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BITSAT Practice Test - 12 - JEE MCQ


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30 Questions MCQ Test - BITSAT Practice Test - 12

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BITSAT Practice Test - 12 - Question 1

Which of the following is used as a coolant in automobile radiator as well as a heater in hot water bags?

Detailed Solution for BITSAT Practice Test - 12 - Question 1

Water has the highest specific heat capacity, so to raise the temperature of water by 1 °C very huge amount of heat is required and hence it is used as a coolant in car radiators and even as heater in hot water bags.

BITSAT Practice Test - 12 - Question 2

If force(F), time(T) and velocity(V) are treated as fundamental quantities then dimensional formula of energy will be

Detailed Solution for BITSAT Practice Test - 12 - Question 2

E = ML2T−2 = MLT−2 × (LT−1) × T = FTV

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BITSAT Practice Test - 12 - Question 3

Two large conducting plates having surface charge densities + σ & –σ are fixed d distance apart. A small test charge q of mass m is attached to two identical springs as shown in the adjacent figure. The charge q is now released from rest with springs in natural length. Then q will [neglect gravity]:

Detailed Solution for BITSAT Practice Test - 12 - Question 3

Given, σ & −σ be the surface charge denisty, electric field between the plates,

Equilibrium position will be at the point where net force is 00.

Consider, Xo be the amplitude.

We know at equilibrium net force is 00.

So,

BITSAT Practice Test - 12 - Question 4

In a radioactive decay process, the negatively charged emitted β- particles are

Detailed Solution for BITSAT Practice Test - 12 - Question 4

In beta minus decay (β ), a neutron is transformed into a proton and an electron is emitted with the nucleus along with an antineutrino.

where  the antineutrino.

BITSAT Practice Test - 12 - Question 5

Function of rectifier is

Detailed Solution for BITSAT Practice Test - 12 - Question 5

A rectifier is an electrical device that converts alternating current to direct current.

Above diagram is the schematic diagram of diode used as half-wave rectifier.

When an A.C. source is given as input signal, for the first half of the signal diode will be in forward bias and the output signal is same as input. On the other hand, for the other half cycle, diode is in reverse bias, and it will behave as insulator in this case, so, no output will be recorded. 

Thus, rectifier converts A.C. input signal into D.C. output signal.

BITSAT Practice Test - 12 - Question 6

A boy throws a ball upwards with velocity u = 15 m s−1. The wind imparts a horizontal acceleration of 3 m s−2 to the left. The angle θ¸ at which the ball must be thrown so that the ball returns to the boy's hand is (use g = 10 m s−2)

Detailed Solution for BITSAT Practice Test - 12 - Question 6


The ball is thrown at an angle θ with the horizontal. Thus,

ux = 15cosθ

uy = 15sinθ

In order to return at the same point, net displacement will be zero. Thus, in the horizontal direction,

and in the vertical direction,

⇒ t = 3sinθ ...(3)

from (1) we get, 

t = 10 cosθ  ...(4)
from Eq. (3) and (4), we get,

BITSAT Practice Test - 12 - Question 7

An electromagnetic wave is propagating along Y-axis. Then

Detailed Solution for BITSAT Practice Test - 12 - Question 7

The direction of propagation of the electromagnetic wave is perpendicular to the variation of the electric field  and as well as to the magnetic field  as shown below.

By Applying the Right-hand rule for vector cross product, the direction of propagation of the wave is 

Here, Electric field oscillations is along Z-axis, Magnetic field oscillations is along X-axis and thus the propagation of electromagnetic wave will be along Y-axis.

BITSAT Practice Test - 12 - Question 8

A long straight wire is placed along the axis of a circular ring of radius R. The mutual inductance of this system is

Detailed Solution for BITSAT Practice Test - 12 - Question 8

Due to current in the wire, the flux through the ring will be zero. Hence the mutual inductance is zero.

BITSAT Practice Test - 12 - Question 9

The effect of rotation of the earth on the value of acceleration due to gravity is

Detailed Solution for BITSAT Practice Test - 12 - Question 9

As g' = g − ω2Rcos2λ

The latitude at a point on the surface of the earth is defined as the angle, which the line joining that point to the centre of the earth makes with equatorial plane.

It is denoted by, λ.

