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BITSAT Practice Test - 14 - JEE MCQ


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30 Questions MCQ Test BITSAT Mock Tests Series & Past Year Papers 2025 - BITSAT Practice Test - 14

BITSAT Practice Test - 14 for JEE 2024 is part of BITSAT Mock Tests Series & Past Year Papers 2025 preparation. The BITSAT Practice Test - 14 questions and answers have been prepared according to the JEE exam syllabus.The BITSAT Practice Test - 14 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for BITSAT Practice Test - 14 below.
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BITSAT Practice Test - 14 - Question 1

A circular coil, of radius R, carries a current I. The expression for the magnetic field due to this coil at its centre is

Detailed Solution for BITSAT Practice Test - 14 - Question 1

The magnetic field due to a circular coil at its centre of radius R and carrying current I is

The direction of magnetic field induction   at the centre is perpendicular to the plane of the circular coil and directed inwards.

BITSAT Practice Test - 14 - Question 2

The acceleration of a particle increases linearly with time t as 6t. If the initial velocity of the particle is zero and the particle starts from the origin, then the distance travelled by the particle in time t will be

Detailed Solution for BITSAT Practice Test - 14 - Question 2

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BITSAT Practice Test - 14 - Question 3

A ray of light travels from a medium of refractive index n1 to a medium of refractive index n2. If angle of incidence is i and the angle of refraction is r. Then sini/sinr is equal to

Detailed Solution for BITSAT Practice Test - 14 - Question 3

According to Snell's law,

BITSAT Practice Test - 14 - Question 4

A block of mass, m = 2 kg is attached to two unstretched springs of force constant, k1 = 100 N m−1 and k2 = 125 N m−1. The block is displaced towards the left through a distance of 10 cm and released. Find the speed of the block as it passes through the mean position.

Detailed Solution for BITSAT Practice Test - 14 - Question 4


When the block is displaced in one direction, the work done by the force is stored in both the springs as potential energy and when it is released, this potential energy is converted into kinetic energy.
Since at mean position, spring will be relaxed, so kinetic energy of block at mean position will be same as initial potential energy stored in springs.
So, the net loss in PE due to displacement of block from mean position = net gain in KE of the block at mean position.

BITSAT Practice Test - 14 - Question 5

A toy car of mass 5 kg moves up a ramp under the influence of force F plotted against displacement x. The maximum height attained by the car is given by,

Detailed Solution for BITSAT Practice Test - 14 - Question 5


Total work done by the force F is the area under the force-displacement curve. This work is responsible to increase kinetic energy of the toy car.
Total work done by the force,

Hence, the kinetic energy of toy car at x = 11 m is,
Work done = Change in kinetic energy = 550 J
By energy conservation,
Potential energy gain by toy car at ymax = Kinetic energy at x = 11 m

BITSAT Practice Test - 14 - Question 6

The graph which represents the relation between the total resistance R of a multi-range moving coil voltmeter and its full-scale deflection V is

Detailed Solution for BITSAT Practice Test - 14 - Question 6

When we convert any galvanometer into voltmeter,

V = Ig(Rg + R)

Let RT be the effective resistance of Rg and R, connected in series.

V = Ig(RT)

where, 

RT = Total resistance of voltmeter

V = Range of voltmeter

Ig = Full scale deflection current
V ∝ RT

For a multi-range voltmeter, as we increase RT, range (V) would also increase.

So, the graph would be a straight line.

BITSAT Practice Test - 14 - Question 7

The mass of a proton is 1836 times more than the mass of an electron. If a subatomic particle of mass (m′) 207 times the mass of an electron is captured by the nucleus, then what will happen to the first ionization potential of H?

Detailed Solution for BITSAT Practice Test - 14 - Question 7

Ionization potential is directly proportional to the mass of the particle.

IE∝ mass of particle (m)

The ionization potential of a subatomic particle is determined by the reduced mass.

Now, calculate the reduced mass:

As we can see that mass of the subatomic particle is more than mass of electron so the ionization potential increases.

BITSAT Practice Test - 14 - Question 8

A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is

Detailed Solution for BITSAT Practice Test - 14 - Question 8

At position x from the mean position in S.H.M.

acceleration, a = xω2

here ω is the angular frequency and A is amplitude.
Given in the problem, v = a

BITSAT Practice Test - 14 - Question 9

In a biprism experiment with sodium light, bands of width 0.0195 cm are observed at 100 cm from slit. On introducing a convex lens 30 cm away from the slit between biprism and screen, two images of the slit are seen 0.7 cm apart at 100 cm distance from the slit. Calculate the wavelength of sodium light.

