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BITSAT Practice Test - 7 - JEE MCQ


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30 Questions MCQ Test - BITSAT Practice Test - 7

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BITSAT Practice Test - 7 - Question 1

A particle begins to move with a tangential acceleration of constant magnitude 0.6ms-2 in a circular path. If it slips when its total acceleration becomes 1 ms−2, then the angle through which it would have turned before it started to slip is

Detailed Solution for BITSAT Practice Test - 7 - Question 1

Let particle move on a circular path with velocity v and tangential acceleration at and radial acceleration ar.  Suppose the particle made angle θ when it slips with acceleration 1ms-2

From the equation,

Equating equation (1) and (2)

BITSAT Practice Test - 7 - Question 2

A solid sphere rolls down without slipping on an inclined plane at angle 60° over a distance of 10 m. The acceleration (in m s−2) is

Detailed Solution for BITSAT Practice Test - 7 - Question 2

Here θ=60°, l=10 m, a=?

For solid sphere K2 = 2/5 R2

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BITSAT Practice Test - 7 - Question 3

A cube has a side of length 1.2 x10-2m. Calculate its volume.

Detailed Solution for BITSAT Practice Test - 7 - Question 3

Volume, V=l3=(1.2 x 10−2m)3
=1.728x10−6 m3

Since length (l) has two significant figures, the volume (V) will also have two significant figures.

Therefore, the correct answer is 1.7 x 10-6m3

BITSAT Practice Test - 7 - Question 4

If frictional force is neglected and girl bends her hand, then (initially girl is rotating on chair)

Detailed Solution for BITSAT Practice Test - 7 - Question 4

In this case angular momentum of the girl will remain conserved as there is no external torque working on it.
I1ω= I2ω2
I2=I1ω1/ω2 
As she folds her hand, angular velocity increases. So Moment of inertia decreases.

Another way to explain it is that moment of inertia is a measure of the resistance offered by the body in its rotational motion. So, when she swirls with her hand stretched, her body offers more resistance than when she folds it. Hence, 

Igirl's hand folded < Igirl's hand stretched

BITSAT Practice Test - 7 - Question 5

The surface considered for Gauss's law is called

Detailed Solution for BITSAT Practice Test - 7 - Question 5

The surface that we choose for the application of Gauss’s law is called Gaussian surface.

BITSAT Practice Test - 7 - Question 6

To use a transistor as an amplifier -

Detailed Solution for BITSAT Practice Test - 7 - Question 6

Above diagram is for npn transistor used as an amplifier in common base configuration.

From the diagram we can see that, emitter- base junction is in forward bias and collector base junction is in reverse bias.

BITSAT Practice Test - 7 - Question 7

A particle of mass M is moving in a circle of fixed radius R in such a way that its centripetal acceleration at time t is given by n2Rt2, where n is a constant. The power delivered to the particle by the force acting on it, is :

Detailed Solution for BITSAT Practice Test - 7 - Question 7

BITSAT Practice Test - 7 - Question 8

The dimensional formula of modulus of rigidity is

Detailed Solution for BITSAT Practice Test - 7 - Question 8

BITSAT Practice Test - 7 - Question 9

If a piece of metal is heated to temperature θ  and then allowed to cool in a room which is at temperature θ0, the graph between the temperature T of the metal and time t will be closest to :

Detailed Solution for BITSAT Practice Test - 7 - Question 9

According to Newton's cooling law, rate of cooling is

Here, θ0= the temperature of the body
Let k be the proportionality constant and θbe the initial temperature.
Therefore, 

Hence, the graph is exponential.

BITSAT Practice Test - 7 - Question 10

In the situation shown in figure all the strings are light and inextensible and pulleys are light. There is no friction at any surface and all block are of cuboidal shape. A horizontal force of magnitude F is applied to right most free end of string in both cases of figure 1 and figure 2 as shown. At the instant shown, the tension in all strings are non zero. Let the magnitude of acceleration of large blocks (of mass M) in figure 1 and figure 2 are a1 and a2 respectively. Then :

Detailed Solution for BITSAT Practice Test - 7 - Question 10

The free diagram for large blocks of figure 1 and figure 2

From FBD it is obvious net force on each block is zero in horizontal direction.

∴ a1 = a2 = 0

BITSAT Practice Test - 7 - Question 11

Two spherical bodies of the same mass M are moving with velocities v1 and v. These collide perfectly inelastically. What is the loss in kinetic energy

Detailed Solution for BITSAT Practice Test - 7 - Question 11

Mass of each spherical body is M.
Velocity of sphere 1 before collision is v1.
Velocity of sphere 2 before collision is v2.
Collision is perfectly inelastic, so both masses move together after collision.

