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BITSAT Practice Test - 11 - JEE MCQ


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30 Questions MCQ Test BITSAT Mock Tests Series & Past Year Papers 2025 - BITSAT Practice Test - 11

BITSAT Practice Test - 11 for JEE 2024 is part of BITSAT Mock Tests Series & Past Year Papers 2025 preparation. The BITSAT Practice Test - 11 questions and answers have been prepared according to the JEE exam syllabus.The BITSAT Practice Test - 11 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for BITSAT Practice Test - 11 below.
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BITSAT Practice Test - 11 - Question 1

Two waves of equal amplitude A, and equal frequency travel in the same direction in a medium. The amplitude of the resultant wave is

Detailed Solution for BITSAT Practice Test - 11 - Question 1

Resultant amplitude of two waves with amplitude A1 and A2 and a phase difference of ϕ is given by.

For the current situation, A1=A2=A.

So, resultant becomes

The maximum amplitude will be possible when the waves have no phase difference(ϕ=0).

The resultant amplitude will be 

The minimum amplitude will be possible when the waves have a phase difference of π.

The resultant amplitude will be 

The amplitude of the resultant wave is therefore between 0 and 2A.

BITSAT Practice Test - 11 - Question 2

The image of an object, formed by a plano - convex lens at a distance of 8 m behind the lens, is real and is one-third the size of the object. The wavelength of light inside the lens is  2/3 times the wavelength in free space. The radius of the curved surface of the lens is

Detailed Solution for BITSAT Practice Test - 11 - Question 2

from lens formula

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BITSAT Practice Test - 11 - Question 3

A galvanometer of resistance 50 Ω is connected to a battery of 3 V, along with a resistance of 2950 Ω in series. A full-scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 divisions, the resistance in series should be

Detailed Solution for BITSAT Practice Test - 11 - Question 3

Current in the galvanometer,

BITSAT Practice Test - 11 - Question 4

The graph between ratio of speed of electron in nth orbit to radius of nth orbit and n for Bohr’s hydrogen atom is

Detailed Solution for BITSAT Practice Test - 11 - Question 4

BITSAT Practice Test - 11 - Question 5

An ammeter and a voltmeter of resistance R are connected in series to an electric cell of negligible internal resistance. Their readings are A and V, respectively. If another resistance R is connected in parallel with the voltmeter, then

Detailed Solution for BITSAT Practice Test - 11 - Question 5

The total resistance of circuit would decrease as resistance R  is connected in parallel. So, the reading of the ammeter would increase, Reading of the voltmeter would decrease as  the final resistance across the voltmeter decreases due to parallel connection of resistance R. In series combination, the potential difference is directly proportional to the resistance.

BITSAT Practice Test - 11 - Question 6

A vertical spring with force constant k is fixed on a table. A ball of mass m at a height h above the free upper end of the spring falls vertically on the spring and at some instant, it compresses the spring by a distance d. The net work done by all the forces up to that instant is

Detailed Solution for BITSAT Practice Test - 11 - Question 6

Situation is shown in figure. When mass m falls vertically on spring, then spring is compressed by distance d

Hence, net work done in the process is W = Potential energy stored in the spring+ Loss of potential energy of mass

BITSAT Practice Test - 11 - Question 7

A particle of mass m and charge q is incident on XZ plane with velocity v in a direction making angle θ with a uniform magnetic field applied along X -axis. The nature of motion performed by the particle is

Detailed Solution for BITSAT Practice Test - 11 - Question 7

Due to the parallel component of velocity to the field, the particle moves in the direction of the field and due to the perpendicular component of velocity, the particle follows circular path so combined path is helical.

BITSAT Practice Test - 11 - Question 8

A transistor is used in the common emitter mode as an amplifier. Then -

A. The base-emitter junction is forward biased

B. The base-emitter junction is reverse biased

C. The input signal is connected in series with the voltage applied to bias the base emitter junction

D. The input signal is connected in series with the voltage applied to bias the base collector junction

Correct options are- 

Detailed Solution for BITSAT Practice Test - 11 - Question 8

Above diagram is of a transistor as an amplifier in common emitter configuration.

We can clearly see from the diagram that base-emitter junction is forward biased and the input signal is supplied from emitter base junction.

BITSAT Practice Test - 11 - Question 9

A steel wire having a radius of 2.0 mm, carrying a load of 4 kg, is hanging from a ceiling. Given that g=3.1π m s−2, what will be the tensile stress that would be developed in the wire?

