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Test: BITSAT Past Year Paper- 2019 - JEE MCQ


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30 Questions MCQ Test BITSAT Mock Tests Series & Past Year Papers 2025 - Test: BITSAT Past Year Paper- 2019

Test: BITSAT Past Year Paper- 2019 for JEE 2024 is part of BITSAT Mock Tests Series & Past Year Papers 2025 preparation. The Test: BITSAT Past Year Paper- 2019 questions and answers have been prepared according to the JEE exam syllabus.The Test: BITSAT Past Year Paper- 2019 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: BITSAT Past Year Paper- 2019 below.
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Test: BITSAT Past Year Paper- 2019 - Question 1

Which one of the following graphs represents the variation of electric field with distance r from the centre of a charged spherical conductor of radius R?

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 1

The charged sphere is a conductor.
Therefore the field inside is zero and outside it is proportional to 1/r2.

Test: BITSAT Past Year Paper- 2019 - Question 2

 are the electric and magnetic field vectors of e.m. waves then the direction of propagation of e.m. wave is along the direction of

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 2

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

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Test: BITSAT Past Year Paper- 2019 - Question 3

The young's modulus of a wire of length L and radius r is Y N/m2. If the length and radius are reduced to L/2 and r/2, then its young's modulus will be

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 3

Young's modulus of wire does not vary with dimention of wire. It is a constant quantity.

Test: BITSAT Past Year Paper- 2019 - Question 4

Twelve resistors each of resistance 16 Ω are connected in the circuit as shown. The net resistance between A and B is

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 4

Redraw the given circuit,

where, R = 16 Ω
Rnet = 4 Ω

Test: BITSAT Past Year Paper- 2019 - Question 5

The time period of a satellite of earth is 5 hours.If the separation between the earth and the satellite is increased to 4 times the previous value, the new time period will become

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 5

Accor ding to Kepler ’s law of plan etary motion, T2 ∝ R 

Test: BITSAT Past Year Paper- 2019 - Question 6

Two trains are moving towards each other with speeds of 20 m/s and 15 m/s relative to the ground. The first train sounds a whistle of frequency 600 Hz. The frequency of the whistle heard by a passenger in the second train before the train meets, is (the speed of sound in air is 340 m/s)

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 6

Test: BITSAT Past Year Paper- 2019 - Question 7

You are asked to design a shaving mirror assuming that a person keeps it 10 cm from his face and views the magnified image of the face at the closest comfortable distance of 25 cm. The radius of curvature of the mirror would then be :

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 7

Concave morror is used as a shaving mirror.



Therefore radius of curvature,
R = 2f = – 60 cm

Test: BITSAT Past Year Paper- 2019 - Question 8

A block is kept on a frictionless inclined surface with angle of inclination ‘α’. The incline is given an acceleration ‘a’ to keep the block stationary.
Then ‘a’  is equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 8

From free body diagram, For block to remain stationary,


⇒ a = g tan α

Test: BITSAT Past Year Paper- 2019 - Question 9

With the increase in temperature, the angle of contact

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 9

On increasing the temperature, angle of contact decreases.

Test: BITSAT Past Year Paper- 2019 - Question 10

Forward biasing is that in which applied voltage

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 10

Forward bias opposes the potential barrier and if the applied voltage is more than knee voltage it cancels the potential barrier.

Test: BITSAT Past Year Paper- 2019 - Question 11

Number of significant figures in expression 

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 11

In multiplication or division the final result should return as many significant figures as there are in the original number with the least significant figures. (Rounding off to three significant digits)

Test: BITSAT Past Year Paper- 2019 - Question 12

The ratio of the specific heats Cp/Cv = γ in terms of degrees of freedom (n) is given by

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 12

Let ‘n’ be the degree of freedom

Test: BITSAT Past Year Paper- 2019 - Question 13

A stone is thrown with a velocity u making an angle θ with the horizontal. The horizontal distance covered by its fall to ground is maximum when the angle θ is equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 13

Since range on horizontal plane is

so it is maximum when, sin 2θ = 1
θ = π/4

Test: BITSAT Past Year Paper- 2019 - Question 14

A ball of mass 150 g, moving with an acceleration 20 m/s2, is hit by a force, which acts on it for 0.1 sec. The impulsive force is

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 14

Force = Mass × acceleration

Impulsive force = F.Δt = 3 × 0.1 = 0.3 N

Test: BITSAT Past Year Paper- 2019 - Question 15

A man drags a block through 10 m on rough surface (µ = 0.5). A force of √3 kN acting at 30° to the horizontal. The work done by applied force is

