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Test: BITSAT Past Year Paper- 2020 - JEE MCQ


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30 Questions MCQ Test BITSAT Mock Tests Series & Past Year Papers 2025 - Test: BITSAT Past Year Paper- 2020

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Test: BITSAT Past Year Paper- 2020 - Question 1

An organ pipe, open from both end produces 5 beats per second when vibrated with a source of frequency 200 Hz. The second harmonic of the same pipes produces 10 beats per second with a source of frequency 420 Hz. The fundamental frequency of organ pipe is

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 1

Let the fundamental frequency of organ pipe be f
Case I :  f  = 200 ± 5 =  205 Hz or 195 Hz

Case II : frequency of 2nd harmonic of organ pipe = 2f (as is clear from the second figure) 2f  = 420 ± 10 or f  = 210 ± 5 or f  = 205 or 215
Hence fundamental frequency of organ pipe  = 205 Hz

Test: BITSAT Past Year Paper- 2020 - Question 2

A vessel of depth 2d cm is half filled with a liquid of refractive index μ1 and the upper half with a liquid of refractive index μ2. The apparent depth of the vessel seen perpendicularly is

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 2

Apparent depth = d/μ1 + d/μ2

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Test: BITSAT Past Year Paper- 2020 - Question 3

The upper half of an inclined plane with inclination f is perfectly smooth while the lower half is rough.A body starting from rest at the top will again come to rest at the bottom if the coefficient of friction for the lower half is given by

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 3

Acceleration of block while sliding down upper half = g sin ф;
retardation of block while sliding down lower half = – (g sin ф - mg cos ф) For the block to come to rest at the bottom, acceleration in I half = retardation in II half. g sin ф = -(g sin ф - mg cos ф)
Þ m = 2 tan ф
Alternative method : According to work-energy theorem, W = ΔK = 0 (Since initial and final speeds are zero)
∴ Work done by friction + Work done by gravity = 0
i.e., 
or 

Test: BITSAT Past Year Paper- 2020 - Question 4

A body of density ρ' is dropped from rest at a height h into a lake of density r where ρ > ρ' neglecting all dissipative forces, calculate the maximum depth to which the body sinks before returning to float on the surface :

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 4

The effective acceleration of the body 

Now, the depth to which the body sinks

Test: BITSAT Past Year Paper- 2020 - Question 5

If the forward voltage in a semiconductor diode is changed from 0.5V to 0.7 V, then the forward current changes by 1.0 mA. The forward resistance of diode junction will be

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 5

Forward resistance

Test: BITSAT Past Year Paper- 2020 - Question 6

The heat generated in a cir cuit is given by Q = I2 Rt,  where I is current, R is resistance and t is time. If the percentage errors in measuring I, R and t are 2%, 1% and 1% respectively, then the maximum error in measuring heat will be

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 6


= 2 x 2% + 1% + 1% = 6%.

Test: BITSAT Past Year Paper- 2020 - Question 7

The r.m.s. velocity of oxygen molecule at 16ºC is 474 m/sec. The r.m.s. velocity in m/s of hydrogen molecule at 127ºC is

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 7


Test: BITSAT Past Year Paper- 2020 - Question 8

A projectile A is thrown at an angle of 30° to the horizontal from point P. At the same time, another projectile B is thrown with velocity v2 upwards from the point Q vertically below the highest point. For B to collide with A, should be

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 8

This happen when vertical velocity of both are same.

Test: BITSAT Past Year Paper- 2020 - Question 9

The coefficient of friction between the rubber tyres and the road way is 0.25. The maximum speed with which a car can be driven round a curve of radius  20 m without skidding is (g = 9.8 m/s2)

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 9

μ mg = mv2 / r or v = 
or v = = 7 m / s

Test: BITSAT Past Year Paper- 2020 - Question 10

A boy pushes a toy box 2.0 m along the floor by means of a force of 10 N directed downward at an angle of 60º to the horizontal. The work done by the boy is

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 10

W = F s cos θ = 10 × 2 cos 60º = 10 J.

Test: BITSAT Past Year Paper- 2020 - Question 11

The engine of a truck moving  along a straight road delivers constant power. The distance travelled by the truck in time t is proportional to

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 11


since both P and m are constants

Test: BITSAT Past Year Paper- 2020 - Question 12

The escape velocity from a planet is ve. A tunnel is dug along a diameter of the planet and a small body is dropped into it at the surface. When the body reaches the centre of the planet, its speed will be

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 12


In tunnel body will perform SHM at centre Vmax = Aω (see chapter on SHM)

Test: BITSAT Past Year Paper- 2020 - Question 13

If the the earth is at one-fourth of its present distance from the sun, the duration of year will be

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 13

The duration of a year is determined by the time it takes for a planet to complete one orbit around the sun. This is governed by Kepler's Third Law of Planetary Motion, which states that the square of the orbital period (T) of a planet is directly proportional to the cube of the semi-major axis (a) of its orbit.

Mathematically, this can be written as:

T² ∝ a³

Let's consider two cases: the current Earth-sun distance and the new distance where Earth is one-fourth of its present distance from the sun.

Case 1 (current distance):
T1² ∝ a1³

Case 2 (new distance, one-fourth the current distance):
T2² ∝ a2³

Since the new distance is one-fourth of the current distance, we can write a2 = 1/4 * a1. Now, we need to find the ratio of T2 to T1.

