Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

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

2

MCQ (Single Correct Answer)

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

3

MCQ (Single Correct Answer)

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^{â€“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

4

MCQ (Single Correct Answer)

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

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