Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

The potential energy function for the force between two atoms in a diatomic molecule is approximately given by $$U\left( x \right) = {a \over {{x^{12}}}} - {b \over {{x^6}}},$$ where $$a$$ and $$b$$ are constants and $$x$$ is the distance between the atoms. If the dissociation energy of the molecule is $$D = \left[ {U\left( {x = \infty } \right) - {U_{at\,\,equilibrium}}} \right],\,\,D$$ is

A

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

B

$${{{b^2}} \over {12a}}$$

C

$${{{b^2}} \over {4a}}$$

D

$${{{b^2}} \over {6a}}$$

Given $$U\left( x \right) = {a \over {{x^{12}}}} - {b \over {{x^6}}}$$

$${U\left( {x = \infty } \right)}$$ = 0

We know $$F = - {{dU} \over {dx}} = - \left[ {{{12a} \over {{x^{13}}}} + {{6b} \over {{x^7}}}} \right]$$

At equilibrium: $${{dU\left( x \right)} \over {dx}} = 0$$

$$ \Rightarrow {{ - 12a} \over {{x^{13}}}} = {{ - 6b} \over {{x^7}}} $$

$$\Rightarrow x = {\left( {{{2a} \over h}} \right)^{{1 \over 6}}}$$

$$\therefore$$ $${U_{at\,\,equilibrium\,}} = {a \over {{{\left( {{{2a} \over b}} \right)}^2}}} - {b \over {\left( {{{2a} \over b}} \right)}}$$

$$ = - {{{b^2}} \over {4a}}$$

$$\therefore$$ $$D = 0 - \left( { - {{{b^2}} \over {4a}}} \right) = {{{b^2}} \over {4a}}$$

$${U\left( {x = \infty } \right)}$$ = 0

We know $$F = - {{dU} \over {dx}} = - \left[ {{{12a} \over {{x^{13}}}} + {{6b} \over {{x^7}}}} \right]$$

At equilibrium: $${{dU\left( x \right)} \over {dx}} = 0$$

$$ \Rightarrow {{ - 12a} \over {{x^{13}}}} = {{ - 6b} \over {{x^7}}} $$

$$\Rightarrow x = {\left( {{{2a} \over h}} \right)^{{1 \over 6}}}$$

$$\therefore$$ $${U_{at\,\,equilibrium\,}} = {a \over {{{\left( {{{2a} \over b}} \right)}^2}}} - {b \over {\left( {{{2a} \over b}} \right)}}$$

$$ = - {{{b^2}} \over {4a}}$$

$$\therefore$$ $$D = 0 - \left( { - {{{b^2}} \over {4a}}} \right) = {{{b^2}} \over {4a}}$$

2

MCQ (Single Correct Answer)

An athlete in the olympic games covers a distance of $$100$$ $$m$$ in $$10$$ $$s.$$ His kinetic energy can be estimated to be in the range

A

$$200J-500J$$

B

$$2 \times {10^5}J - 3 \times {10^5}J$$

C

$$20,000J - 50,000J$$

D

$$2,000J - 5,000J$$

The average speed of the athelete

$$v = {{100} \over {10}} = 10m/s\,\,\,\,$$ $$\therefore$$ $$K.E. = {1 \over 2}m{v^2}$$

If mass of athlete is $$40$$ $$kg$$ then, $$K.E.$$ $$ = {1 \over 2} \times 40 \times {\left( {10} \right)^2} = 2000J$$

If mass of athlete is $$100$$ $$kg$$ then, $$K.E.$$ $$ = {1 \over 2} \times 100 \times {\left( {10} \right)^2} = 5000J$$

His kinetic energy can be in the range = 2000 J to 5000 J.

$$v = {{100} \over {10}} = 10m/s\,\,\,\,$$ $$\therefore$$ $$K.E. = {1 \over 2}m{v^2}$$

If mass of athlete is $$40$$ $$kg$$ then, $$K.E.$$ $$ = {1 \over 2} \times 40 \times {\left( {10} \right)^2} = 2000J$$

If mass of athlete is $$100$$ $$kg$$ then, $$K.E.$$ $$ = {1 \over 2} \times 100 \times {\left( {10} \right)^2} = 5000J$$

His kinetic energy can be in the range = 2000 J to 5000 J.