For the poles λ = 90° and for equator λ = 0°.
(i) Substituting λ = 90° in the above expression,

we get,

gpole = g − ω2Rcos290°

∴ gpole = g

i.e., there is no effect of rotational motion of the earth on the value of g at the poles.
(ii) Substituting λ = 0° in the above expression,

we get,

gequator = g − ω2Rcos2

∴ gequator = g − ω2R
i.e., the effect of rotation of the earth on the value of g at the equator is maximum.

BITSAT Practice Test - 12 - Question 10

A variable force, given by the 2-dimensional vector  acts on a particle. The force is in newton and x is in metre. What is the change in the kinetic energy of the particle as it moves from the point with coordinates (2, 3) to (3, 0)? (The coordinates are in metres)

Detailed Solution for BITSAT Practice Test - 12 - Question 10



= 27 + 0 – (8 + 12) = 27 – 20 = +7 J
According to work energy theorem,
Change in the kinetic energy
= Work done, W
= +7 J

BITSAT Practice Test - 12 - Question 11

The current i in the network is

Detailed Solution for BITSAT Practice Test - 12 - Question 11

Both the diodes are in reverse biased.

BITSAT Practice Test - 12 - Question 12

The surface that is chosen for the application of Gauss's law is called

Detailed Solution for BITSAT Practice Test - 12 - Question 12

The surface that we choose for the application of Gauss’s law is called Gaussian surface. A Gaussian surface is a closed surface in three-dimensional space through which the flux of a vector field is calculated, usually the gravitational field, the electric field, or magnetic field.

BITSAT Practice Test - 12 - Question 13

The inductance of a closed-packed coil of 400 turns is 8 mH. A current of 5 mA is passed through it. The magnetic flux through each turn of the coil is

Detailed Solution for BITSAT Practice Test - 12 - Question 13

Given,

Number of turns in the coil, N = 400

The inductance of the coil, L = 8 mH

Current in the coil, i = 5 mA

Let the magnetic flux through each coil be ϕ, then

BITSAT Practice Test - 12 - Question 14

When the temperature increases, the viscosity of,

Detailed Solution for BITSAT Practice Test - 12 - Question 14

Viscosity is the property of fluids by the virtue of which it opposes any relative motion between the layers of fluid or between any object and fluid surrounding it. It is analogous to friction for fluids.
It depends strongly on temperature. In liquids, it usually decreases with increasing temperature, whereas, in gases, viscosity increases with increasing temperature.

BITSAT Practice Test - 12 - Question 15

Assertion: A body whatever its motion is always at rest in reference frame which is fixed to the body itself.
Reason: The relative velocity of a body with respect to itself is always zero.

Detailed Solution for BITSAT Practice Test - 12 - Question 15

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

BITSAT Practice Test - 12 - Question 16

A bus starts to move with an acceleration of 1 m/s2. A man, who is 48 m behind the bus, runs to catch it with a constant velocity of 10 m/s. In how much time will he catch the bus?

Detailed Solution for BITSAT Practice Test - 12 - Question 16

Let in time 't', the bus travel a distance 'x' and the person be behind the bus by a distance'd'.
In order to catch the bus, the person has to travel x + d in the same time.
x + d = ut




BITSAT Practice Test - 12 - Question 17

Blocks A and B of masses 2m and m are connected as shown in the figure. The spring has negligible mass. The string is suddenly cut. The respective magnitudes of acceleration of masses 2m and m at that instant are

Detailed Solution for BITSAT Practice Test - 12 - Question 17

The equilibrium condition FBD of 2m

When the system is equilibrium, the spring force = 3 mg.
When the string is cut, the net force on block A = 3 mg - 2 mg = mg.
Hence, the acceleration of this block at this instant is

When the string is cut, the block B falls freely with an acceleration equal to g.

BITSAT Practice Test - 12 - Question 18

A solid cylinder of mass M and radius R is mounted on a frictionless horizontal axle, so that it can freely rotate about this axis. A string of negligible mass is wrapped round the cylinder and a body of mass m is hung from the string as shown in the figure. The mass is released from rest.

The acceleration with which the mass will fall is

Detailed Solution for BITSAT Practice Test - 12 - Question 18


From eqs. (1) and (2), we get
which is choice (d).

BITSAT Practice Test - 12 - Question 19

In case of a spherical shell having radius R and mass M, which of the following graphs between gravitational intensity and distance is true?

Detailed Solution for BITSAT Practice Test - 12 - Question 19

Inside the shell, there is no mass. So, gravitational field is zero. Outside the shell, gravitational field is inversely proportional to the square of the distance.