Detailed Solution for BITSAT Practice Test - 11 - Question 9

BITSAT Practice Test - 11 - Question 10

In a Young's experiment, two coherent sources are placed 0.90 mm apart and the fringes are observed one metre away. If it produces the second dark fringe at a distance of 1 mm from the central fringe, the wavelength of monochromatic light used would be

Detailed Solution for BITSAT Practice Test - 11 - Question 10

Distance of nth dark fringe from central fringe

BITSAT Practice Test - 11 - Question 11

A 20 cm long string, having a mass of 1.0 g, is fixed at both the ends. The tension in the string is 0.5 N. the string is set into vibration using an external vibrator of frequency 100 Hz. Find the separation (in cm) between the successive nodes on the string

Detailed Solution for BITSAT Practice Test - 11 - Question 11

The speed of wave on a string is given by

So, the wavelength of the wave

The distance between successive nodes on the string

BITSAT Practice Test - 11 - Question 12

Three points charges +q, +2q and −4q where q = 0.1 μC, are placed at the vertices of an equilateral triangle of side 10 cm as shown in figure. The potential energy of the system is

Detailed Solution for BITSAT Practice Test - 11 - Question 12

BITSAT Practice Test - 11 - Question 13

An induced emf has

Detailed Solution for BITSAT Practice Test - 11 - Question 13

From Lenz's law, the direction of induced emf in a circuit is such that it opposes the magnetic flux that produces it.

So, if the magnetic flux linked with a closed circuit increases the induced current flows in a direction so to develop a magnetic flux in the opposite direction of original flux.

If the magnetic flux linked with a closed circuit decreases then the induced current flows in the same direction of original flux. So, the induced emf has not direction of its own.

BITSAT Practice Test - 11 - Question 14

To get an output 1 from the circuit shown in the figure, the input must be

Detailed Solution for BITSAT Practice Test - 11 - Question 14

The Boolean expression for the given combination is

output Y=(A+B).C
 
The truth table is

Hence A=1, B=0, C=1

BITSAT Practice Test - 11 - Question 15

Good absorbers of heat are

Detailed Solution for BITSAT Practice Test - 11 - Question 15

The good absorbers of heat are also good emitters of heat. This is based on Kirchhoff's law of radiation. This law states that at a given temperature the coefficient of absorption of a body is equal to its coefficient of emission. For example a blackbody is good absorber as well as emitter of heat and the coefficient of absorption and emission of a blackbody are also equal.

BITSAT Practice Test - 11 - Question 16

Two cars A and B are traveling in the same direction with velocities v1 and v2 (v1 > v2), respectively. When the car A is at a distance d behind the car B, the driver of the car A applies the brakes, producing uniform retardation, a. There will be no collision, when

Detailed Solution for BITSAT Practice Test - 11 - Question 16

Their final relative velocity = 0
Relative velocity of A w.r.t B is u = v- v2
Relative acceleration = -a

BITSAT Practice Test - 11 - Question 17

Bullets of mass 0.03 kg each hit a plate at the rate of 200 bullets per second with a velocity of 50 ms-1 and reflect back with a velocity of 30 ms-1. The average force (in Newton) acting on the plate is

Detailed Solution for BITSAT Practice Test - 11 - Question 17

Change of momentum of one bullet = m(v - u)
= 0.03 x {50 - (-30)}
= 2.4 kg ms-1
Average force = rate of change of momentum of 200 bullets
= 200 x 2.4 = 480 N,
which is choice (4).

BITSAT Practice Test - 11 - Question 18

A block of mass m slides down a wedge of mass M as shown. The whole system is at rest, when the height of the block is h above the ground. The wedge surface is smooth and gradually flattens. There is no friction between wedge and ground.

If there is no friction anywhere, the speed of the wedge, as the block leaves the wedge, is

Detailed Solution for BITSAT Practice Test - 11 - Question 18

P1 = P2
o = mv1 - MV2
mV1 = MV2
k1 + v1 = k2 + v2

BITSAT Practice Test - 11 - Question 19

If Earth is supposed to be a sphere of radius 'R' and if 'g30°' is the value of acceleration due to gravity at a latitude of 30°, while 'g' is at the equator, the value of '(g - g30°)' is

Detailed Solution for BITSAT Practice Test - 11 - Question 19

The variation of acceleration due to gravity with latitude is given by,

BITSAT Practice Test - 11 - Question 20

A thick uniform rubber rope of density 1.5 g cm-3 and Young's modulus 5 x106 Nm-2 has a length of 8 m. When hung from the ceiling of a room, the increase in length of the rope due to its own weight will be

Detailed Solution for BITSAT Practice Test - 11 - Question 20

The weight of the rope can be assumed to act at its mid-point. Now, the extension l is proportional to original length L. If the weight of the rope acts at its mid-point, the extension will be that produced by half the rope. So, replacing L by L/2 in the expression for Young's modulus, we have

BITSAT Practice Test - 11 - Question 21

In forced oscillation of a particle, the amplitude is maximum for a frequency  of the force while the energy is maximum for a frequency ω2 of the force. Then,

Detailed Solution for BITSAT Practice Test - 11 - Question 21

When the frequency of the driving force is equal to the natural frequency i.e. at resonance, the amplitude is maximum.
At resonance, average power absorbed by the forced oscillator is also maximum.
So, both ω1 and ω2 are same at resonant frequencies.

