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

A particle which is experiencing a force, given by $$\overrightarrow F = 3\widehat i - 12\widehat j,$$ undergoes a displacement of $$\overrightarrow d = 4\overrightarrow i $$ particle had a kinetic energy of 3 J at the beginning of the displacement, what is its kinetic energy at the end of the displacement ?

A

9 J

B

10 J

C

12 J

D

15 J

Work done = $$\overrightarrow F \cdot \overrightarrow d $$

$$=$$ 12 J

work energy theorem

w_{net} $$=$$ $$\Delta $$K.E.

12 $$=$$ K_{f} $$-$$ 3

K_{f} = 15 J

$$=$$ 12 J

work energy theorem

w

12 $$=$$ K

K

2

A body of mass 1 kg falls freely from a height of 100 m, on a platform mass 3 kg which is mounted on a spring having spring constant k = 1.25 $$ \times $$ 10^{6} N/m. The body sticks to the platform and the spring's maximum compression is found to be x. Given that g = 10 ms^{–2}
, the value of x will be close to :

A

8 cm

B

4 cm

C

40 cm

D

80 cm

velocity of 1 kg block just before it collides with 3kg block

= $$\sqrt {2gh} = \sqrt {2000} $$ m/s

Applying momentum conversation just before and just after collision.

1 $$ \times $$ $$\sqrt {2000} $$ = 4v $$ \Rightarrow $$ v = $${{\sqrt {2000} } \over 4}$$ m/s

initial compression of spring

1.25 $$ \times $$ 10^{6} x_{0} = 30 $$ \Rightarrow $$ x_{0} $$ \approx $$ 0

applying work energy theorem,

W_{g} + W_{sp} = $$\Delta $$KE

$$ \Rightarrow $$ 40 $$ \times $$ x + $${1 \over 2}$$ $$ \times $$ 1.25 $$ \times $$ 10^{6} (0^{2} $$-$$ x^{2})

= 0 $$-$$ $${1 \over 2}$$ $$ \times $$ 4 $$ \times $$ v^{2}

solving x $$ \approx $$ 4 cm

= $$\sqrt {2gh} = \sqrt {2000} $$ m/s

Applying momentum conversation just before and just after collision.

1 $$ \times $$ $$\sqrt {2000} $$ = 4v $$ \Rightarrow $$ v = $${{\sqrt {2000} } \over 4}$$ m/s

initial compression of spring

1.25 $$ \times $$ 10

applying work energy theorem,

W

$$ \Rightarrow $$ 40 $$ \times $$ x + $${1 \over 2}$$ $$ \times $$ 1.25 $$ \times $$ 10

= 0 $$-$$ $${1 \over 2}$$ $$ \times $$ 4 $$ \times $$ v

solving x $$ \approx $$ 4 cm

3

For the given cyclic process CAB as shown for a gas, the work done is :

A

1 J

B

10 J

C

5 J

D

30 J

Since P$$-$$V indicator diagram is given, so work done by gas is area under the cyclic diagram.

$$ \therefore $$ $$\Delta $$W = Work done by gas = $${1 \over 2}$$ $$ \times $$ 4 $$ \times $$ 5 J

= 10 J

$$ \therefore $$ $$\Delta $$W = Work done by gas = $${1 \over 2}$$ $$ \times $$ 4 $$ \times $$ 5 J

= 10 J

4

There is a uniform spherically symmetric surface charge density at a distance R_{0} from the origin. The charge distribution is initially at rest and starts expanding because of mutual repulsion. The figure that represents best the speed V (R(t)) of the distribution as a function of its instantaneous radius R (t) is :

A

B

C

D

At any instant 't'

Total energy of charge distribution is constant

i.e. $${1 \over 2}m{V^2} + {{K{Q^2}} \over {2R}} = 0 + {{K{Q^2}} \over {2{R_0}}}$$

$$ \therefore $$ $${1 \over 2}m{V^2} = {{K{Q^2}} \over {2{R_0}}} - {{K{Q^2}} \over {2R}}$$

$$ \therefore $$ V = $$\sqrt {{2 \over m}{{K{Q^2}} \over 2}.\left( {{1 \over {{R_0}}} - {1 \over R}} \right)} $$

$$ \therefore $$ V = $$\sqrt {{{K{Q^2}} \over m}\left( {{1 \over {{R_0}}} - {1 \over R}} \right)} $$ = C$$\sqrt {{1 \over {{R_0}}} - {1 \over R}} $$

Also the slope of v-s curve will go on decreasing

Total energy of charge distribution is constant

i.e. $${1 \over 2}m{V^2} + {{K{Q^2}} \over {2R}} = 0 + {{K{Q^2}} \over {2{R_0}}}$$

$$ \therefore $$ $${1 \over 2}m{V^2} = {{K{Q^2}} \over {2{R_0}}} - {{K{Q^2}} \over {2R}}$$

$$ \therefore $$ V = $$\sqrt {{2 \over m}{{K{Q^2}} \over 2}.\left( {{1 \over {{R_0}}} - {1 \over R}} \right)} $$

$$ \therefore $$ V = $$\sqrt {{{K{Q^2}} \over m}\left( {{1 \over {{R_0}}} - {1 \over R}} \right)} $$ = C$$\sqrt {{1 \over {{R_0}}} - {1 \over R}} $$

Also the slope of v-s curve will go on decreasing

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