Let us assume that
velocity of masses after collision is v'.

Now, from the law of conservation of linear momentum, net force on the system is zero,

Let us assume that 
initial kinetic energy of system is Ki
final kinetic energy of system is Kf


Now, loss of kinetic energy is 
ΔK = Ki - Kf

Now, on putting the corresponding values in the above equation 

BITSAT Practice Test - 7 - Question 12

A pipe open at both the ends produce a note of fundamental frequency v1. When the pipe is kept with (3/4)th of its length in water, it produces a note of fundamental frequency v2. The ratio of v1/v2 is 

Detailed Solution for BITSAT Practice Test - 7 - Question 12

BITSAT Practice Test - 7 - Question 13

In Young's double-slit experiment, the distance between slits is d/3 and the distance between screen and slits is 3D, then fringe width is

Detailed Solution for BITSAT Practice Test - 7 - Question 13

BITSAT Practice Test - 7 - Question 14

A beam of light of wavelength 400 nm and power 1.55 mW is directed at the cathode of a photoelectric cell. If only 10% of the incident photons effectively produce photoelectrons, then find current due to these electrons.
(given, hc=1240eV -nm, e= 1.6x10-19C

Detailed Solution for BITSAT Practice Test - 7 - Question 14

The energy of incident photon of wavelength 400 nm is

∴ Number of such photons in a beam of light of power 1.55 mW

This implies, number of photons that were used to produce photoelectron per second = 10% of 3.125 x 1015 = 3.125 x 14 per second
Hence, the current due to such photoelectron per second
=3.125 x 1014 x 16 x 10-19

= 5 x 10-5 A = 50μA

BITSAT Practice Test - 7 - Question 15

In an experiment, a boy plots a graph between (vmax)2 and (amax)2 for a simple pendulum for different values of (small) amplitudes, where vmax and amax is the maximum velocity and the maximum acceleration respectively. He found the graph to be a straight line with a negative slope, making an angle of 30° when the experiment was conducted on the earth surface. When the same experiment was conducted at a height h above the surface, the line was at an angle of 60°. The value of h is [radius of the earth = R]

Detailed Solution for BITSAT Practice Test - 7 - Question 15

For a simple pendulum,

Where ω is the angular frequency.
g is the acceleration due to gravity.
l is the length of the pendulum.

BITSAT Practice Test - 7 - Question 16

If a vector  is perpendicular to the vector  then the value of α is

Detailed Solution for BITSAT Practice Test - 7 - Question 16

When two vectors are perpendicular to each other, then their dot product is zero.

BITSAT Practice Test - 7 - Question 17

A horizontal force of 10 N is necessary to just hold a stationary block against a wall. The coefficient of friction between the block and the wall is 0.2. The weight of the block is

Detailed Solution for BITSAT Practice Test - 7 - Question 17

BITSAT Practice Test - 7 - Question 18

Spheres A and B of equal radius and of masses 2 m and m, respectively, are moving towards each other and strike directly. The speeds of A and B before the collision are 3u and u, respectively. The collision is such that B experiences an impulse of 4mcu, where c is constant.

What is the coefficient of restitution?

Detailed Solution for BITSAT Practice Test - 7 - Question 18

Magnitude of the Impulse received | ΔI| = 4mcu

V1 and V2 are velocities after collision.
Magnitude of impulse is the same on both A and B.
For body A:
2mv1 - 6mu = -4mcu
2mv1 = 6mu - 4cum
V1 = 3u - 2cu(→)
For body B:
mv2 - (-mu) = 4mcu
mv2 = 4mcu - mu
v2 = 4cu - u(→) - speed of B

BITSAT Practice Test - 7 - Question 19

A body is projected vertically upward from the surface of the Earth with a velocity equal to half the escape velocity. If R is the radius of the Earth, then the maximum height attained by the body is

Detailed Solution for BITSAT Practice Test - 7 - Question 19

If the body comes to rest at a height r from the centre of the earth, its final energy will be given be 

Equating (i) and (ii) we get r=4R/3. Maximum height attained 

BITSAT Practice Test - 7 - Question 20

If a wire having initial diameter of 2 mm produces the longitudinal strain of 0.1%, then the final diameter of wire is
(σ = 0.5)