Detailed Solution for BITSAT Practice Test - 14 - Question 9

Lens will form image of two sources at 70 cm from lens (i.e. screen)
From magnification of lens 

On solving, d = 0.3 cm
Now, fringes width β = λD/d

BITSAT Practice Test - 14 - Question 10

Two masses of M and 4M are moving with equal kinetic energy. The ratio of their linear momentum is

Detailed Solution for BITSAT Practice Test - 14 - Question 10

Two masses are moving with equal kinetic energy.

The ratio of linear momentum is

BITSAT Practice Test - 14 - Question 11

In the figure, the blocks A, B and C each of mass m have acceleration a1, a2 and a3, respectively. F1 and F2 are external forces of magnitude 2mg and mg respectively. Then

Detailed Solution for BITSAT Practice Test - 14 - Question 11

For A, T = f = 2mg

2mg − mg = ma1

∴ a1 = g

For B,

From the force diagram shown in the figure,

2mg − mg = 3ma2

a2 = g/3

For C,

∴ 3mg − mg = 2ma3

⇒ a3 = g

So, a1 = a3 > a2

BITSAT Practice Test - 14 - Question 12

A whistle of frequency 540 Hz rotates in a horizontal circle of radius 2 m at an angular speed of 15 rad s−1. The highest frequency heard by a listener at rest at large distance with respect to the centre of circle (velocity of sound in air = 330 m s−1)

Detailed Solution for BITSAT Practice Test - 14 - Question 12

Velocity of source
vs = rω = 2 × 15 = 30ms−1
The highest frequency heard by the stationary listener

or 

BITSAT Practice Test - 14 - Question 13

Consider two observers moving with respect to each other at a speed v along a straight line. They observe a block of mass moving over a distance l on a rough surface. The following quantities will be same as observed by the two observers

Detailed Solution for BITSAT Practice Test - 14 - Question 13

Velocity and displacement of block will be different from both the observers. But the force of friction observed by both the observes will be same. That is why acceleration of the block will be same.

BITSAT Practice Test - 14 - Question 14

A charge Q is placed at a distance a/2 above the centre of a square surface of side length a. The electric flux through the square surface due to the charge would be?

Detailed Solution for BITSAT Practice Test - 14 - Question 14


Charged particle can be considered at centre of a cube of side a, and given surface represents its one face. So, flux 

BITSAT Practice Test - 14 - Question 15

Find the elongation of the steel and brass wire in the fig. unloaded length of steel wire = 1.5 m, unloaded length of brass wire 0.3 cm, diameter of each wire = 0.3 cm. Young's modulus for steel = 20 × 1010 Pa and that for brass = 9.0 × 1010 Pa.

Detailed Solution for BITSAT Practice Test - 14 - Question 15

For steel wire: Y = 20 × 1010 Pa; L = 1.5 m ; d = 0.3 cm = 0.3 × 10−2m
Therefore, area of cross-section of the steel wire,

The stretching force for the steel wire,
F = 4.0 + 6.0 = 10 kgf = 10 × 9.8 N
If l is extension in the steel wire, then

For brass wire: Y = 9.0 × 1010 Pa ; L = 1m;

The stretching force for the brass wire,

F = 6 kgf = 6 × 9.8 N

BITSAT Practice Test - 14 - Question 16

If a = 18√v (where 'a' and 'v' are acceleration and velocity at any instant, respectively), then the acceleration when the time t = 1 second is

Detailed Solution for BITSAT Practice Test - 14 - Question 16




At t = 1 sec,

BITSAT Practice Test - 14 - Question 17

The following figure is part of a horizontally stretched net. Section 'AB' is stretched with a force of 10 N. The tensions in the sections 'BC' and 'BF', respectively, are

Detailed Solution for BITSAT Practice Test - 14 - Question 17

From the figure,

T1 cos30° = T2 cos30°
∴ T1 = T2 = T (Let)
Again, Tsin 30° + T2 sin 30° = 10
2T sin 30° = 10
2T × 1/2 = 10
T = 10 N
∴ Tension in section BC and BF are 10 N and 10 N.

BITSAT Practice Test - 14 - Question 18

A cricket ball of mass 200 gms moving with a velocity of 20 m/sec is brought to rest by a player in 0.1 sec. The average force applied by the player is

Detailed Solution for BITSAT Practice Test - 14 - Question 18

Average force = Change in momentum/Time interval

BITSAT Practice Test - 14 - Question 19

A hollow sphere of mass M and radius R is initially at rest on a horizontal rough surface. It moves under the action of a constant horizontal force F as shown in the figure.

The frictional force between the sphere and the surface is

Detailed Solution for BITSAT Practice Test - 14 - Question 19


Let a and α be the linear and angular accelerations of the sphere respectively.
For translational motion,
F + f = Ma ...(1)
The magnitude of the net torque acting on the sphere = FR - fR.
Hence, for rotational motion the equation is

  • FR - fR = Iα = Ia/R (∵ a = αR)

For a hollow sphere, I = 2/3 MR2. Hence
FR - fR = 2/3 MR2 ×  MRa
F - f = 2/3 Ma ...(2)
From Eqs. (1) and (2) we get f = .
Hence the correct choice is (4).