BITSAT Practice Test - 11 - Question 22

A train blowing its whistle moves with a constant velocity u away from an observer on the ground. The ratio of the actual frequency of the whistle to that measured by the observer is found to be 1.2. If the train is at rest and the observer moves away from it at the same velocity, then the ratio would be

Detailed Solution for BITSAT Practice Test - 11 - Question 22

If the train is going away from the observer, the apparent frequency is

In the second case, the apparent frequency is

Now, from (i) we have

Using this in (ii), we get

Hence the correct choice is (2).

BITSAT Practice Test - 11 - Question 23

In a Carnot engine, efficiency is 40% at hot reservoir temperature T. For efficiency to be 50%, what shall be the temperature of the hot reservoir?

Detailed Solution for BITSAT Practice Test - 11 - Question 23

The efficiency of a heat engine is defined as the ratio of work done to the heat supplied, i.e.,

where T2 is temperature of sink,
and T1 is temperature of hot reservoir.

BITSAT Practice Test - 11 - Question 24

The pressure and density of a diatomic gas (y = 7/5) change adiabatically from  then what should be the value of 

Detailed Solution for BITSAT Practice Test - 11 - Question 24

In an adiabatic process.

Thus, from Eq. (i), we have

BITSAT Practice Test - 11 - Question 25

The figure shows five capacitors connected across a 12 V power supply. What is the charge on the 2μF capacitor?

Detailed Solution for BITSAT Practice Test - 11 - Question 25

If capacitors 1μF, 2μF and 3μF are in parallel, then their total capacitance is 6μF.
Thus, we have three capacitors in series, each of capacitance 6 across the 12 V power supply.
So, the potential drop across each is 12/3 = 4 V.
This is also the potential across 1μF capacitor, 2μF capacitor and 3μF capacitor because they are in parallel.
Therefore, charge on 2μF capacitor = 2μF x 4 V = 8μC.
Hence, the correct choice is (2).

BITSAT Practice Test - 11 - Question 26

The figure below shows four cells E, F, G and H of emfs 2 V, 1 V, 3 V and 1 V and internal resistances 2Ω , 1Ω , 3Ω  and 1Ω , respectively.

The potential difference between points B and D is

Detailed Solution for BITSAT Practice Test - 11 - Question 26

Using Kirchoff's law for the loop BADB, we get

And we have, I1 = I2 + I3 ---- 3

From 1, 2 and 3, we have

I3 = -1/13 A, which means B is at high potential as compared to A

Hence, potential difference across B and D = 2/13 V

BITSAT Practice Test - 11 - Question 27

A moving coil galvanometer consists of a coil of N turns and area A, suspended by a thin phosphor bronze strip in radial magnetic field B. The moment of inertia of the coil about the axis of rotation is l and C is the torsional constant of the phosphor bronze strip. When a current i is passed through the coil, it deflects through an angle θ (in radians).
When a charge Q is passed almost instantly through the coil, the angular speed ω acquired by the coil is

Detailed Solution for BITSAT Practice Test - 11 - Question 27

if ω is the angular speed acquired by the coil when a charge Q is passed through it for very short time Δt, then 

BITSAT Practice Test - 11 - Question 28

The magnetic field in a region is given by . A square loop of side d is placed with its edges along the x and y-axis. The loop is moved with a constant velocity . The emf induced in the loop is

Detailed Solution for BITSAT Practice Test - 11 - Question 28

Total flux linked with the coil at time t = 0 is

Flux linked with the coil at time t = 1 is

Now, induced emf = Change in flux in unit time

BITSAT Practice Test - 11 - Question 29

 fossil bone has a C14: C12 ratio, which is [1/16] of that in a living animal bone. If the half-life of C14 is 5730 years, then the age of the fossil bone is

Detailed Solution for BITSAT Practice Test - 11 - Question 29

BITSAT Practice Test - 11 - Question 30

If the potential energy of electron in a hydrogen atom is -Ke2/r, then its kinetic energy is

Detailed Solution for BITSAT Practice Test - 11 - Question 30

PE = -2 KE

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