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Test: BITSAT Past Year Paper- 2018 - JEE MCQ


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30 Questions MCQ Test - Test: BITSAT Past Year Paper- 2018

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Test: BITSAT Past Year Paper- 2018 - Question 1

Four point charges –Q, –q, 2q and 2Q are placed, one at each corner of the square. The relation between Q and q for which the potential at the centre of the square is zero is :​

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 1

Let the side length of square be 'a' then potential at centre O is


= – Q – q + 2q + 2Q = 0 ⇒ Q + q = 0 (Given) Q
= – q

Test: BITSAT Past Year Paper- 2018 - Question 2

Two long parallel wires carry equal current i flowing in the same direction are at a distance 2d apart. The magnetic field B at a point lying on the perpendicular line joining the wires and at a distance x from the midpoint is –

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 2

The magnetic field due to two wires at P


Both the magnetic fields act in opposite direction.

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Test: BITSAT Past Year Paper- 2018 - Question 3

In the circuit shown, the symbols have their usual meanings. The cell has emf E. X is initially joined to Y for a long time. Then, X is joined to Z. The maximum charge on C at any later time will be

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 3

Current in inductor = E/R
∴  its energy = 1/2 LE2/R2
Same energy is later stored in capacitor

Test: BITSAT Past Year Paper- 2018 - Question 4

A point object O is placed in front of a glass rod having spherical end of radius of curvature 30 cm. The image would be formed at

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 4

Using, μ/v - 1/u = μ - 1/R

Or 1.5/v - 1/-15 = 1.5 - 1/+30
∴ v = - 30cm

Test: BITSAT Past Year Paper- 2018 - Question 5

In Young’s double slit experiment, λ = 500nm, d = 1mm, D = 1m. Minimum distance from the central maximum for which intensity is half of the maximum intensity is

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 5

Test: BITSAT Past Year Paper- 2018 - Question 6

What is the voltage gain in a common emitter amplifier, where input resistance is 3 Ω and load resistance 24 Ω, b = 0.6 ?

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 6

Voltage gain,  Av = β RL/Ri = 0.6 x 24/3 = 4.8

Test: BITSAT Past Year Paper- 2018 - Question 7

The acceleration due to gravity on the surface of the moon is 1/6 that on the surface of earth and the diameter of the moon is one-fourth that of earth. The ratio of escape velocities on earth and moon will be​

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 7

Test: BITSAT Past Year Paper- 2018 - Question 8

Given The magnitude of their resultant is

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 8

Test: BITSAT Past Year Paper- 2018 - Question 9

A particle of mass m executes simple harmonic motion with amplitude a and frequency n. The average kinetic energy during its motion from the position of equilibrium to the end is

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 9

The kinetic energy of a particle executing S.H.M. is given by
K = 1/2 ma2 ω2 sin2ωt
= 1/2 mω2a2(1/2) (∵ < sin2 θ > = 1/2)
= 1/4 mω2a2 = 1/4 ma2 (2πv)2 (∵ ω = 2πv)
or, <K> = πma2v2

Test: BITSAT Past Year Paper- 2018 - Question 10

The dipole moment of the given charge distribution is

Test: BITSAT Past Year Paper- 2018 - Question 11

At a place, if the earth's horizontal and vertical components of magnetic fields are equal, then the angle of dip will be

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 11

tan θ = By/B= 1 ∴ θ = 45°

Test: BITSAT Past Year Paper- 2018 - Question 12

The third line of Balmer series of an ion equivalnet to hydrogen atom has wavelength of 108.5 nm.The ground state energy of an electron of this ion will be

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 12

For third line of Balmer series n1 = 2, n2 = 5

On putting values Z = 2 From
E

Test: BITSAT Past Year Paper- 2018 - Question 13

The binding energy per nucleon of 10X is 9 MeV and that of 11X is 7.5 MeV where X represents an element. The minimum energy required to remove a neutron from 11X is

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Test: BITSAT Past Year Paper- 2018 - Question 14

If C, the velocity of light, g the acceleration due to gravity and P the atmospheric pressure be the fundamental quantities in MKS system, then the dimensions of length will be same as that of​

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 14

Test: BITSAT Past Year Paper- 2018 - Question 15

Figure shows a capillary rise H. If the air is blown through the horizontal tube in the direction as shown then rise in capillary tube will be​

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 15

Due to increase in velocity, pressure will be low above the surface of water.

Test: BITSAT Past Year Paper- 2018 - Question 16

A boy running on a horizontal road at 8 km/h finds the rain falling vertically. He increases his speed to 12 km/h and finds that the drops makes 30° with the vertical. The speed of rain with respect to the road is

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 16

Test: BITSAT Past Year Paper- 2018 - Question 17

A hunter aims his gun and fires a bullet directly at a monkey on a tree. At the instant the bullet leaves the barrel of the gun, the monkey drops.
Pick the correct statement regarding the situation.