BITSAT Practice Test - 12 - Question 20

A wooden block, with a coin placed on its top, floats in water as shown in the figure. The distance h and l are shown there. After sometime, the coin falls into the water. Then,
:

Detailed Solution for BITSAT Practice Test - 12 - Question 20

As the block moves up with the fall of coin, l decreases, similarly h will also decrease because when the coin falls in water, it displaces water equal to its own volume only. The fall in level of water is higher from upward displacement of block. Hence, option b is correct.

BITSAT Practice Test - 12 - Question 21

If a source emitting waves of frequency f moves towards an observer with a velocity v/4 and the observer moves away from the source with a velocity v/6. The apparent frequency as heard by the observer will be
(v = Velocity of sound)

Detailed Solution for BITSAT Practice Test - 12 - Question 21

BITSAT Practice Test - 12 - Question 22

The temperature-entropy diagram of a reversible engine cycle is given in the figure. Its efficiency is

Detailed Solution for BITSAT Practice Test - 12 - Question 22


Q1 = T0S0 + 1/2 T0S0 = 3/2 T0S0
Q2 = T0(2S0 - S0) = T0S0 and Q3 = 0

BITSAT Practice Test - 12 - Question 23

The point charges 3μC and 4μC are placed at a separation of 7 m. The medium between them is of two type as shown in figure. What is the electric force acting between them?

Detailed Solution for BITSAT Practice Test - 12 - Question 23

Force between 2 charges in a medium = F' = 
∴ F' = 
In the given question, total effective distance = 
Therefore, force F = 

= 1.73 × 10-4 C

BITSAT Practice Test - 12 - Question 24

In the Wheatstone's bridge shown below, in order to balance the bridge, we must have

Detailed Solution for BITSAT Practice Test - 12 - Question 24

BITSAT Practice Test - 12 - Question 25

A thin wire loop carrying a current I is placed in a uniform magnetic field B pointing out of the plane of the coil as shown in the figure. The loop will tend to

Detailed Solution for BITSAT Practice Test - 12 - Question 25

The force experienced by the current element is 
dl is the small current element of the loop and the force is acting perpendicular to the magnetic field as well as current element. So, the loop will expand.
Fleming's left hand rule: At any point, the force exerted by the magnetic field on the current element is directed outwards, and hence loop will expand.
Thus, the correct choice is (c).

BITSAT Practice Test - 12 - Question 26

A galvanometer of 50 Ω resistance has 25 divisions. A current of 4 x 10-4 A gives a deflection of one division. To convert this galvanometer into a voltmeter having a range of 25 V, it should be connected with a resistance of

Detailed Solution for BITSAT Practice Test - 12 - Question 26

Ig = Current for full-scale deflection = 4 x 10-4 x 25 = 10-2 ohm
R =  ohm in series

BITSAT Practice Test - 12 - Question 27

The voltage 7.25 sin (300t) is applied to a series RLC circuit with R = 120 ohms, L = 0.140 H and C = 1.45 μF. What is the impedance Z and the phase angle θ?

Detailed Solution for BITSAT Practice Test - 12 - Question 27

The impedance of the RLC circuit is Z = [(R2 + (ωL - 1/ωC)2)]1/2
= [(120)2 + (42 - 2298.85)2]1/2 = 2260 Ω
The phase angle is given by tan θ = (R/(ωL - 1/ωC)) or, tan θ = (120/(42 - 2298.85)) = (-120/2256.85) or, θ = -87o

BITSAT Practice Test - 12 - Question 28

A person can see clearly only up to a distance of 30 cm. He wants to read a book placed at a distance of 50 cm from his eyes. What is the power of the lens he requires for his spectacles?

Detailed Solution for BITSAT Practice Test - 12 - Question 28

The person is myopic.
This defect is corrected by using a concave lens.
Putting u = -50 cm and v = -30 cm in the lens formula  gives f = -75 cm.
Therefore, the power of the lens = -1.33 D

BITSAT Practice Test - 12 - Question 29

When a certain metallic surface is illuminated with monochromatic light of wavelength λ, the stopping potential for photoelectric current is 3V0. When the same surface is illuminated with light of wavelength 2λ, the stopping potential is V0. The threshold wavelength for this surface for photoelectric effect is

Detailed Solution for BITSAT Practice Test - 12 - Question 29

BITSAT Practice Test - 12 - Question 30

In an n-p-n transistor amplifier, the collector current is 9 mA. If 90% of the electrons from the emitter reach the collector, then

Detailed Solution for BITSAT Practice Test - 12 - Question 30

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