Joint Entrance Examination

Graduate Aptitude Test in Engineering

Strength of Materials Or Solid Mechanics

Structural Analysis

Construction Material and Management

Reinforced Cement Concrete

Steel Structures

Geotechnical Engineering

Fluid Mechanics and Hydraulic Machines

Hydrology

Irrigation

Geomatics Engineering Or Surveying

Environmental Engineering

Transportation Engineering

Engineering Mathematics

General Aptitude

1

If the Earth has no rotational motion, the weight of a person on the equator is W. Determine the speed with which the earth would have to rotate about its axis so that the person at the equator will weigh $${3 \over 4}$$ W. Radius of the Earth is 6400 km and g=10 m/s^{2}.

A

1.1 $$ \times $$ 10^{−3} rad/s

B

0.83 $$ \times $$ 10^{−3} rad/s

C

0.63 $$ \times $$ 10^{−3} rad/s

D

0.28 $$ \times $$ 10^{−3} rad/s

Initially when earth is not rotating then weight of the person is $$w$$.

When earth rotares about it's axis then weight = $${{3\omega } \over 4}$$

Then,

g' = g $$-$$ $$\omega $$^{2}R cos^{2}$$\theta $$

$$ \Rightarrow $$ $$\,\,\,$$ $${{3g} \over 4}$$ = g $$-$$ $$\omega $$^{2}R cos^{2} 0^{o}

$$ \Rightarrow $$ $$\,\,\,$$ $$\omega $$^{2}R = $${g \over 4}$$

$$\,\,\,$$ $$\omega $$ = $$\sqrt {{g \over {4R}}} $$

$$ \Rightarrow $$ $$\omega = \sqrt {{{10} \over {4 \times 6400 \times {{10}^3}}}} $$

= 0.63 $$ \times $$ 10^{$$-$$3} rad/s

When earth rotares about it's axis then weight = $${{3\omega } \over 4}$$

Then,

g' = g $$-$$ $$\omega $$

$$ \Rightarrow $$ $$\,\,\,$$ $${{3g} \over 4}$$ = g $$-$$ $$\omega $$

$$ \Rightarrow $$ $$\,\,\,$$ $$\omega $$

$$\,\,\,$$ $$\omega $$ = $$\sqrt {{g \over {4R}}} $$

$$ \Rightarrow $$ $$\omega = \sqrt {{{10} \over {4 \times 6400 \times {{10}^3}}}} $$

= 0.63 $$ \times $$ 10

2

The mass density of a spherical body is given by

$$\rho $$ (r) = $${k \over r}$$ for r $$ \le $$ R and $$\rho $$ (r) = 0 for r > R,

where r is the distance from the centre.

The correct graph that describes qualitatively the acceleration, a, of a test particle as a function of r is :

$$\rho $$ (r) = $${k \over r}$$ for r $$ \le $$ R and $$\rho $$ (r) = 0 for r > R,

where r is the distance from the centre.

The correct graph that describes qualitatively the acceleration, a, of a test particle as a function of r is :

A

B

C

D

3

A particle is moving with a uniform speed in a circular orbit of radius R in a central force inversely
proportional to the n^{th} power of R. If the period of rotation of the particle is T, then :

A

T $$ \propto $$ R^{n/2}

B

T $$ \propto $$ R^{3/2} for any n

C

T $$ \propto $$ R^{n/2 +1}

D

T $$ \propto $$ R^{(n+1)/2}

We know, Central force in circular motion, F = $$m{\omega ^2}R$$

According to the question,

$$F \propto {1 \over {{R^n}}}$$

$$\therefore$$ $$m{\omega ^2}R$$ $$ \propto {1 \over {{R^n}}}$$

$$ \Rightarrow m{\omega ^2}R = {k \over {{R^n}}}$$

$$ \Rightarrow {\omega ^2} = {k \over {m{R^{n + 1}}}}$$

$$\therefore$$ $$\omega \propto {1 \over {{R^{{{n + 1} \over 2}}}}}$$ .......(1)

And we know, $$T = {{2\pi } \over \omega }$$

$$\therefore$$ $$T \propto {1 \over \omega }$$ ...... (2)