3

MCQ (Single Correct Answer)

A block of mass $$0.50$$ $$kg$$ is moving with a speed of $$2.00$$ $$m{s^{ - 1}}$$ on a smooth surface. It strike another mass of $$1.0$$ $$kg$$ and then they move together as a simple body. The energy loss during the collision is

A

$$0.16J$$

B

$$1.00J$$

C

$$0.67J$$

D

$$0.34$$ $$J$$

Let $$m$$ = 0.50 kg and $$M$$ = 1.0 kg

Initial kinetic energy of the system when 1 kg mass is at rest,

$$K.{E_i} = {1 \over 2}m{u^2} + {1 \over 2}M{\left( 0 \right)^2}$$

$$ = {1 \over 2} \times 0.5 \times 2 \times 2 + 0 = 1J$$

For collision, applying conservation of linear momentum

$$\,\,\,\,\,\,\,\,\,\,\,\,m \times u = \left( {m + M} \right) \times v$$

$$\therefore$$ $$0.5 \times 2 = \left( {0.5 + 1} \right) \times v \Rightarrow v = {2 \over 3}m/s$$

Final kinetic energy of the system is

$$K.{E_f} = {1 \over 2}\left( {m + M} \right){v^2}$$

$$ = {1 \over 2}\left( {0.5 + 1} \right) \times {2 \over 3} \times {2 \over 3} = {1 \over 3}J$$

$$\therefore$$ Energy loss during collision

$$ = \left( {1 - {1 \over 3}} \right)J = 0.67J$$

Initial kinetic energy of the system when 1 kg mass is at rest,

$$K.{E_i} = {1 \over 2}m{u^2} + {1 \over 2}M{\left( 0 \right)^2}$$

$$ = {1 \over 2} \times 0.5 \times 2 \times 2 + 0 = 1J$$

For collision, applying conservation of linear momentum

$$\,\,\,\,\,\,\,\,\,\,\,\,m \times u = \left( {m + M} \right) \times v$$

$$\therefore$$ $$0.5 \times 2 = \left( {0.5 + 1} \right) \times v \Rightarrow v = {2 \over 3}m/s$$

Final kinetic energy of the system is

$$K.{E_f} = {1 \over 2}\left( {m + M} \right){v^2}$$

$$ = {1 \over 2}\left( {0.5 + 1} \right) \times {2 \over 3} \times {2 \over 3} = {1 \over 3}J$$

$$\therefore$$ Energy loss during collision

$$ = \left( {1 - {1 \over 3}} \right)J = 0.67J$$

4

MCQ (Single Correct Answer)

A $$2$$ $$kg$$ block slides on a horizontal floor with a speed of $$4m/s.$$ It strikes a uncompressed spring, and compress it till the block is motionless. The kinetic friction force is $$15N$$ and spring constant is $$10, 000$$ $$N/m.$$ The spring compresses by

A

$$8.5cm$$

B

$$5.5cm$$

C

$$2.5cm$$

D

$$11.0cm$$

Let the block compress the spring by $$x$$ before coming to rest.

Initial kinetic energy of the block $$=$$ (potential energy of compressed spring) $$+$$ work done due to friction.

$${1 \over 2} \times 2 \times {\left( 4 \right)^2} = {1 \over 2} \times 10000 \times {x^2} + 15 \times x$$

$$10,000{x^2} + 30x - 32 = 0$$

$$ \Rightarrow 5000{x^2} + 15x - 16 = 0$$

$$\therefore$$ $$x = {{ - 15 \pm \sqrt {{{\left( {15} \right)}^2} - 4 \times \left( {5000} \right)\left( { - 16} \right)} } \over {2 \times 5000}}$$

$$\,\,\,\,\, = 0.055m = 5.5cm.$$

Initial kinetic energy of the block $$=$$ (potential energy of compressed spring) $$+$$ work done due to friction.

$${1 \over 2} \times 2 \times {\left( 4 \right)^2} = {1 \over 2} \times 10000 \times {x^2} + 15 \times x$$

$$10,000{x^2} + 30x - 32 = 0$$

$$ \Rightarrow 5000{x^2} + 15x - 16 = 0$$

$$\therefore$$ $$x = {{ - 15 \pm \sqrt {{{\left( {15} \right)}^2} - 4 \times \left( {5000} \right)\left( { - 16} \right)} } \over {2 \times 5000}}$$

$$\,\,\,\,\, = 0.055m = 5.5cm.$$

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