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BITSAT Practice Test - 16 - JEE MCQ


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30 Questions MCQ Test - BITSAT Practice Test - 16

BITSAT Practice Test - 16 for JEE 2024 is part of JEE preparation. The BITSAT Practice Test - 16 questions and answers have been prepared according to the JEE exam syllabus.The BITSAT Practice Test - 16 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for BITSAT Practice Test - 16 below.
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BITSAT Practice Test - 16 - Question 1

An electric motor operates at 20 r.p.s. What will be the approximate power delivered by the motor, if it supplies a torque of 75 N-m?

BITSAT Practice Test - 16 - Question 2

Which of th following transitions in a hydrogen atom emits the photon of highest frequency?

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BITSAT Practice Test - 16 - Question 3

The ratio of the longest to shortest wavelengths in Brackett series of hydrogen spectra is

Detailed Solution for BITSAT Practice Test - 16 - Question 3

 

BITSAT Practice Test - 16 - Question 4

Potential difference across the terminals of a battery is 50 V, when 11 A current is drawn. And it is 60 V, when 1 A current is drawn. The e.m.f. and internal resistance of battery is

Detailed Solution for BITSAT Practice Test - 16 - Question 4

e = v + Ir
e = emf
V = potential difference
r = internal resistance
Here e = 50 + 11r .....(1)
and e = 60 + 1r .....(2)
Solving these equations (1) and (2) for e and r we get
e = 61V and r = 1 Ω

BITSAT Practice Test - 16 - Question 5

The unit of specific conductivity is

BITSAT Practice Test - 16 - Question 6

When an inert gas is filled in the place vacuum in a photo cell, then

Detailed Solution for BITSAT Practice Test - 16 - Question 6
In the presence of inert gas photoelectrons emitted by cathode ionise the gas by collision and hence the current increases.
BITSAT Practice Test - 16 - Question 7

Light of frequency 4v₀ is incident on the metal of the threshold frequency v₀. The maximum kinetic energy of the emitted photoelectrons is

BITSAT Practice Test - 16 - Question 8

An L.C. circuit contains 10 mH inductor and a 25 μ F capacitor. The resistance of the circuit is negligible. The energy stored in the circuit is completely magnetic at time (in milli seconds) being measured from the instant when the circuit is closed

Detailed Solution for BITSAT Practice Test - 16 - Question 8

 

BITSAT Practice Test - 16 - Question 9

Minimum no. of 8 μF and 250 V capacitors, used to make a combination of 16 μF and 1000 V, is

Detailed Solution for BITSAT Practice Test - 16 - Question 9

To get 1000 volts from 250 volts, we have to connect Four capacitors in series. 

For such a series connection, effective capacitance is  2 micro farad, because 1/Ceffective = 1/C1 + 1/C2 +1/C3 +1/C4 = 1/8 + 1/8 + 1/8 + 1/8 = 1/2.
 Therefore Ceffective  = 2 microfarad. Hence the series combination is equivalent of 2 microfarads, 1000 Volts.

Now we have to connect 8 of these four capacitor units in parallel to get 16 micro farads capacitor because in parallel combination, effective capacitance is equal to summation of each capacitor.  Hence capacitors required will be:

4 numbers for each series combinations * 8 such combinations = 32 capacitors. 

BITSAT Practice Test - 16 - Question 10

The ratio of charge to potential of a capacitor is known as its

BITSAT Practice Test - 16 - Question 11

To aviod slipping while walking on ice, one should take smaller steps because of the

BITSAT Practice Test - 16 - Question 12

The acceleration due to gravity on the planet A is 9 times the acceleration due to gravity on planet B. A man jumps to a height of 2 m on the surface of A. What will be the height of jump by the same person on the planet B?

BITSAT Practice Test - 16 - Question 13

When a body is heated, which colour corresponds to highest temperature

BITSAT Practice Test - 16 - Question 14

A vessel contains a mixture of one mole of oxygen and two moles of nitrogen at 300 K. The ratio of the average rotational kinetic energy per O₂ molecule to that per N₂ molecule is

BITSAT Practice Test - 16 - Question 15

When a charged particle moves perpendicular to a magnetic field, then

BITSAT Practice Test - 16 - Question 16

The force between two short bar magnets with magnetic moments M₁ and M₂ whose centres are r meter apart is 8 N, when their axes are in same line. If the separation is increased to 2 r, the force between them is reduced to

Detailed Solution for BITSAT Practice Test - 16 - Question 16

Correct Answer :- d

Explanation : In magnetic dipole, force ∝ 1/r4

Hence, new force = 8/24

= 8/16

= 0.5N

BITSAT Practice Test - 16 - Question 17

Oil spreads over the surface of water, whereas water does not spread over the surface of oil, because

BITSAT Practice Test - 16 - Question 18

A 20 cm long capillary tube is dipped in water. The water rises up to 8 cm. If entire arrangement is put in a freely elevator, the length of water column in the capillary tube will be

BITSAT Practice Test - 16 - Question 19

The breaking stress of a wire depends upon

BITSAT Practice Test - 16 - Question 20

A particle covers 150 m in 8th second when starts from rest, its acceleration is

BITSAT Practice Test - 16 - Question 21

The resultant of two forces, one double the other in magnitude, is perpendicular to the smaller of the two forces. The angle between the two forces is

BITSAT Practice Test - 16 - Question 22

Two vectors A̅ and B̅ are such that A̅ + B̅ = A̅ - B̅. Then

Detailed Solution for BITSAT Practice Test - 16 - Question 22

Correct Answer :- d

Explanation : A + B = A - B

cancelling A both sides, we get

B = - B

Taking -B to other side, we get

2B = 0

B = 0

BITSAT Practice Test - 16 - Question 23

A car of mass 12kg at rest explodes into two pieces of masses 4kg and 8kg which move in opposite direction. If the velocity of 8kg piece is 6 m/s, the kinetic energy of other piece is

BITSAT Practice Test - 16 - Question 24

The image formed by an objective lens of a compound microscope is

BITSAT Practice Test - 16 - Question 25

A chimpanzee, swinging on a swing in a sitting position, stands up suddenly. The time period will

BITSAT Practice Test - 16 - Question 26

An instantaneous displacement of a simple harmonic oscillator is x = A cos (ωt + π/4 ). Its speed will be maximum at time

Detailed Solution for BITSAT Practice Test - 16 - Question 26

The velocity is the time derivative of displacement: v = dy/dt = -Aω (sinωt + π/4 ).
Its maximum magnitude equal to Aω is obtained when ωt = π/4 , from which t = π/4ω

BITSAT Practice Test - 16 - Question 27

A ray of light is incident on a plane mirror at an angle of 60o . The angle of deviation produced by the mirror is

BITSAT Practice Test - 16 - Question 28

Two spheres of masses 2M and M are initially at rest at a distance R apart. Due to mutual force of attraction, they approach each other. When they are at separation R/2, the acceleration of the centre of mass of spheres would be

BITSAT Practice Test - 16 - Question 29

A soild sphere, disc and solid cylinder all of the same mass and made of the same material are allowed to roll down (from rest) on an inclined plane, then

BITSAT Practice Test - 16 - Question 30

A small steel sphere is tied to a string and is whirled in a horizontal circle with a uniform angular velocity ω₁. The string is suddenly pulled so that radius of the circle is halved. If ω₂ is new angular velocity. then

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