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 17


t = OC/u cos θ = x/u cos θ

AC = x tan q
BC = distance travelled by bullet in time t, vertically.
y = u sin θ t – 1/2 gt2
AB = x tan θ – (u sin θt – 1/2 gt2)
= x tan θ – (usinθ × x/u cos θ  – 1/2 gt2)
(∴ bullet will always hit the monkey)

Test: BITSAT Past Year Paper- 2018 - Question 18

A particle of mass m1 moving with velocity v collides with a mass m2 at rest, then they get embedded. Just after  collision, velocity of the system

Test: BITSAT Past Year Paper- 2018 - Question 19

The ratio of the specific heats of a gas is Cp/Cv = 1.66, then the gas may be

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 19

Let ‘n’ be the degree of freedom

⇒ n = 3 ⇒ gas must be monoatomic.

Test: BITSAT Past Year Paper- 2018 - Question 20

Two oscillators ar e started simultaneously in same phase. After 50 oscillations of one, they get out of phase by p, that is half oscillation. The percentage difference of frequencies of the two oscillators is nearest to

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 20

Phase change π in 50 oscillations.
Phase change 2π in 100 oscillations.
So frequency different ~ 1 in 100.

Test: BITSAT Past Year Paper- 2018 - Question 21

A juggler keeps on moving four balls in the air throwing the balls after intervals. When one ball leaves his hand (speed = 20 ms–1) the position of other balls (height in m) will be (Take g = 10 ms–2)

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 21

Time taken by same ball to return to the hands of juggler = 2u/g = 2 x 20/10 = 4 S. So he is throwing the balls after each 1 s. Let at some instant he is throwing ball number 4. Before 1 s of it he throws ball. So height of ball 3:
h3 = 20 × 1 – 1/2 10(1)2 = 15
Before 2s, he throws ball 2. So height of ball 2:
h2 = 20 × 2 – 1/2 10(2)2 = 20 m
Before 3 s, he throws ball 1. So height of ball 1:
h1 = 20 × 3 – 1/2 10(3)2 = 15 m

Test: BITSAT Past Year Paper- 2018 - Question 22

If a stone of mass 0.05 kg is thrown out a window of a train moving at a constant speed of 100 km/ h then magnitude of the net force acting on the stone is

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 22

After the stone is thrown out of the moving train, the only force acting on it is the force of gravity i.e. its weight.
∴ F = mg = 0.05 × 10 = 0.5 N.

Test: BITSAT Past Year Paper- 2018 - Question 23

A body of mass M hits normally a rigid wall with velocity V and bounces back with the same velocity. The impulse experienced by the body is

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 23

Impulse experienced by the body = MV – (–MV) = 2MV.

Test: BITSAT Past Year Paper- 2018 - Question 24

A hoop rolls down an inclined plane. The fraction of its total kinetic energy that is associated with rotational motion is

Test: BITSAT Past Year Paper- 2018 - Question 25

Infinite number of masses, each 1 kg are placed along the x-axis at x = ± 1m, ± 2m, ± 4m, ± 8m, ± 16m ..... the magnitude of the resultant gravitational potential in terms of gravitational constant G at the orgin (x = 0) is

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 25

Test: BITSAT Past Year Paper- 2018 - Question 26

Water of volume 2 litre in a container is heated with a coil of 1 kW at 27°C. The lid of the container is open and energy dissipates at rate of 160 J/s.
In how much time temperature will rise from 27°C to 77°C? [Given specific heat of water is 4.2 kJ/kg]

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 26

Heat gained by the water = (Heat supplied by the coil) – (Heat dissipated to environment)
⇒ mc Δθ = PCoil t – PLoss t
⇒ 2 × 4.2 × 103 × (77 – 27) = 1000 t – 160 t
⇒ t = 4.2x105/840 = 500 s = 8 min 20 s

Test: BITSAT Past Year Paper- 2018 - Question 27

In the following P-V diagram of an ideal gas, two adiabates cut two isotherms at T1 = 300K and T2 = 200K. The value of VA = 2 unit, VB = 8 unit, VC = 16 unit. Find the value of VD.

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 27

Test: BITSAT Past Year Paper- 2018 - Question 28

The mass of H2 molecule is 3.32 × 10–24 g. If 1023 hydrogen molecules per second strike 2 cm2 of wall at an angle of 45o with the normal, while moving with a speed of 105 cm/s, the pressure exterted on the wall is nearly.

Test: BITSAT Past Year Paper- 2018 - Question 29

The wavelength of two waves are 50 and 51 cm respectively. If the temperature of the room is 20°C then what will be the number of beats produced per second by these waves, when the speed of sound at 0°C is 332 m/s?

Detailed Solution for Test: BITSAT Past Year Paper- 2018 - Question 29

λ1= 50 cm. λ2= 51 cm.

⇒ v2 = 319.23.
v= v2= 319.23/0.50 = 640 Hz.
v2 = v22 = 319.23/51x10-2 = 625.94 =  62 HZ
No. of beats = v2 - v= 14 Hz

Test: BITSAT Past Year Paper- 2018 - Question 30

The figure shows the interference pattern obtained in a double-slit experiment using light of wavelength 600nm. 1, 2, 3, 4 and 5 are marked on five fringes.The third order bright fringe is

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