JEE Exam  >  JEE Tests  >  BITSAT Physics Test - 7 - JEE MCQ

BITSAT Physics Test - 7 - JEE MCQ


Test Description

30 Questions MCQ Test - BITSAT Physics Test - 7

BITSAT Physics Test - 7 for JEE 2024 is part of JEE preparation. The BITSAT Physics Test - 7 questions and answers have been prepared according to the JEE exam syllabus.The BITSAT Physics Test - 7 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for BITSAT Physics Test - 7 below.
Solutions of BITSAT Physics Test - 7 questions in English are available as part of our course for JEE & BITSAT Physics Test - 7 solutions in Hindi for JEE course. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free. Attempt BITSAT Physics Test - 7 | 30 questions in 40 minutes | Mock test for JEE preparation | Free important questions MCQ to study for JEE Exam | Download free PDF with solutions
BITSAT Physics Test - 7 - Question 1

Which among the following has the dimension of Volume [L3]? Provided that:
n = number of particle per unit volume
q = charge
T = Absolute temperature
KB = Boltzmann's Constant
ε = permittivity of dielectric material

Detailed Solution for BITSAT Physics Test - 7 - Question 1


For other options:

Therefore, it is the correct option.

BITSAT Physics Test - 7 - Question 2

Three blocks A, B and C are attached to an inextensible light string. The string is passed over two pullies as shown in the figure. Block A is moving downwards with a constant acceleration of a. What will be the value of a if the mass of the blocks A, B and C are m, m and 3m, respectively and the plane of incline is frictionless.

Detailed Solution for BITSAT Physics Test - 7 - Question 2


For the block A

For the bock B

For the Block C

Adding 1, 2 and 3, we have

1 Crore+ students have signed up on EduRev. Have you? Download the App
BITSAT Physics Test - 7 - Question 3

A block of mass m is residing on an inclined plane of angle θ. A force F is applied on the block in horizontal direction. The coefficient of friction between the block and the inclined surface is μ, for the block to be in static condition, the value of θ must be

Detailed Solution for BITSAT Physics Test - 7 - Question 3

For the block to be in static condition:

Putting the value of N in the first equation:

This is our required solution.

BITSAT Physics Test - 7 - Question 4

A block of 5 kg is attached to a massless spring with spring constant k = 4 N/m. The whole arrangement is placed on a plane inclined at an angle 60o. If the coefficient of friction is 0.5, the energy stored in the spring will be

Detailed Solution for BITSAT Physics Test - 7 - Question 4

Let the block moves a distance x down the inclined plane.


Using Work Energy principle, Energy stored in spring = Decrease in the gravitational potential energy - Work done by the friction

Energy stored in the spring

BITSAT Physics Test - 7 - Question 5

A uniform rod of mass 10 kg and length 1 meter is hinged from one end and is at rest on a smooth plane. An impulse p is applied on the free end of the rod and the rod is rotated 90o from the initial position in 8 seconds. The value of the impulse applied is

Detailed Solution for BITSAT Physics Test - 7 - Question 5

We know that;

Also,
Angular Impulse = Change in the angular momentum

This is our required solution.

BITSAT Physics Test - 7 - Question 6

Two balls A and B are thrown from the top of a hill of 30-meter height. Ball B is thrown sometimes after the ball A. Ball B is thrown horizontally and ball A is thrown at an angle of 30o with the vertical. If the initial velocity of both the balls are same and is equal to 20 m/s, at what point does the two balls would collide in the air during their flight?

Detailed Solution for BITSAT Physics Test - 7 - Question 6

u = 20 m/s
Therefore,

Both the balls meet at a point let's say P.
This means horizontal displacement of both the balls are equal at point P.
Therefore,

Height h;

By putting the value of tA and tB from the above equation, we get:
tA = 7 sec
tB = 3.5 sec
Therefore, x = 70 m
h = 61.25
y = h - 30 m
y = 31.25
Therefore, point of collision is (70, 31.25)

BITSAT Physics Test - 7 - Question 7

A block of mass m is in rest at a point P on a vertical semi-circular track. The point P is at an angle α as shown in the figure. If the value of the coefficient of friction between the block and the track is μ, the maximum value of α in terms of μ will given by

Detailed Solution for BITSAT Physics Test - 7 - Question 7

Draw the FBD of the block:

 

From the FBD:
μN = mg sin α
N = mg cos α
Or
μ mg cos α = mg sinα

BITSAT Physics Test - 7 - Question 8

Two bodies of mass m and 2m are moving with a velocity of 2v and v, respectively. Mass m is moving along the x axis and mass 2m is moving along the y axis. After sometime, mass 2m hits the body of mass m. The collision is purely inelastic. The percentage loss of the energy in the collision will be

Detailed Solution for BITSAT Physics Test - 7 - Question 8

During the collision, the momentum of the system will remain constant.
Let the final velocity of the two bodies after the collision is v'.
Then


This is our required solution.

BITSAT Physics Test - 7 - Question 9

Work done in moving a 5 kg body in xy plane is given by (5x + 20y) J. At time t = 0, the velocity of the body is zero and displacement x = 0. What is the magnitude of the velocity of the particle at t = 2 sec?

