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Test: BITSAT Past Year Paper- 2012 - JEE MCQ


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30 Questions MCQ Test - Test: BITSAT Past Year Paper- 2012

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Test: BITSAT Past Year Paper- 2012 - Question 1

What is the moment of inertia of a solid sphere of density r and radius ρ about its diameter?

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 1

For solid sphere

Test: BITSAT Past Year Paper- 2012 - Question 2

A body moves with uniform acceleration, then which of the following graph is correct ?

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 2

An object is said to be moving with a uniform acceleration, if its velocity changes by equal amount in equal intervals of time. The velocity-time graph of uniformly accelerated motion is a straight line inclined to time axis.
Acceleration of an object in a uniformly accelerated motion in one dimension is equal to the slope of the velocity-time graph with time axis.

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Test: BITSAT Past Year Paper- 2012 - Question 3

A projectile can have the same range R for two angles of projection. If t1 and t2 be the times of flight in two cases, then what is the product of two times of flight?

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 3



where R is the range.
Hence t1 t2 ∝ R

Test: BITSAT Past Year Paper- 2012 - Question 4

A horizontal overhead powerline is at height of 4m from the ground and carries a current of 100A from east to west. The magnetic field directly below it on the ground is (μ0 = 4π × 10–7 Tm A–1)

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 4

The magnetic field is 

According to right hand palm rule, the magnetic field is directed towards south.

Test: BITSAT Past Year Paper- 2012 - Question 5

A man of mass 100 kg. is standing on a platform of mass 200 kg. which is kept on a smooth ice surface. If the man starts moving on the platform with a speed 30 m/sec relative to the platform then calculate with what velocity relative to the ice the platform will recoil?

Test: BITSAT Past Year Paper- 2012 - Question 6

If the unit of force and length be each increased by four times, then the unit of energy is increased by

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 6

Since unit of energy = (unit of force).(unit of length) so if we increase unit of length and force, each by four times, then unit of energy will increase by sixteen times.

Test: BITSAT Past Year Paper- 2012 - Question 7

Which of the following must be known in order to determine the power output of an automobile?

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 7

Power is defined as the rate of doing work.
For the automobile, the power output is the amount of work done (overcoming friction) divided by the length of time in which the work was done.

Test: BITSAT Past Year Paper- 2012 - Question 8

If the force is given by F = at + bt2 with t as time. The dimensions of a and b are

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 8

Dimension of at = Dimension of F


Dimension of bt2 = Dimension of F

Test: BITSAT Past Year Paper- 2012 - Question 9

A wheel of radius R rolls on the ground with a uniform velocity v. The relative acceleration of topmost point of the wheel with respect to the bottom most point is

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 9

As  aCM = 0   [vCM = constant],
Tangential acceleration of each point

Test: BITSAT Past Year Paper- 2012 - Question 10

If the radius of the earth were to shrink by one per cent, its mass remaining the same, the value of g on the earth’s surface would

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 10

Test: BITSAT Past Year Paper- 2012 - Question 11

The Young’s modulus of a perfectly rigid body is

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 11

For a perfectly rigid body strain produced is zero for the given force applied, so Y = stress/strain = ∞

Test: BITSAT Past Year Paper- 2012 - Question 12

An ice block floats in a liquid whose density is less than water. A part of block is outside the liquid. When whole of ice has melted, the liquid level will

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 12

Ice is lighter than water. When ice melts, the volume occupied by water is less than that of ice. Due to which the level of water goes down.

Test: BITSAT Past Year Paper- 2012 - Question 13

A large drop of oil (density 0.8 g/cm3 and viscosity η0) floats up through a column of another liquid (density 1.2 g/cm3 and viscosity ηL).
Assuming that the two liquids do not mix, the velocity with which the oil drop rises will depend on:

Test: BITSAT Past Year Paper- 2012 - Question 14

A solid body of constant heat capacity 1 J/°C is being heated by keeping it in contact with reservoirs in two ways:
(i) Sequen tially keepin g in con tact with 2 reservoirs such that each reservoir supplies same amount of heat.
(ii) Sequen tially keepin g in con tact with 8 reservoirs such that each reservoir supplies same amount of heat.
In both the cases body is brought from initial temperature 100°C to final temperature 200°C.
Entropy change of the body in the two cases respectively is :

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 14

The entropy change of the body in the two cases is same as entropy is a state function.

