Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

When a rubber-band is stretched by a distance $$x$$, it exerts restoring force of magnitude $$F = ax + b{x^2}$$ where $$a$$ and $$b$$ are constants. The work done in stretching the unstretched rubber-band by $$L$$ is :

A

$$a{L^2} + b{L^3}$$

B

$${1 \over 2}\left( {a{L^2} + b{L^3}} \right)$$

C

$${{a{L^2}} \over 2} + {{b{L^3}} \over 3}$$

D

$${1 \over 2}\left( {{{a{L^2}} \over 2} + {{b{L^3}} \over 3}} \right)$$

Given Restoring force, F = ax + bx^{2}

Work done in stretching the rubber-band by a distance $$dx$$ is

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,dW = F\,dx = \left( {ax + b{x^2}} \right)dx$$

Intergrating both sides,

$$W = \int\limits_0^L {axdx + \int\limits_0^L {b{x^2}dx}}$$

= $$\left[ {a{{{x^2}} \over 2} + b{{{x^3}} \over 3}} \right]_0^L$$

= $${{a{L^2}} \over 2} + {{b{L^3}} \over 3}$$

Work done in stretching the rubber-band by a distance $$dx$$ is

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,dW = F\,dx = \left( {ax + b{x^2}} \right)dx$$

Intergrating both sides,

$$W = \int\limits_0^L {axdx + \int\limits_0^L {b{x^2}dx}}$$

= $$\left[ {a{{{x^2}} \over 2} + b{{{x^3}} \over 3}} \right]_0^L$$

= $${{a{L^2}} \over 2} + {{b{L^3}} \over 3}$$

2

MCQ (Single Correct Answer)

This question has Statement $$1$$ and Statement $$2.$$ Of the four choices given after the Statements, choose the one that best describes the two Statements.

If two springs $${S_1}$$ and $${S_2}$$ of force constants $${k_1}$$ and $${k_2}$$, respectively, are stretched by the same force, it is found that more work is done on spring $${S_1}$$ than on spring $${S_2}$$.
**STATEMENT 1:** If stretched by the same amount work done on $${S_1}$$, Work done on $${S_1}$$ is more than $${S_2}$$
**STATEMENT 2:** $${k_1} < {k_2}$$

A

Statement 1 is false, Statement 2 is true

B

Statement 1 is true, Statement 2 is false

C

Statement 1 is true, Statement 2 is true, Statement 2 is the correct explanation for Statement 1

D

Statement 1 is true, Statement 2 is true, Statement 2 is not the correct explanation for Statement 1

We know force (F) = kx

$$W = {1 \over 2}k{x^2}$$

$$W =$$ $${{{{\left( {kx} \right)}^2}} \over {2k}}$$ $$\,\,\,$$

$$\therefore$$ $$W = {{{F^2}} \over {2k}}$$ [ as $$F=kx$$ ]

When force is same then,

$$W \propto {1 \over k}$$

Given that, $${W_1} > {W_2}$$

$$\therefore$$ $${k_1} < {k_2}$$

**Statement-2 is true.**

For the same extension, x_{1}
= x_{2}
= x

Work done on spring S_{1} is W_{1} = $${1 \over 2}{k_1}x_1^2 = {1 \over 2}{k_1}{x^2}$$

Work done on spring S_{2} is W_{2} = $${1 \over 2}{k_2}x_2^2 = {1 \over 2}{k_2}{x^2}$$

$$ \therefore $$ $${{{W_1}} \over {{W_2}}} = {{{k_1}} \over {{k_2}}}$$

As $${k_1} < {k_2}$$ then $${W_1} < {W_2}$$

**So, Statement-1 is false.**

$$W = {1 \over 2}k{x^2}$$

$$W =$$ $${{{{\left( {kx} \right)}^2}} \over {2k}}$$ $$\,\,\,$$

$$\therefore$$ $$W = {{{F^2}} \over {2k}}$$ [ as $$F=kx$$ ]

When force is same then,

$$W \propto {1 \over k}$$

Given that, $${W_1} > {W_2}$$

$$\therefore$$ $${k_1} < {k_2}$$

For the same extension, x

Work done on spring S

Work done on spring S

$$ \therefore $$ $${{{W_1}} \over {{W_2}}} = {{{k_1}} \over {{k_2}}}$$

As $${k_1} < {k_2}$$ then $${W_1} < {W_2}$$

3

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

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

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

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

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

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

4

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

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

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