Detailed Solution for BITSAT Physics Test - 7 - Question 9

As we have

This is our required solution.

BITSAT Physics Test - 7 - Question 10

A sphere of radius 100 mm is rolling without slipping on a block which is moving with a constant velocity of 20 m/s. If the linear velocity of the centre of mass of the sphere for a stationary observer is 26 m/s, the angular velocity of the sphere is

Detailed Solution for BITSAT Physics Test - 7 - Question 10

For rolling without slipping, the relative velocity of the point of contact of the sphere and the block should be zero.
Therefore,

This is our required solution.

BITSAT Physics Test - 7 - Question 11

Three sources of equal intensities with frequencies 400, 401 and 402 vibrations per second are sounded together. The number of beats per second is

Detailed Solution for BITSAT Physics Test - 7 - Question 11

Beats are the periodic and repeating functions heard in the intensity of sound, when two sound waves of very similar frequency interface with one another.
Beats = Difference in frequencies.
Maximum number of beats = 402 − 400 = 2. 

BITSAT Physics Test - 7 - Question 12

At a moment in a progressive wave, the phase of a particle executing SHM is π/3. Then the phase of the particle 15 cm 15 cm ahead and at time T/2 will be, if the wavelength is 60 cm

Detailed Solution for BITSAT Physics Test - 7 - Question 12

Phase of second particle = ϕ

BITSAT Physics Test - 7 - Question 13

A person standing at a distance of 6 m from a source of sound receives sound wave in two ways, one directly from the source and others after reflection from a rigid boundary as shown in the diagram. The maximum wavelength for which, the person will receive maximum sound intensity, is

Detailed Solution for BITSAT Physics Test - 7 - Question 13


⇒  Δx = (5 + 5) − 6 = 4 m
For maxima, Δx = nλ ⇒ 4 = nλ

nmin = 1
⇒ λmax = 4m

BITSAT Physics Test - 7 - Question 14

Two trains A and B are moving with speeds 20 m/s and 30 m/s respectively in the same direction on the same straight track, with B ahead of A. The engines are at the front ends. The engine of trains A blows a long whistle.

Assume that the sound of the whistle is composed of components varying in frequency from f1 = 800 Hz to f= 1120 Hz, as shown in the figure. The spread in the frequency (highest frequency - lowest frequency) is thus 320 Hz. The speed of sound in still air is 340 m/s.
The distribution of the sound intensity of the whistle as observed by the passengers in train A is best represented by

Detailed Solution for BITSAT Physics Test - 7 - Question 14

The engine as well as the passengers in train A are moving with the same velocity, therefore the distribution of sound intensity in the whistle remains unchanged.

BITSAT Physics Test - 7 - Question 15

Two instruments having stretched strings are being played in union. When the tension of one of the instruments is increased by 1%, 3 beats are produced in 2 s. The initial frequency of vibration of each wire is

Detailed Solution for BITSAT Physics Test - 7 - Question 15

Frequency

BITSAT Physics Test - 7 - Question 16

In Melde's experiment, the string vibrates in 4 loops when a 50g weight is placed in the pan of weight 15g. To make the string to vibrates in 6 loops the weight that has to be removed from the pan is

Detailed Solution for BITSAT Physics Test - 7 - Question 16

The transverse vibrations of a string are determined by Melde's method.
The frequency of vibration of a string of length l, mass per unit length m and vibration in p loops under tension T is given by

Or

If n, l and m are constant
Hence,


So, weight removed from the pan
= 65 - 29
= 36 g
= 0.036 kg-wt

BITSAT Physics Test - 7 - Question 17

A string of length l = 10 cm is fixed at one end and free at other. The maximum possible wavelength that can be generated in the string is

Detailed Solution for BITSAT Physics Test - 7 - Question 17

The maximum wavelength is equal to four times the length of the string, λ/4 = 10 cm
⇒ λ = 40 cm

BITSAT Physics Test - 7 - Question 18

When two wave of almost equal frequencies n1 and n2 are produced simultaneously, then the time interval between successive maxima is

Detailed Solution for BITSAT Physics Test - 7 - Question 18

Number of beats s−1 = n1 − n2 
∴ Time interval between two successive beats/successive maxima 

BITSAT Physics Test - 7 - Question 19

In the production of beats by two waves of same amplitude and nearly same frequency, the maximum intensity to each of the constituent waves is

Detailed Solution for BITSAT Physics Test - 7 - Question 19

Given that amplitude A, and frequency of both wave the same.
We know that, the intensity I ∝ A2,
⇒I ∝ K A2 ......(1)
In constructive interference, both amplitudes of the wave added and intensity will be maximum. Amax = 2A.
Therefore, intensity equation I ∝ 4A2 ,
⇒ I = 4A2 ......(2)
So, Beats formation in case of constructive interference for which the maximum intensity is 4 times.

