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BITSAT Practice Test - 13 - JEE MCQ


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30 Questions MCQ Test BITSAT Mock Tests Series & Past Year Papers 2025 - BITSAT Practice Test - 13

BITSAT Practice Test - 13 for JEE 2024 is part of BITSAT Mock Tests Series & Past Year Papers 2025 preparation. The BITSAT Practice Test - 13 questions and answers have been prepared according to the JEE exam syllabus.The BITSAT Practice Test - 13 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for BITSAT Practice Test - 13 below.
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BITSAT Practice Test - 13 - Question 1

If the source of light used in a Young's double slit experiment is changed from red to violet,

Detailed Solution for BITSAT Practice Test - 13 - Question 1

where, D=distance between slit and screen,
d = distance of separation between the slits.
λ = wavelength of light used.
Let λR = wavelength of red light
and λ= wavelength of violet light.

We know, 
λR > λV
So, if we use violet light, fringe width decrease and fringes will come close to each other.

BITSAT Practice Test - 13 - Question 2

A body in laboratory takes 4 minutes to cool from 61℃ to 59℃. If the laboratory temperature is 30℃, then the time taken by it to cool from 51℃  to 49℃ is

Detailed Solution for BITSAT Practice Test - 13 - Question 2

According to Newton's law of cooling,

By newton's law of cooling,

On calculating,

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BITSAT Practice Test - 13 - Question 3

The second overtone of an open pipe A and a closed pipe B have the same frequencies at a given temperature. Both the pipes contain air. The ratio of fundamental frequency of A to that of B is,

Detailed Solution for BITSAT Practice Test - 13 - Question 3

Let, v be the velocity of a wave.
Length of open pipe A be l1.
 Length of closed pipe B be l2.

Formula used:

nth overtone of an open pipe f= (n+1)f0, where for open pipe fundamental frequency f= v/2l
nth overtone of a closed pipe fn = (2n+1)f0, where for closed pipe fundamental frequency f= v/4l.

Now, solving the problem


Since, it is given that frequencies are same,

Now, the ratio of fundamental frequencies,

BITSAT Practice Test - 13 - Question 4

A particle of mass 5 kg moving in the x−y plane has its potential energy given by U = (−7x+24y) Joule, x and y
being in meter. Initially at t =0 ,the particle is located at the origin (0, 0) and moving with a velocity of . Find the magnitude of force acting on the particle.

Detailed Solution for BITSAT Practice Test - 13 - Question 4

BITSAT Practice Test - 13 - Question 5

In a double-slit experiment, the separation between the slits is d=0.25  cm and the distance of the screen D=100  cm from the slits. If the wavelength of light used is  and I0 is the intensity of the central bright fringe, the intensity at a distance x=4 x 10−5  m from central maximum is

Detailed Solution for BITSAT Practice Test - 13 - Question 5

BITSAT Practice Test - 13 - Question 6

A rocket is fired perpendicular to the surface of the earth with a speed equal to 50% of the escape velocity from earth surface. The maximum height reached by it in terms of radius R of the earth is :

Detailed Solution for BITSAT Practice Test - 13 - Question 6

When the body's kinetic energy approaches zero, it reaches its maximum height.

Equating maximum height and kinetic energy 

BITSAT Practice Test - 13 - Question 7

The escape velocity from the earth is 11 km s−1. The escape velocity from a planet having twice the radius and same mean density as that of the earth is

Detailed Solution for BITSAT Practice Test - 13 - Question 7

Escape velocity,

∴ve∝R if ρ=constant.

Since the planet is having double radius in comparision to earth, therefore escape velocity becomes twice ie, 22 km s−1 .

BITSAT Practice Test - 13 - Question 8

In Bernoulli's theorem, which of the following is conserved?

Detailed Solution for BITSAT Practice Test - 13 - Question 8

Here,
P− P2 is the pressure difference between points A and B,
v1 is the velocity of the fluid at point A,  
vis the velocity of the fluid at point B and
h− h1 is the height difference between them.

The sum of energy per unit volume due to pressure (Pressure energy), energy per unit volume due to motion(Kinetic energy) and energy per unit volume due to gravity(Gravitational potential energy) remains constant, i.e., energy per unit volume remains constant, or we can say that in Bernoulli's theorem the energy remains conserved.

BITSAT Practice Test - 13 - Question 9

The vector product of the force (F) and distance (r) from the centre of action represents

Detailed Solution for BITSAT Practice Test - 13 - Question 9

Torque is a measure of how much a force acting on an object causes that object to rotate. The object rotates about an axis (O). The distance from O is r,
where forces acts, hence torque τ=r x F. It is a vector quantity and points from axis of rotation to the point where the force acts.

