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Test: BITSAT Past Year Paper- 2011 - JEE MCQ


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30 Questions MCQ Test - Test: BITSAT Past Year Paper- 2011

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Test: BITSAT Past Year Paper- 2011 - Question 1

A passenger in a open car travelling at 30 m/s throws a ball out over the bonnet. Relative to the car the initial velocity of the ball is 20 m/s at 60° to the horizontal. The angle of projection of the ball with respect to the horizontal road will be

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 1

Test: BITSAT Past Year Paper- 2011 - Question 2

A particle is moving in a straight line with initial velocity and uniform acceleration a. If the sum of the distance travelled in tth and (t + 1)th seconds is 100 cm, then its velocity after t seconds, in cm/s, is

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Test: BITSAT Past Year Paper- 2011 - Question 3

The two vectors are drawn from a common point and , then angle between is –
(1) 90° if C2 = A2 + B2
(2) greater than  90° if C2 < A2 + B2
(3) greater than  90° if C2 > A2 + B2
(4) less than  90° if C2 > A2 + B2
Correct options are –

Test: BITSAT Past Year Paper- 2011 - Question 4

If then find the dimensions of q. Where T is the time period of bar of mass M, length L and Young modulus Y.

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 4

, writing dimensions of both the sides, we get 
or q = [L4]

Test: BITSAT Past Year Paper- 2011 - Question 5

An object experiences a net force and accelerates from rest to its final position in 16s. How long would the object take to reach the same final velocity from rest if the force was four times larger ?

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 5

When the force is increased by a factor of four the acceleration is also increased by the factor of 4, using Vf = Vi + at. We can see that the time it takes to reach the same Vfrom rest will be reduced by the factor of 4.

Test: BITSAT Past Year Paper- 2011 - Question 6

Three blocks of masses m1, m2 and m3 are connected by massless strings, as shown, on a frictionless table. They are pulled with a force T3 = 40 N. If m1 = 10 kg, m2 = 6 kg and m3 = 4kg, the tension T2 will be

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 6

For equilibrium of all 3 masses,

For equilibrium of m1 & m2
T2 = (m1 + m2) × a   or, 2
Given m1 = 10 kg, m2 = 6 kg, m3 = 4 kg, T3 = 40 N

Test: BITSAT Past Year Paper- 2011 - Question 7

A massless platform is kept on a light elastic spring as shown in fig. When a sand particle of mass 0.1 kg is dropped on the pan from a height of 0.24 m, the particle strikes the pan and spring is compressed by 0.01 m. From what height should the particle be dropped to cause a compression of 0.04 m.

Test: BITSAT Past Year Paper- 2011 - Question 8

A constant torque of 31.4 N-m is exerted on a pivoted wheel. If angular acceleration of wheel is 4 p rad/s2, then the moment of inertia of the wheel is

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 8

Test: BITSAT Past Year Paper- 2011 - Question 9

A man of mass m starts falling towards a planet of mass M and radius R. As he reaches near to the surface, he realizes that he will pass through a small hole in the planet. As he enters the hole, he sees that the planet is really made of two pieces a spherical shell of negligible thickness of mass 2M/3 and a point mass M/3 at the centre.
Change in the force of gravity experienced by the man is

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 9

Change in force of gravity

(only due to mass M/3 due to shell gravitational field is zero (inside the shell))
= 2GMm/3R2

Test: BITSAT Past Year Paper- 2011 - Question 10

Geo-stationary satellite is one which

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 10

Geo-stationary satellites are also called synchronous satellite. They always remain about the same path on equater, i.e., it has a period of exactly one day (86400 sec) So orbit radius comes out to be 42400 km, which is nearly equal to the circumference of earth. So height of Geostationary satellite from the earth surface is 42,400 – 6400 = 36,000 km.

Test: BITSAT Past Year Paper- 2011 - Question 11

Two wires are made of the same material and have the same volume. However wire 1 has crosssectional area A and wire 2 has cross-sectional area 3A. If the length of wire 1 increases by Dx on applying force F, how much force is needed to stretch wire 2 by the same amount?

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 11


As shown in the figure, the wires will have the same Young’s modulus (same material) and the length of the wire of area of crosssection 3A will be  l/3 (same volume as wire 1).

⇒ F' = 9F

Test: BITSAT Past Year Paper- 2011 - Question 12

An iron rod of length 2m and cross-sectional area of 50 mm2 stretched by 0.5 mm, when a mass of 250 kg is hung from its lower end. Young’s modulus of iron rod is

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 12

Test: BITSAT Past Year Paper- 2011 - Question 13

Viscosity is the property of a liquid due to which it :

Test: BITSAT Past Year Paper- 2011 - Question 14

The radiation emitted by a perfectly black body is proportional to

Test: BITSAT Past Year Paper- 2011 - Question 15

A copper sphere cools from 62°C to 50°C in 10 minutes and to 42°C in the next 10 minutes.Calculate the temperature of the surroundings.

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 15

By Newton's law of cooling,

A sphere cools from 62°C to 50°C in 10 min.