BITSAT Physics Test - 7 - Question 20

In the experiment for the determination of the speed of sound in air using the resonance column , it resonates in the fundamental mode, with a tuning fork when the length is 0.1 m . When this length is changed to 0.35 m, the same tuning fork resonates with the first overtone. Calculate the end correction.

Detailed Solution for BITSAT Physics Test - 7 - Question 20

Let Δl be the end correction. Given that,
Fundamental tone for a length 0.1m = first overtone for the length 0.35m

Solving this equations we get Δl = 0.025m = 2.5 cm

BITSAT Physics Test - 7 - Question 21

A body of mass m=3.513 kg  is moving along the x− axis with a speed of 5.00 ms−1. The magnitude of its momentum (according to significant figures) is recorded as

Detailed Solution for BITSAT Physics Test - 7 - Question 21

Given, m = 3.513 kg and v = 5.00 ms−1.
So, momentum, p = mv = 17.565
As the number of significant digits in m is 4 and v is 3, so, p must have 3 significant digits p = 17.6 kgms−1 

BITSAT Physics Test - 7 - Question 22

1 ns is defined as

Detailed Solution for BITSAT Physics Test - 7 - Question 22

The second is the duration of 9192631770 period of the radiation corresponding to the transition between the two hyperfine levels of the ground state of Cesium-133 atom. Therefore, 1 ns is 10−9 s of Cs-clock of 9192631770 oscillations.

BITSAT Physics Test - 7 - Question 23

Unit of electric flux is

Detailed Solution for BITSAT Physics Test - 7 - Question 23

If E is the intensity of electric field over a small area element dS and θ is angle between E and outdrawn normal to area element. Therefore, electric flux through this element is
dϕE = (dS)(E cosθ)

Hence,

= volt − metre

BITSAT Physics Test - 7 - Question 24

While measuring length of an object it was observed that the zero of the vernier lies between 1.4 and 1.5 of the main scale and the fifth vernier division coincides with a main scale division. If the length of the object measured is l, then the value of (l − 1.4) in terms of the least count C of the instrument is

Detailed Solution for BITSAT Physics Test - 7 - Question 24

Let x = l − 1.4
Then, 1.4 + x + 5v = 1.4 + 5s, where s denotes the length of one main scale division, and v denotes the length of one vernier scale division.
Hence, x = 5s − 5v = 5(1s − 1v) = 5 C.

BITSAT Physics Test - 7 - Question 25

The least count of the main scale of a screw gauge is 1 mm. The minimum number of divisions on its circular scale required to measure 5 μm diameter of a wire is:

Detailed Solution for BITSAT Physics Test - 7 - Question 25

Least count of screw gauge = 5 μm

BITSAT Physics Test - 7 - Question 26

The radius of the proton is about 10−15m. The radius of the observable universe is 1026m. identify the distance which is half-way between these two extremes on a logarithmic scale.

Detailed Solution for BITSAT Physics Test - 7 - Question 26

BITSAT Physics Test - 7 - Question 27

A new unit of length is chosen such that the speed of light in vacuum is unity. What is the distance between the sun and the earth in terms of this new unit, if light takes 8 min and 20 sec to cover this distance?

Detailed Solution for BITSAT Physics Test - 7 - Question 27

The speed of light in vacuum (c) = 1 (new unit of length s–1)
Time taken by light to reach the earth
t = 8 min + 20 s
= (8 × 60 + 20) s = 500 s
∴ Distance between the sun and the earth = Speed of light × Time
x = c × t 
= 1 (new unit of length s–1) × 500 s
= 500 new unit of length

BITSAT Physics Test - 7 - Question 28

The legs of a spherometer are 5 cm apart. There are 10 division cm−1 on linear scale and circular scale has 100 100 divisions. The height h of a convex mirror measured in 2 MSD + 37 circular scale divisions. Find the radius of curvature of a convex mirror.

Detailed Solution for BITSAT Physics Test - 7 - Question 28

BITSAT Physics Test - 7 - Question 29

In an experiment for the determination of the speed of sound in air using the resonance method, the length of the air column that resonates in the fundamental mode with a tuning fork is 0.1 m. When this length is changed to 0.35 m , the same tuning fork resonates with the first overtone. Calculate the end correction.

Detailed Solution for BITSAT Physics Test - 7 - Question 29

BITSAT Physics Test - 7 - Question 30

The diameter of a cylinder is measured using a vernier callipers with no zero error. It is found that the  zero of the vernier scale lies between 5.10 cm and 5.15 cm of the main scale. The vernier scale has 50 divisions equivalent to 2.45 cm. The 24th division of the vernier scale exactly coincides with one of the main scale divisions. The diameter of the cylinder is 

Detailed Solution for BITSAT Physics Test - 7 - Question 30

1 MSD = 0.05 ​

Diameter of the cylinder
= MSR + L.C. × 24
= 5.124 cm

Information about BITSAT Physics Test - 7 Page
In this test you can find the Exam questions for BITSAT Physics Test - 7 solved & explained in the simplest way possible. Besides giving Questions and answers for BITSAT Physics Test - 7, EduRev gives you an ample number of Online tests for practice

Top Courses for JEE

Download as PDF

Top Courses for JEE