Detailed Solution for BITSAT Practice Test - 7 - Question 20

When a deforming force is applied at the free end of a suspend wire of length l and radius R, then its length increases by dl, but its radius decreases by dR.
Now two types of strains are produced by a single force

Then the Poisson's ratio

= 0.0005 mm
∴ Final radius R - ΔR
= 1 mm - 0.0005 mm
= 0.9995 mm
∴ The final diameter = 2 x 0.9995 mm
= 1.9990 mm

BITSAT Practice Test - 7 - Question 21

A point particle of mass 0.1 kg is executing SHM of amplitude 0.1 m. When the particle passes through the mean position, its KE is 8 x 10-3 J. What is the equation of motion of this particle if the initial phase of oscillation is 45o?

Detailed Solution for BITSAT Practice Test - 7 - Question 21

BITSAT Practice Test - 7 - Question 22

A stone hangs from the free end of a sonometer wire whose vibrating length, when tuned to a tuning fork, is 40 cm. When the stone hangs wholly immersed in water, the resonant length is reduced to 30 cm. The relative density of the stone is

Detailed Solution for BITSAT Practice Test - 7 - Question 22

 where M = mass of stone. If ρ is the density of the stone and V its volume, then m = ρV.
When the stone is wholly immersed in water of density ρ', the effective weight of the stone

Given l = 40 cm and l' = 30 cm.
Also v = v', which gives

Hence the correct choice is (2).

BITSAT Practice Test - 7 - Question 23

The figure below shows the P - V diagram for a fixed mass of an ideal gas undergoing cyclic process ABCA. If the temperature at A is T, then what is the temperature at C?

Detailed Solution for BITSAT Practice Test - 7 - Question 23

Volume at A = Volume at C = V, but pressure at A = 2P and pressure at C = P. According to Charles' law, at constant volume, P is directly proportional to T. Hence, the temperature at C will become half of temperature at A, i.e. T/2, which is choice (4).

BITSAT Practice Test - 7 - Question 24

The potential at a point P, which is forming a corner of a square of side 93 mm with charges q1 = 33 nC, q2 = -51 nC and q3 = 47 nC located at the other three corners is nearly

Detailed Solution for BITSAT Practice Test - 7 - Question 24

Total Potential at the point p is 

 

BITSAT Practice Test - 7 - Question 25

Consider a thin square sheet of side L and thickness t, made of a material of resistivity ρ. The resistance between two opposite faces shown by the shaded areas in the figure is

Detailed Solution for BITSAT Practice Test - 7 - Question 25

Here, area of cross section A = L × t

Hence, resistance is inversely proportional to the thickness 't' and is independent of L.

BITSAT Practice Test - 7 - Question 26

A square wire of each side L carries a current i. What is the magnetic field at the mid-point of the square?

Detailed Solution for BITSAT Practice Test - 7 - Question 26

Magnetic field due of one side of the square is

Resultant magnetic firld at centre due to four sides is

BITSAT Practice Test - 7 - Question 27

If a resistance of 100 Ω, an inductance of 0.5 H and a capacitance of 10 x 10−6 F are connected in series with a 50 Hz AC supply, then what will be the impedance?

Detailed Solution for BITSAT Practice Test - 7 - Question 27

The impedance of an A.C. circuit containing resistance, inductance and capacitance is

Here : R = 100Ω,
L = 0.5 H
C = 10 x 10-6 F
f = 50 Hz

BITSAT Practice Test - 7 - Question 28

In a single-slit diffraction experiment, the width of the slit is made double its original width. The central maximum of the diffraction pattern will become

Detailed Solution for BITSAT Practice Test - 7 - Question 28

The angular width of the central maximum is 2λ/a, where a is the width of the slit. If the value of a is doubled, the angular width of the central maximum decreases to half its earlier value. This implies that the central maximum becomes much sharper. Furthermore if a is doubled, the intensity of the central maximum becomes four times. Thus the central maximum becomes much sharper (narrow) and brighter.

BITSAT Practice Test - 7 - Question 29

Assuming human pupil to have a radius of 0.25 cm and a comfortable viewing distance of 25 cm, the minimum separation between two objects that human eye can resolve at 500 nm wavelength is

Detailed Solution for BITSAT Practice Test - 7 - Question 29

BITSAT Practice Test - 7 - Question 30

A radioactive nuclide can decay simultaneously by two different processes, which have individual decay constants  and λ2, respectively. The effective decay constant of the nuclide is given by

Detailed Solution for BITSAT Practice Test - 7 - Question 30

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