at x = 200 cm is +5 cm. What is the wavelength of the wave?

Detailed Solution for BITSAT Practice Test - 15 - Question 23

BITSAT Practice Test - 15 - Question 24

The figure below shows the P-V diagram for a fixed mass of an ideal gas undergoing cyclic process ABCA. AB represents an isothermal process.

Which of the following graphs represents the P-T diagram of the cyclic process?

Detailed Solution for BITSAT Practice Test - 15 - Question 24

BITSAT Practice Test - 15 - Question 25

A heater is designed to operate with a power of 1000 W in a 100 V line. It is connected in combination with a resistance of 10 Ω  and a resistance R to a 100 V line as shown in the figure. What should be the value of R so that the heater operates with a power of 62.5 W?

Detailed Solution for BITSAT Practice Test - 15 - Question 25

BITSAT Practice Test - 15 - Question 26

A coil having N turns is wound tightly in the form of a spiral, with inner and outer radii a and b, respectively. When a current I passes through the coil, the magnetic field at the centre is

Detailed Solution for BITSAT Practice Test - 15 - Question 26

Consider circular loop of Radius r with thickness 'dr'
The number of loops in ehickness dr is

BITSAT Practice Test - 15 - Question 27

A rectangular loop carrying a current i is situated near a long straight wire such that the wire is parallel to one of the sides of the loop. If a steady current is established in the wire, as shown in the figure, the loop will

Detailed Solution for BITSAT Practice Test - 15 - Question 27

Referring to the figure, the forces acting on arms BC and AD are equal and opposite. The force on arm AB is given by

which is directed towards the wire. The force on arm CD is given by

which is directed away from the wire. Since F1 > F2, the loop will move towards the wire. Hence the correct choice is (3).

BITSAT Practice Test - 15 - Question 28

A transformer has 200 windings in the primary and 400 windings in the secondary. The primary is connected to an AC supply of 110 V and a current of 10 A flows in it. The voltage across the secondary and the current in it respectively are

Detailed Solution for BITSAT Practice Test - 15 - Question 28

BITSAT Practice Test - 15 - Question 29

A point object is placed at a distance of 12 cm on the axis of a convex lens of focal length 10 cm. On the other side of the lens, a convex mirror is placed at a distance of 10 cm from the lens such that the image formed by the combination coincides with the object itself. What is the focal length of the convex mirror?

Detailed Solution for BITSAT Practice Test - 15 - Question 29

The incident ray OA is refracted along BD by the lens. Singe the image coincides with the object. the refracted ray BD must retrace its path after reflection from the mirror, i.e., it falls normally on the mirror.
Let C be the point on the axis where BD produced meets the axis. It is clear that C is the centre of curvature of the mirror. Also is the imago of the object O formed by the lens alone. Thus
f = +10 cm and u = - 12 cm
The distance v of the image is obtained from the lens formula

which gives v = - 60 cm
Since x = 10 cm. R = v - x = 60 - 10 = 50 cm.

Hence, the focal length of the mirror = R/2 = 25cm

BITSAT Practice Test - 15 - Question 30

A radioactive sample S1 having an activity of 5 μCi has twice the number of nuclei as another sample S2, which has an activity of 10μCi  The half lives of S1 and S2 can be

Detailed Solution for BITSAT Practice Test - 15 - Question 30

Of the given options, option (1) satisfies the above condition.

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