Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

When a particle of mass m moves on the x-axis in a potential of the form V(x) = kx^{2}, it performs simple harmonic motion. The corresponding time period is proportional to $$\sqrt {{m \over k}} $$, as can be seen easily using dimensional analysis. However, the motion of a particle can be periodic even when its potential energy increases on both sides of x = 0 in a way different from kx^{2} and its total energy is such that the particle does not escape to infinity. Consider a particle of mass m moving on the x-axis. Its potential energy is V(x) = $$\alpha$$x^{4} ($$\alpha$$ > 0) for | x | near the origin and becomes a constant equal to V_{0} for (see figure).

If the total energy of the particle is E, it will perform periodic motion only if

A

E < 0

B

E > 0

C

V_{0} > E > 0

D

E > V_{0}

The kinetic energy of the particle cannot be negative. The total energy E is the sum of the kinetic energy K(x) and potential energy V(x) i.e.,

E = K(x) + V(x). ....... (1)

From the given figure, V(x) $$\ge$$ 0 for all x. If E $$\le$$ 0 for some x then kinetic energy K(x) = E $$-$$ V(x) $$\le$$ 0 for those x and hence the motion of the particle is not allowed for those x. On the other hand, if E $$\ge$$ V_{0} then K(x) = E $$-$$ V(x) $$\ge$$ 0 for all x and particle is allowed to move for all x, including infinity (in this case the particle will escape to infinity). Thus, for the particle to have a periodic motion, V_{0} > E > 0. In this case, particle is allowed at points where

K(x) = E $$-$$ V(x) = E $$-$$ $$\alpha$$x^{4} $$\ge$$ 0 i.e.,

| x | $$\le$$ (E/$$\alpha$$)^{1/4}.

2

MCQ (Single Correct Answer)

A simple pendulum has time period T_{1}. The point of suspension is now moved upward according to the relation y = Kt^{2}, (K = 1 m/s^{2}) where y is the vertical displacement. The time period now become T_{2}. The ratio of $${{T_1^2} \over {T_2^2}}$$ is (g = 10 m/s^{2})

A

$${5 \over 6}$$

B

$${6 \over 5}$$

C

1

D

$${4 \over 5}$$

3

MCQ (Single Correct Answer)

A particle executes simple harmonic motion between x = - A to x = + A. The time taken for it to go from 0 to $${A \over 2}$$ is T_{1} and to go from $${A \over 2}$$ to A is T_{2}. Then

A

T_{1} < T_{2}

B

T_{1} > T_{2}

C

T_{1} = T_{2}

D

T_{1} = 2T_{2}

4

MCQ (Single Correct Answer)

The period of oscillation of a simple pendulum of length $$L$$ suspended from the roof of a vehicle which moves without friction down an inclined plane of inclination $$\alpha$$, is given by

A

$$2\pi \sqrt {{L \over {g\cos \alpha }}} $$

B

$$2\pi \sqrt {{L \over {g\sin \alpha }}} $$

C

$$2\pi \sqrt {{L \over g}} $$

D

$$2\pi \sqrt {{L \over {g\tan \alpha }}} $$

On those following papers in MCQ (Single Correct Answer)

Number in Brackets after Paper Indicates No. of Questions

IIT-JEE 2012 Paper 1 Offline (1)

IIT-JEE 2011 Paper 2 Offline (2)

IIT-JEE 2011 Paper 1 Offline (3)

IIT-JEE 2010 Paper 1 Offline (3)

IIT-JEE 2005 Screening (1)

IIT-JEE 2001 Screening (1)

IIT-JEE 2000 Screening (1)

IIT-JEE 1999 Screening (1)

IIT-JEE 1988 (1)

Units & Measurements

Motion

Laws of Motion

Work Power & Energy

Simple Harmonic Motion

Impulse & Momentum

Rotational Motion

Gravitation

Properties of Matter

Heat and Thermodynamics

Waves

Wave Optics

Geometrical Optics

Electrostatics

Current Electricity

Magnetism

Electromagnetic Induction

Alternating Current

Dual Nature of Radiation

Atoms and Nuclei