BITSAT Practice Test - 13 - Question 10

When a tuning fork of frequency 341 Hz is sounded with another tuning fork, six beats per second are heard. When the second tuning fork is loaded with wax and sounded with the first tuning fork, the number of beats is two per second. The natural frequency of the second tuning fork is

Detailed Solution for BITSAT Practice Test - 13 - Question 10

Let frequency of second fork =x
We know that beat frequency =|f− f2|

On loading the fork with wax, its frequency decreases. So beat frequency will increase if,x=335 Hz and decrease if, x=347 Hz. According to the given condition, beat frequency decreases, so x=347 Hz

BITSAT Practice Test - 13 - Question 11

A force of 1200 N acts on a 0.5 kg steel ball as result of collision lasting 25 ms. If the force is in a direction opposite to the initial velocity of 14 ms then the final speed of the steel ball would be

Detailed Solution for BITSAT Practice Test - 13 - Question 11

Initial velocity of steel ball u=14 m s−1
Mass of steel ball m=0.5
Force acting on the steel ball from opposite direction.

F = −1200 N
Time duration for collision,

t = 25 ms = 25 x 10−3 s
Let final velocity of ball=v
Applying Newton's II law of motion to steel ball ie,

[−ve sign shows that now ball will move in the direction of force]

BITSAT Practice Test - 13 - Question 12

Figure shows, a glass prism ABC (refractive index 1.5), immersed in water (refractive index 43). A ray of light incident normally on face AB. If it is totally reflected at face AC then

Detailed Solution for BITSAT Practice Test - 13 - Question 12

The ray falls normally on AB is refracted undeviated into the prism and incident on face AC at an angle of incidence i=θ.

So, for total reflection at AC face,

BITSAT Practice Test - 13 - Question 13

A hot liquid is kept in a big room. The logarithm of the numerical value of the temperature difference between the liquid and the room is plotted against time. The plot will be very nearly,

Detailed Solution for BITSAT Practice Test - 13 - Question 13

Using Newtons law of cooling 

where 

integrating above equation we have , 

Clearly above relation is linear, so graph will be straight line.

BITSAT Practice Test - 13 - Question 14

Three identical balls, ball I, ball II and ball III are placed on a smooth floor on a straight line at a separation of 10 m
between them as shown in the figure. Initially, the balls are stationary. Ball I is given a velocity of 10 m s−1 towards ball II. The collision between ball I and II is inelastic with coefficient of restitution 0.5 but collision between ball II
and III is perfectly elastic. What is the time interval between two consecutive collisions between ball I and II ?

Detailed Solution for BITSAT Practice Test - 13 - Question 14

Let the velocity of ball I and ball III after collision be v1 and v2.
v− v= 0.5 x 10  ...(i)
mv+ mv= m x 10   ...(ii)
⇒   v+ v1=10
Solving equations (i) and (ii),

v1=2.5 m s−1 
v2=7.5 m s−1

Ball II after moving 10 m collides with ball III elastically and stops. But ball I moves towards ball II. Time taken between two consecutive collisions,

BITSAT Practice Test - 13 - Question 15

A semiconductor has an electron concentration of 8×1013 per cmand a hole concentration of 5 x 1012 per cm3. The electron mobility is 25000 cm2 V−1 s−1 
 and the hole mobility is 100 cm2 V. s−1 . Then,

Detailed Solution for BITSAT Practice Test - 13 - Question 15

BITSAT Practice Test - 13 - Question 16

A particle moves in xy plane according to the law x = a sin ωt and y = a(1 - cos ωt), where a and w are constant. The particle traces

Detailed Solution for BITSAT Practice Test - 13 - Question 16

x = a sin t ...(i)
y = a(1 - cos t) ...(ii)
x2 = a2 sin2 ωt = a2(1 - cos2 ωt)
x2 = a2 - a2 cos2 t
a2 cos2 ωt = a2 - x2 ...(iii)
y = a(1 - cos t) = a - a cos t
a - y = a cos t
(a - y)2 = (a cos t)2
a2 + y2 - 2ay = a2 cos2 ωt ...(iv)
Put the value from equation (iii),
a2 + y2 - 2ay = a2 - x2
x2 + y2 - 2ay = 0
This is the equation of a circle.

BITSAT Practice Test - 13 - Question 17

In the given figure, the block of mass M is at rest on the floor. The acceleration with which a boy of mass m should climb along the rope of negligible mass so as to lift the block from the floor, is

Detailed Solution for BITSAT Practice Test - 13 - Question 17

BITSAT Practice Test - 13 - Question 18

A uniform metal chain is placed on a rough table, such that one end of it hangs down over the edge of the table. When one-third of its length hangs over the edge, the chain starts sliding. Then, the coefficient of static friction is

Detailed Solution for BITSAT Practice Test - 13 - Question 18

Weight of the hanging portion of the chain = mg/3
Weight of the part of the chain on the table = 2mg/3

For the hanging part of the chain,


For the part of the chain on the table, we have

and we have

From 1, 2, 3 and 4, we have

BITSAT Practice Test - 13 - Question 19

A mass of 1 kg suspended by a thread deviates through an angle of 30°. Find the tension of the thread at the moment the weight passes through the position of equilibrium?