Now, sphere cools from 50°C to 42°C in next 10 min.

Dividing eqn. (2) by (3) we get,

Hence θ0 = 26°C

Test: BITSAT Past Year Paper- 2011 - Question 16

An air bubble of volume v0 is released by a fish at a depth h in a lake. The bubble rises to the surface. Assume constant temperature and standard atmospheric pressure above the lake.
The volume of the bubble just before touching the surface will be (density) of water is ρ

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 16

As the bubble rises the pr essur e gets reduced for constant temperature, if P is the standard atmospheric pressure, then (P + ρgh ) V0 = PV
or, 

Test: BITSAT Past Year Paper- 2011 - Question 17

The molecules of a given mass of gas have a root mean square velocity of 200m s–1 at 27°C and 1.0 × 105 N m–2 pressure. When the temperature is 127°C and the pressure 0.5 × 105 Nm–2, the root mean square velocity in ms–1, is

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 17

Test: BITSAT Past Year Paper- 2011 - Question 18

Which of the following expressions corresponds to simple harmonic motion along a straight line, where x is the displacement and a, b, c are positive constants?

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 18

In linear S.H.M., the restoring force acting on particle should always be proportional to the displacement of the particle and directed towards the equilibrium position. i.e., F ∝ x
or F = –bx where b is a positive constant.

Test: BITSAT Past Year Paper- 2011 - Question 19

A mass m is suspended from a spring of force constant k and just touches another identical spring fixed to the floor as shown in the figure. The time period of small oscillations is

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 19

When the spring undergoes displacement in the downward direction it completes one half oscillation while it completes another half oscillation in the upward direction. The total time period is:

Test: BITSAT Past Year Paper- 2011 - Question 20

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- 2011 - Question 20

For open pipe, n = V2l, where n0 is the fundamental frequency of open pipe.

As freq. of 1st overtone of open pipe = freq. of 1st overtone of closed pipe.

= 41.25 cm

Test: BITSAT Past Year Paper- 2011 - Question 21

In an uniformly charged sphere of total charge  Q and radius R, the electric field E is plotted as function of distance from the centre. The graph which would correspond to the above will be

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 21

Electric field inside the uniformly charged sphere varies linearly, while outside the sphere, it varies as inverse square of distance,which is correctly represented in option (c).

Test: BITSAT Past Year Paper- 2011 - Question 22

A charge Q1 exerts some force on a second charge Q2. If a 3rd charge Q3 is brought near, then the force of Q1 exerted on Q2

Test: BITSAT Past Year Paper- 2011 - Question 23

A hollow metal sphere of radius 5 cm is charged such that the potential on its surface is 10 V. The potential at a distance of 2 cm from the centre of the sphere is

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 23

Potential at any point inside the sphere = potential at the surface of the sphere = 10V.

Test: BITSAT Past Year Paper- 2011 - Question 24

If the potential of a capacitor having capacity 6 mF is increased from 10 V to 20 V, then increase in its energy will be

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 24

Capacitan ce of capacitor (C) = 6 m F = 6 × 10–6 F; Initial potential (V1) = 10 V and final potential (V2) = 20 V.
The increase in energy (ΔU)

= 1/2 x (6x10-6) x [(20)2 - (10)2]
= (3x10-6) x 300 = 9 x 10-4 J.

Test: BITSAT Past Year Paper- 2011 - Question 25

Calculate the effective resistance between A and B in following network.

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 25

Equivalent resistance = (5 + 10 + 15) | | (10 + 20 + 30)
So, 

Test: BITSAT Past Year Paper- 2011 - Question 26

A steady current is set up in a cubic network composed of wires of equal resistance and length d as shown in figure. What is the magnetic field at the centre P due to the cubic network

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 26

By symmetry, the magnetic field at the centre P is zero.

Test: BITSAT Past Year Paper- 2011 - Question 27

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- 2011 - Question 27

Torque, 

Test: BITSAT Past Year Paper- 2011 - Question 28

A metal rod of length 1 m is rotated about one of its ends in a plane right angles to a field of inductance 2.5 × 10–3 Wb/m². If it makes 1800 revolutions/min. Calculate induced e.m.f. between its ends.

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 28

Given: l = 1m, B = 5 × 10–3 Wb/m²
f = 1800/60 = 30 rotations/sec In one rotation, the moving rod of the metal traces a circle of radius r = l
∴ Area swept in one rotation = πr2

= Bfπr2 = (5 × 10–3) × 3.14 × 30 × 1 = 0.471 V

Test: BITSAT Past Year Paper- 2011 - Question 29

Which one of the following curves represents the variation of impedance (Z) with frequency f in series LCR circuit?

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 29

Test: BITSAT Past Year Paper- 2011 - Question 30

An electromagnetic wave passes through space and its equation is given by E = E0 sin (ωt – kx) where E is electric field. Energy density of electromagnetic wave in space is

Detailed Solution for Test: BITSAT Past Year Paper- 2011 - Question 30

Energy density

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