From (1) and (2) we can conclude that,

$$T \propto {R^{{{n + 1} \over 2}}}$$

According to the question,

$$F \propto {1 \over {{R^n}}}$$

$$\therefore$$ $$m{\omega ^2}R$$ $$ \propto {1 \over {{R^n}}}$$

$$ \Rightarrow m{\omega ^2}R = {k \over {{R^n}}}$$

$$ \Rightarrow {\omega ^2} = {k \over {m{R^{n + 1}}}}$$

$$\therefore$$ $$\omega \propto {1 \over {{R^{{{n + 1} \over 2}}}}}$$ .......(1)

And we know, $$T = {{2\pi } \over \omega }$$

$$\therefore$$ $$T \propto {1 \over \omega }$$ ...... (2)

From (1) and (2) we can conclude that,

$$T \propto {R^{{{n + 1} \over 2}}}$$

4

A particle is moving in a circular path of radius $$a$$ under the action of an attractive potential $$U = - {k \over {2{r^2}}}$$ Its total energy is:

A

$$ - {3 \over 2}{k \over {{a^2}}}$$

B

Zero

C

$$ - {k \over {4{a^2}}}$$

D

$$ {k \over {2{a^2}}}$$

We know, Total energy = Kinetic energy + Potential energy

Potential energy given as $$U = - {k \over {2{r^2}}}$$

We need to find Kinetic Energy.

As Force acting on the particle (F) = $$ - {{dU} \over {dr}}$$

$$ \Rightarrow F = - {d \over {dr}}\left( {{{ - k} \over {2{r^2}}}} \right)$$

$$= {k \over 2} \times \left( { - 2} \right) \times {r^{ - 3}}$$

$$ = - {k \over {{r^3}}}$$

Because of this force particle is having circular motion so it will provide possible centripetal force.

$$\left| F \right| = {{m{v^2}} \over r}$$

$$ \Rightarrow {{m{v^2}} \over r} = {k \over {{r^3}}}$$

$$ \Rightarrow $$ $$m{v^2} = {k \over {{r^2}}}$$

We know kinetic energy of particle, K = $${1 \over 2}m{v^2}$$ = $${k \over {2{r^2}}}$$

As Total energy = Kinetic energy + Potential energy

So Total energy = $${k \over {2{r^2}}}$$ $$ - {k \over {2{r^2}}}$$ = 0

Potential energy given as $$U = - {k \over {2{r^2}}}$$

We need to find Kinetic Energy.

As Force acting on the particle (F) = $$ - {{dU} \over {dr}}$$

$$ \Rightarrow F = - {d \over {dr}}\left( {{{ - k} \over {2{r^2}}}} \right)$$

$$= {k \over 2} \times \left( { - 2} \right) \times {r^{ - 3}}$$

$$ = - {k \over {{r^3}}}$$

Because of this force particle is having circular motion so it will provide possible centripetal force.

$$\left| F \right| = {{m{v^2}} \over r}$$

$$ \Rightarrow {{m{v^2}} \over r} = {k \over {{r^3}}}$$

$$ \Rightarrow $$ $$m{v^2} = {k \over {{r^2}}}$$

We know kinetic energy of particle, K = $${1 \over 2}m{v^2}$$ = $${k \over {2{r^2}}}$$

As Total energy = Kinetic energy + Potential energy

So Total energy = $${k \over {2{r^2}}}$$ $$ - {k \over {2{r^2}}}$$ = 0

Number in Brackets after Paper Name Indicates No of Questions

AIEEE 2002 (4) *keyboard_arrow_right*

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Atoms and Nuclei *keyboard_arrow_right*

Electronic Devices *keyboard_arrow_right*

Communication Systems *keyboard_arrow_right*

Practical Physics *keyboard_arrow_right*

Dual Nature of Radiation *keyboard_arrow_right*

Units & Measurements *keyboard_arrow_right*

Motion *keyboard_arrow_right*

Laws of Motion *keyboard_arrow_right*

Work Power & Energy *keyboard_arrow_right*

Simple Harmonic Motion *keyboard_arrow_right*

Impulse & Momentum *keyboard_arrow_right*

Rotational Motion *keyboard_arrow_right*

Gravitation *keyboard_arrow_right*

Properties of Matter *keyboard_arrow_right*

Heat and Thermodynamics *keyboard_arrow_right*

Waves *keyboard_arrow_right*

Vector Algebra *keyboard_arrow_right*

Ray & Wave Optics *keyboard_arrow_right*

Electrostatics *keyboard_arrow_right*

Current Electricity *keyboard_arrow_right*

Magnetics *keyboard_arrow_right*

Alternating Current and Electromagnetic Induction *keyboard_arrow_right*