BITSAT Practice Test - 14 - Question 20

A body of mass m is raised to a height h above the surface of Earth of mass M and radius R until its gravitational potential energy increases by 1/3 mgR. The value of h is

Detailed Solution for BITSAT Practice Test - 14 - Question 20


BITSAT Practice Test - 14 - Question 21

A capillary tube of radius r is immersed in water and water rises in it to a height h. The mass of water in the capillary tube is 5 g. Another capillary tube of radius 2r is immersed in water. The mass of water, that will rise in this tube, is

Detailed Solution for BITSAT Practice Test - 14 - Question 21

Mass of water in first tube is

Now, surface tension 
where h' is the height to which water rises in the second tube and r' its radius.
Since r' = 2r, h' = h/2.
Therefore, the mass of water in the second capillary tube is

BITSAT Practice Test - 14 - Question 22

Two harmonic motions are represented by the equations y= 10 sin  and y2 = 5 . Their amplitudes are in the ratio of

Detailed Solution for BITSAT Practice Test - 14 - Question 22

y1 = 10 sin (3)
y2 = 5 (sin 3)
= 10
= 10 sin (3)
A1 : A2 = 1 : 1

BITSAT Practice Test - 14 - Question 23

When sounded together, a column of air and a tuning fork produce 4 beats per second. The tuning fork gives the lower note. The temperature of air is 15°C. When the temperature falls to 10°C, the two produce 3 beats per second. Find the frequency of the fork.

Detailed Solution for BITSAT Practice Test - 14 - Question 23

Let v = Frequency of the tuning fork
v1 = Frequency of the air column
At 15°C, v1 = v + 4
At 10°C, v1 = v + 3
We know, velocity u = v
 u15 = (v + 4)
and u10 = (v + 3)
 
Also,  …(2)
From (1) and (2), we get

Using binomial expansion and neglecting terms, we get


Or, 1 + 
Or, (v + 3) = 113.2
 v = 110.2 Hz  110 Hz

BITSAT Practice Test - 14 - Question 24

A 2 μF, capacitor C1 is charged to a voltage 100 V and a 4 μF capacitor C2 is charged to a voltage 50 V. The capacitors are then connected in parallel. What is the loss of energy due to parallel connection?

Detailed Solution for BITSAT Practice Test - 14 - Question 24

Loss in energy when two capacitors are connected in parallel is


BITSAT Practice Test - 14 - Question 25

The following figure shows a spherical Gaussian surface and a charge distribution (magnitude of all the given point charges is different). When calculating the flux of electric field through the Gaussian surface, the electric field will be due to

Detailed Solution for BITSAT Practice Test - 14 - Question 25

The electric flux is given by the surface integral ∫ E.ds. Here the electric field E is due to all the charges, both inside and outside the Gaussian surface. Hence the correct choice is (d)

BITSAT Practice Test - 14 - Question 26

The length of a wire of a potentiometer is 100 cm and the emf of its stand and cell is E volt. It is employed to measure the emf of a battery whose internal resistance is 0.5 W. If the balance point is obtained at l = 30 cm from the positive end, then the emf of the battery is

Detailed Solution for BITSAT Practice Test - 14 - Question 26

Potential gradient of the potentiometer wire is
 V/cm
Emf of the cell is
, l = 30 cm

BITSAT Practice Test - 14 - Question 27

A wire of length L metres carrying a current I amperes is bent in the form of a circle. The magnitude of the magnetic moment is

Detailed Solution for BITSAT Practice Test - 14 - Question 27

Magnetic moment m = AI = r2I, where r is the radius of the circular loop.
Now, the circumference of the circle = length of the wire, i.e.
2πr = L
or r2 = 
Therefore, m = πr2I = 
Hence the correct choice is (d)

BITSAT Practice Test - 14 - Question 28

In a series circuit, L, C and R are connected with an alternating voltage source of frequency f. The current leads the voltage by 45o. The value of C is

Detailed Solution for BITSAT Practice Test - 14 - Question 28

As the current is leading the voltage, then we have



BITSAT Practice Test - 14 - Question 29

When a thin transparent plate of thickness t and refractive index m is placed in the path of one of the two interfering waves of light, then the path difference changes by

Detailed Solution for BITSAT Practice Test - 14 - Question 29

BITSAT Practice Test - 14 - Question 30

The output current (I) versus time (t) curve of a rectifier is shown in the figure. The average value of the output current in this case is

Detailed Solution for BITSAT Practice Test - 14 - Question 30

Average value of the current, 

As .
Thus, 

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