Divide the equations for Case 2 by Case 1:

(T2² / T1²) = (a2³ / a1³)
(T2² / T1²) = ((1/4 * a1)³ / a1³)
(T2² / T1²) = (1/64)

Now, take the square root of both sides:

T2 / T1 = √(1/64)
T2 / T1 = 1/8

So, the duration of the year when the Earth is at one-fourth of its present distance from the sun would be one-eighth of the present year

Test: BITSAT Past Year Paper- 2020 - Question 14

A vessel with water is placed on a weighing pan and it reads 600 g. Now a ball of mass 40 g and density 0.80 g cm–3 is sunk into the water with a pin of negligible volume, as shown in figure keeping it sunk. The weighing pan will show a reading

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 14

Volume of ball = 
Downthrust on water = 50 g.
Therefore reading is 650 g.

Test: BITSAT Past Year Paper- 2020 - Question 15

In an adiabatic process, the pressure is increased by If γ = 3/2, then the volume decreases by
nearly 

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 15

PV3/2 = K

Test: BITSAT Past Year Paper- 2020 - Question 16

The equation of a projectile is The angle of projection is given by

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 16

Comparing the given equation with we get tan θ = √3

Test: BITSAT Past Year Paper- 2020 - Question 17

Frequency of oscillation  is proportional to

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 17


Let mass is displaced towards left by x then force on mass = – kx – 2kx = – 3kx [negative sign is taken because force is opposite to the direction of motion]


Thus it is propotional to 

Test: BITSAT Past Year Paper- 2020 - Question 18

Two wires A and B of the same material, having radii in the ratio 1 : 2 and carry currents in the ratio 4 : 1. The ratio of drift speed of electrons in A and B is

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 18

Current flowing through the conductor, I = n e v A. Hence

Test: BITSAT Past Year Paper- 2020 - Question 19

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

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 19

 Velocity, = -Aω sin (ωt + π/4)
Velocity will be maximum, when
ωt + π/4 = π/2 or ωt = π/2 – π/4 = π/4 or t = π/4ω

Test: BITSAT Past Year Paper- 2020 - Question 20

The energy of electron in the n th orbit of hydrogen atom is expressed as 
The shortest and longest wavelength of Lyman series will be

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 20



and

Test: BITSAT Past Year Paper- 2020 - Question 21

In the circuit given below, the charge in μC, on the capacitor  having 5 μF is

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 21

Potential difference across the branch de is 6 V. Net capacitance of de branch is 2.1 µF
So, q = CV
⇒ q = 2.1 × 6 µC ⇒ q = 12.6 µ C Potential across 3 µF capacitance is

Potential across 2 and 5 combination in parallel is 6 – 4.2 = 1.8 V
So, q' = (1.8) (5) = 9 µC

Test: BITSAT Past Year Paper- 2020 - Question 22

A crystal has a coefficient of expansion 13×10– 7 in one direction and 231 × 10–7 in every direction at right angles to it. Then the cubical coefficient of expansion is

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 22

γ = ∝1 + ∝2 + ∝3
= 13 x 10-7 + 231 x 10-7 + 231 x 10-7 = 475 x l0-7

Test: BITSAT Past Year Paper- 2020 - Question 23

A solid cylinder and a hollow cylinder both of the same mass and same external diameter are released from the same height at the same time on an inclined plane. Both roll down without slipping. Which one will reach the bottom first?

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 23

Solid cylin der reaches the bottom first because for solid cylinder and for hollow cylinder,  Acceleration down the inclined plane Solid cylin der h as greater acceleration. It reaches the bottom first.

Test: BITSAT Past Year Paper- 2020 - Question 24

A whistle of frequency 1000 Hz is sounded on a car travelling towards a cliff with velocity of 18 m s–1 normal to the cliff. If c = 330 m s–1, then the apparent frequency of the echo as heard by the car driver is nearly

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 24

(a) By the concept of accoustic, the observer and source are moving towards each other, each with a velocity of 18 m s–1.

Test: BITSAT Past Year Paper- 2020 - Question 25

A thin sheet of glass (μ = 1.5) of thickness 6 micron introduced in the path of one of the interfering beams in a double slit experiment shifts the central fringe to a position previously occupied by fifth bright fringe. Then the wavelength of light used is

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 25

nλ = (μ - 1)t;

Test: BITSAT Past Year Paper- 2020 - Question 26

M.I of a circular loop of radius R about the axis in figure is​

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 26

Use theorem of parallel axes.

Test: BITSAT Past Year Paper- 2020 - Question 27

Three charge q, Q and 4q are placed in a straight line of length l at points distant 0, 1/2
 and l respectively from one end. In order to make the net froce on q zero, the charge Q must be equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 27

(Fnet )q =0

Test: BITSAT Past Year Paper- 2020 - Question 28

In series combination of R, L and C with an A.C. source at resonance, if R = 20 ohm, then impedence Z of the combination is

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 28

We know at resonance, reactance (resistance due to inductor and capacitor) be zero (0).
At resonance, Impedance (Z) = Resistance (R)
Therefore, Z=20 ohm

Test: BITSAT Past Year Paper- 2020 - Question 29

An electron moves in a circular arc of radius 10 m at a contant speed of 2 × 107 ms–1 with its plane of motion normal to a magnetic flux density of 10–5 T. What will be the value of specific charge of the electron?

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 29

Bqv = mv2 /r  or q/m = v /rB.

Test: BITSAT Past Year Paper- 2020 - Question 30

From a 200 m high tower, one ball is thrown upwards with speed of 10 ms-1 and another is thrown vertically downwards at the same speeds simultaneously. The time difference of their reaching the ground will be nearest to

Detailed Solution for Test: BITSAT Past Year Paper- 2020 - Question 30

The ball thrown upward will lose velocity in 1s. It return back to thrown point in another 1 s with the same velocity as second. Thus the difference will be 2 s.

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