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 15

Test: BITSAT Past Year Paper- 2019 - Question 16

A force of  acts on a body for 4 second, produces a displacement of  The power used is

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 16

Test: BITSAT Past Year Paper- 2019 - Question 17

The Earth is assumed to be a sphere of radius R. A platform is arranged at a height R from the surface of the Earth. The escape velocity of a body from this platform is fv, where v is its escape velocity from the surface of the Earth. The value of f is

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 17

Test: BITSAT Past Year Paper- 2019 - Question 18

Kepler ’s second law regarding constancy of areal velocity of a planet is a consequence of the law of conservation of

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 18

dA/dt = L/2m = Constant

Test: BITSAT Past Year Paper- 2019 - Question 19

Water is flowing through a horizonal tube having cross-sectional areas of its two ends being A and A' such that the ratio A/A' is 5. If the pressure difference of water between the two ends is 3 × 105 N m–2, the velocity of water with which it enters the tube will be (neglect gravity effects)

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 19

According to Bernoulli’s theorem

According to the condition,


From equation of continuity,

From equation (i)

Test: BITSAT Past Year Paper- 2019 - Question 20

A thermodynamic system is taken from state A to B along ACB and is brought back to A along BDA  as shown in the PV diagram. The net work done during the complete cycle is given by the area

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 20

Work done = Area under curve ACBDA

Test: BITSAT Past Year Paper- 2019 - Question 21

A boat crosses a river from port A to port B, which are just on the opposite side. The speed of the water is Vw and that of boat is VB relative to still water. Assume Vw = 2Vw. What is the time taken by the boat, if it has to cross the river directly on the AB line [D = width of the river]

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 21

From figure, VB sin θ = VW 

Time taken to cross the river.

Test: BITSAT Past Year Paper- 2019 - Question 22

Two springs, of force constants k1 and k2 are connected to a mass m as shown. The frequency of oscillation of the mass is f. If both k1 and k2 are made four times their original values, the frequency of oscillation becomes

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 22

The two springs are in parallel.
∴ Effective spring constant, k = k1 + k2 Now, frequency of oscillation is given by


When both k1 and k2 are made four times their original values, the new frequency is given by


Test: BITSAT Past Year Paper- 2019 - Question 23

When a potential difference V is applied across a conductor at a temperature T, the drift velocity of electrons is proportional to

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 23

Drift velocity,

Test: BITSAT Past Year Paper- 2019 - Question 24

The amplitude of a damped oscillator becomes (1/3)rd in 2 secon ds. If its amplitude after 6 seconds is 1/n times the original amplitude, the value of n is 

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 24

Amplitude of a damped oscillator
A = A0e–bt/2m
Case 1 :-

Test: BITSAT Past Year Paper- 2019 - Question 25

The angular speed of the electron in the nth orbit of Bohr hydrogen atom is

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 25

Angular speed of electron in the nth orbit of Bohr H-atom is inversely proportional to n3 

Test: BITSAT Past Year Paper- 2019 - Question 26

In the given figure, the charge on 3 µF capacitor is

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 26

C = equivalent capacitance

Charge on each capacitor in series circuit will be same.
∴ q = CV = (1 × 10-6) × 10 = 10μC
∴ Charge across 3µF capacitor will be 10µC.

Test: BITSAT Past Year Paper- 2019 - Question 27

Two bodies A and B are placed in an evacuated vessel maintained at a temperature of 27ºC. The temperature of A is 327ºC and that of B is 227ºC.
The ratio of heat loss from A and B is about

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 27

Test: BITSAT Past Year Paper- 2019 - Question 28

If a rigid body is rotating about an axis with a constant velocity, then

Test: BITSAT Past Year Paper- 2019 - Question 29

The fundamental frequency of an open organ pipe is 300 Hz. The first overtone of this pipe has same  frequency as first overtone of a closed organ pipe. If speed of sound is 330 m/s, then the length of closed organ pipe is

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 29

For open pipe, n1 = v/2ℓ , where n1 is the fundamental frequency of open pipe. length of open pipe is,

Ist overtone of open pipe, 
Ist overtone of closed pipe

where, ℓ’ = length of closed pipe
As freq. of 1st overtone of open pipe = freq. of 1st overtone of closed pipe

Test: BITSAT Past Year Paper- 2019 - Question 30

In Young's experiment, the distance between the slits is reduced to half and the distance between the slit and screen is doubled, then the fringe width

Detailed Solution for Test: BITSAT Past Year Paper- 2019 - Question 30

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