Detailed Solution for BITSAT Practice Test - 13 - Question 19

By the conservation of energy

BITSAT Practice Test - 13 - Question 20

Two masses M and m (with M > m) are connected by means of a pulley as shown in the figure. The system is released. At the instant when mass M has fallen through a distance h, the velocity of mass m will be

Detailed Solution for BITSAT Practice Test - 13 - Question 20

If mass m falls through a distance h, mass m rises up through the same distance h.
Let v be the common velocity of the masses when this happens.
Now, loss in PE = gain in KE, i.e.

BITSAT Practice Test - 13 - Question 21

Two identical solid copper spheres of radius R are placed in contact with each other. The gravitational attraction between them is proportional to

Detailed Solution for BITSAT Practice Test - 13 - Question 21

Gravitational force between two similar spheres of mass M and radius R in contact with each other,


Distance between the centres of the two spheres is 2 R.

Mass = Density x Volume of the sphere

BITSAT Practice Test - 13 - Question 22

A pendulum made of a uniform wire of cross-sectional area A has time period T. When an additional mass M is added to its bob, the time period changes to TM. If the Young's modulus of the material of the wire is Y, then 1/Y is equal to
(g = gravitational acceleration)

Detailed Solution for BITSAT Practice Test - 13 - Question 22

BITSAT Practice Test - 13 - Question 23

For a particle executing simple harmonic motion, the displacement x is given by x = A cosωt. Identify the graphs which represent the variation of potential energy (PE) as a function of time t and displacement x (see figure).

Detailed Solution for BITSAT Practice Test - 13 - Question 23

Given x = A cos ωt. As a function of x, the PE is given by

At x = 0, PE = 0.
Hence, the correct graph is III.
As a function of t, the PE is given by


At t = 0, PE is maximum equal to 

Hence, the correct graph is I.
Thus, the correct choice is (1).

BITSAT Practice Test - 13 - Question 24

1 mole of a gas with γ = 7/5 is mixed with 1 mole of a gas with γ = 5/3, then the value of γ for the resulting mixture is

Detailed Solution for BITSAT Practice Test - 13 - Question 24

BITSAT Practice Test - 13 - Question 25

The figure below shows a circuit with two cells in opposition to each other. One cell has an emf of 6 V and internal resistance of 2Ω and the other cell has an emf of 4 V and internal resistance of 8Ω. The potential difference across the terminals X and Y is

Detailed Solution for BITSAT Practice Test - 13 - Question 25

Since the two cells are in opposition, the effective voltage = 6 - 4 = 2 V. The current in the circuit is

∴ Terminal voltage of 6 V cell = 6 - 2  0.2 = 5.6 V
Terminal voltage of 4 V cell = 4 + 8  0.2 = 5.6 V
Therefore, the potential difference across terminals X and Y is 5.6 V. Hence, the correct option is (2).

BITSAT Practice Test - 13 - Question 26

The potential difference between the terminals of a cell in open circuit is 2.2 volt with resistance of 5 ohm across the terminals of a cell, the terminal potential difference is 1.8 volt. The internal resistance of the cell is

Detailed Solution for BITSAT Practice Test - 13 - Question 26

BITSAT Practice Test - 13 - Question 27

The flux linked with a coil at any instant t is given by
φ = 10t2 - 50t + 250
The induced emf at t = 3 s is

Detailed Solution for BITSAT Practice Test - 13 - Question 27

According to Lenz's law,

BITSAT Practice Test - 13 - Question 28

In the given circuit, the current drawn from the source is

Detailed Solution for BITSAT Practice Test - 13 - Question 28

Impedence of the circuit,

Peak value of the current,


Hence, r.m.s. value of the current is

BITSAT Practice Test - 13 - Question 29

A soap film of refractive index 4/3 is illuminated by white light incident at an angle of 45°. The transmitted light is examined in a spectroscope and a bright band is found to be at 6000 . Find the thickness of the film.

Detailed Solution for BITSAT Practice Test - 13 - Question 29

For transmitted light, bright fringe is formed if 2nt cos r = mλ
Or t = (mλ)/(2n cos r) ....................(1)

BITSAT Practice Test - 13 - Question 30

A radioactive element of mass number 208 at rest disintegrates by emitting an α-particle. If E is the energy of the emitted α-particle, then the energy of disintegration is

Detailed Solution for BITSAT Practice Test - 13 - Question 30

MAss number of daughter nucleus (M) = 208- 4 = 204
Now, total energy of distintegration = energy of daughter nucleus + energy of α-particle

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