Joint Entrance Examination

Graduate Aptitude Test in Engineering

1

MCQ (Single Correct Answer)

The motion of a mass on a spring, with spring constant K is as shown in figure.

The equation of motion is given by

x(t) = A sin$$\omega$$t + B cos$$\omega$$t with $$\omega$$ = $$\sqrt {{K \over m}} $$

Suppose that at time t = 0, the position of mass is x(0) and velocity v(0), then its displacement can also be represented as x(t) = C cos($$\omega$$t $$-$$ $$\phi$$), where C and $$\phi$$ are :

The equation of motion is given by

x(t) = A sin$$\omega$$t + B cos$$\omega$$t with $$\omega$$ = $$\sqrt {{K \over m}} $$

Suppose that at time t = 0, the position of mass is x(0) and velocity v(0), then its displacement can also be represented as x(t) = C cos($$\omega$$t $$-$$ $$\phi$$), where C and $$\phi$$ are :

A

$$C = \sqrt {{{2v{{(0)}^2}} \over {{\omega ^2}}} + x{{(0)}^2}} ,\phi = {\tan ^{ - 1}}\left( {{{x(0)\omega } \over {2v(0)}}} \right)$$

B

$$C = \sqrt {{{v{{(0)}^2}} \over {{\omega ^2}}} + x{{(0)}^2}} ,\phi = {\tan ^{ - 1}}\left( {{{x(0)\omega } \over {v(0)}}} \right)$$

C

$$C = \sqrt {{{v{{(0)}^2}} \over {{\omega ^2}}} + x{{(0)}^2}} ,\phi = {\tan ^{ - 1}}\left( {{{v(0)} \over {x(0)\omega }}} \right)$$

D

$$C = \sqrt {{{2v{{(0)}^2}} \over {{\omega ^2}}} + x{{(0)}^2}} ,\phi = {\tan ^{ - 1}}\left( {{{v(0)} \over {x(0)\omega }}} \right)$$

$$C\cos \phi = x(0)$$

$$tC\omega \sin \phi = v(0)$$

$${\left[ {{{v(0)} \over \omega }} \right]^2} + {[x(0)]^2} = {C^2}$$

$$\tan \phi = {{v(0)} \over {x(0)\omega }}$$

$$tC\omega \sin \phi = v(0)$$

$${\left[ {{{v(0)} \over \omega }} \right]^2} + {[x(0)]^2} = {C^2}$$

$$\tan \phi = {{v(0)} \over {x(0)\omega }}$$

2

MCQ (Single Correct Answer)

A sound wave of frequency 245 Hz travels with the speed of 300 ms^{$$-$$1} along the positive x-axis. Each point of the wave moves to and from through a total distance of 6 cm. What will be the mathematical expression of this travelling wave?

A

Y(x, t) = 0.03 [ sin 5.1x $$-$$ (0.2 $$\times$$ 10^{3})t ]

B

Y(x, t) = 0.03 [ sin 5.1x $$-$$ (1.5 $$\times$$ 10^{3})t ]

C

Y(x, t) = 0.06 [ sin 5.1x $$-$$ (1.5 $$\times$$ 10^{3})t ]

D

Y(x, t) = 0.06 [ sin 0.8x $$-$$ (0.5 $$\times$$ 10^{3})t ]

$$Y = A\sin (kx - \omega t)$$

$$A = {6 \over 2}$$ = 3cm = 0.03 m

$$\omega = 2\pi f = 2\pi \times 245$$

$$\omega = 1.5 \times {10^3}$$

$$k = {\omega \over v} = {{1.5 \times {{10}^3}} \over {300}}$$

$$k = 5.1$$

$$y = 0.03\sin (5.1x - (1.5 \times {10^3})t)$$

$$A = {6 \over 2}$$ = 3cm = 0.03 m

$$\omega = 2\pi f = 2\pi \times 245$$

$$\omega = 1.5 \times {10^3}$$

$$k = {\omega \over v} = {{1.5 \times {{10}^3}} \over {300}}$$

$$k = 5.1$$

$$y = 0.03\sin (5.1x - (1.5 \times {10^3})t)$$

3

MCQ (Single Correct Answer)

For what value of displacement the kinetic energy and potential energy of a simple harmonic oscillation become equal ?

A

x = $${A \over 2}$$

B

x = $$\pm$$ A

C

x = $$\pm$$ $${A \over {\sqrt 2 }}$$

D

x = 0

KE = PE

$${1 \over 2}k({A^2} - {X^2}) = {1 \over 2}K{X^2}$$

$$ \Rightarrow $$ $${A^2} - {X^2} = {X^2}$$

$$ \Rightarrow $$ $$2{X^2} = {A^2}$$

$$ \Rightarrow $$ $${X^2} = {{{A^2}} \over {\sqrt 2 }}$$

$$ \Rightarrow $$ $$X = \pm {A \over {\sqrt 2 }}$$

$${1 \over 2}k({A^2} - {X^2}) = {1 \over 2}K{X^2}$$

$$ \Rightarrow $$ $${A^2} - {X^2} = {X^2}$$

$$ \Rightarrow $$ $$2{X^2} = {A^2}$$

$$ \Rightarrow $$ $${X^2} = {{{A^2}} \over {\sqrt 2 }}$$

$$ \Rightarrow $$ $$X = \pm {A \over {\sqrt 2 }}$$

4

MCQ (Single Correct Answer)

A tuning fork A of unknown frequency produces 5 beats/s with a fork of known frequency 340 Hz. When fork A is field, the beat frequency decreases to 2 beats/s. What is the frequency of fork A?

A

335 Hz

B

345 Hz

C

338 Hz

D

342 Hz

Initially beat frequency = 5Hz

so, $$\rho$$_{A} = 340 $$ \pm $$ 5 = 345 Hz, or 335 Hz

after filing frequency increases slightly so, new value of frequency of A > $$\rho$$_{A}

Now, beat frequency = 2Hz

$$ \Rightarrow $$ new $$\rho$$_{A} = 340 $$ \pm $$ 2 = 342 Hz, or 338 Hz

hence, original frequency of A is $$\rho$$_{A} = 335 Hz

so, $$\rho$$

after filing frequency increases slightly so, new value of frequency of A > $$\rho$$

Now, beat frequency = 2Hz

$$ \Rightarrow $$ new $$\rho$$

hence, original frequency of A is $$\rho$$

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Units & Measurements

Motion

Laws of Motion

Work Power & Energy

Simple Harmonic Motion

Impulse & Momentum

Rotational Motion

Gravitation

Properties of Matter

Heat and Thermodynamics

Waves

Vector Algebra

Ray & Wave Optics

Electrostatics

Current Electricity

Magnetics

Alternating Current and Electromagnetic Induction

Atoms and Nuclei

Dual Nature of Radiation

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