Test: BITSAT Past Year Paper- 2012 - Question 15

Which of the following process is possible according to the first law of thermodynamics?

Test: BITSAT Past Year Paper- 2012 - Question 16

For an isothermal expansion of a perfect gas, the value of ΔP/P is equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 16

Differentiate PV = constant w.r.t V

Test: BITSAT Past Year Paper- 2012 - Question 17

A sample of ideal monoatomic gas is taken round the cycle ABCA as shown in the figure. The work done during the cycle is

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 17

ΔW = area under the p – V curve

Test: BITSAT Past Year Paper- 2012 - Question 18

The average translational kinetic energy of O2 (molar mass 32) molecules at a particular temperature is 0.048 eV. The translational kinetic energy of N2 (molar mass 28) molecules in eV at the same temperature is

Test: BITSAT Past Year Paper- 2012 - Question 19

For a gas if ratio of specific heats at constant pressure and volume is γ then value of degrees of freedom is

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 19

Test: BITSAT Past Year Paper- 2012 - Question 20

One end of a long metallic wire of length L tied to the ceiling. The other end is tied with a massless spring of spring constant K. A mass hangs freely from the free end of the spring. The area of cross section and the young’s modulus of the wire are A and Y respectively. If the mass slightly pulled down and released, it will oscillate with a time period T equal to :

Test: BITSAT Past Year Paper- 2012 - Question 21

The transverse displacement y(x, t) of a wave on a string is given by

This represents a

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 21


It is a function of type  y = f (ωt + kx)
∴ y (x, t) represents wave travelling along –x direction.

Test: BITSAT Past Year Paper- 2012 - Question 22

A sound source is moving towards stationary listener with 1/10th of the speed of sound. The ratio of apparent to read frequency is

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Test: BITSAT Past Year Paper- 2012 - Question 23

In a region of space having a uniform electric field E, a hemispherical bowl of radius r is placed.The electric flux ϕ through the bowl is

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 23

ϕ = E(ds) cos θ = E(2πr2) cos 0° = 2πr2 E.

Test: BITSAT Past Year Paper- 2012 - Question 24

The electric field intensity just sufficient to balance the earth’s gravitational attraction on an electron will be: (given mass and charge of an electron respectively are 9.1 × 10–31 kg and 1.6 × 10–19 C.)

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 24

– eE = mg

Test: BITSAT Past Year Paper- 2012 - Question 25

Two capacitors C1 and C2 are charged to 120 V and 200 V respectively. It is found that by connecting them together the potential on each one can be made zero. Then

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 25

For potential to be made zero, after connection

⇒ 3C1 = 5C2 

Test: BITSAT Past Year Paper- 2012 - Question 26

Three voltmeters A, B and C having resistances R, 1.5 R and 3R, respectively, are connected as shown. When some potential difference is applied between X and Y, the voltmeter readings are VA, VB and VC respectively. Then –

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 26

VA = IR

∴ VA = V= VC

Test: BITSAT Past Year Paper- 2012 - Question 27

The range of the particle when launched at an angle of 15º with the horizontal is 1.5 km. What is the range of the projectile when launched at an angle of 45º to the horizontal.

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 27

Test: BITSAT Past Year Paper- 2012 - Question 28

If m is magnetic moment and B is the magnetic field, then the torque is given by

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 28

Test: BITSAT Past Year Paper- 2012 - Question 29

Magnetic moment of bar magnet is M. The work done to turn the magnet by 90° of magnet in direction of magnetic field B will be

Detailed Solution for Test: BITSAT Past Year Paper- 2012 - Question 29

Work done, W = MB (1 – cos θ)
θ = 90°
W = MB

Test: BITSAT Past Year Paper- 2012 - Question 30

The laws of electromagnetic induction have been used in the construction of a

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