Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

A pendulum made of a uniform wire of cross sectional area $$A$$ has time period $$T.$$ When an additional mass $$M$$ is added to its bob, the time period changes to $${T_{M.}}$$ If the Young's modulus of the material of the wire is $$Y$$ then $${1 \over Y}$$ is equal to :

($$g=$$ $$gravitational$$ $$acceleration$$)

($$g=$$ $$gravitational$$ $$acceleration$$)

A

$$\left[ {1 - {{\left( {{{{T_M}} \over T}} \right)}^2}} \right]{A \over {Mg}}$$

B

$$\left[ {1 - {{\left( {{T \over {{T_M}}}} \right)}^2}} \right]{A \over {Mg}}$$

C

$$\left[ {{{\left( {{{{T_M}} \over T}} \right)}^2} - 1} \right]{A \over {Mg}}$$

D

$$\left[ {{{\left( {{{{T_M}} \over T}} \right)}^2} - 1} \right]{{Mg} \over A}$$

As we know, time period, $$T = 2\pi \sqrt {{\ell \over g}} $$

When a additional mass $$M$$ is added then

$${T_M} = 2\pi \sqrt {{{\ell + \Delta \ell } \over g}} $$

$${{{T_M}} \over T} = \sqrt {{{\ell + \Delta \ell } \over \ell }} $$

or, $$\,\,{\left( {{{{T_M}} \over T}} \right)^2} = {{\ell + \Delta \ell } \over \ell }$$

or, $$\,\,{\left( {{{{T_M}} \over T}} \right)^2} = 1 + {{Mg} \over {Ay}}$$

$$\left[ \, \right.$$ as $$\left. {\Delta \ell = {{Mg\ell } \over {Ay}}\,} \right]$$

$$\therefore$$ $${1 \over y} = \left[ {{{\left( {{{{T_M}} \over T}} \right)}^2} - 1} \right]{A \over {Mg}}$$

When a additional mass $$M$$ is added then

$${T_M} = 2\pi \sqrt {{{\ell + \Delta \ell } \over g}} $$

$${{{T_M}} \over T} = \sqrt {{{\ell + \Delta \ell } \over \ell }} $$

or, $$\,\,{\left( {{{{T_M}} \over T}} \right)^2} = {{\ell + \Delta \ell } \over \ell }$$

or, $$\,\,{\left( {{{{T_M}} \over T}} \right)^2} = 1 + {{Mg} \over {Ay}}$$

$$\left[ \, \right.$$ as $$\left. {\Delta \ell = {{Mg\ell } \over {Ay}}\,} \right]$$

$$\therefore$$ $${1 \over y} = \left[ {{{\left( {{{{T_M}} \over T}} \right)}^2} - 1} \right]{A \over {Mg}}$$

2

MCQ (Single Correct Answer)

A particle moves with simple harmonic motion in a straight line. In first $$\tau s,$$ after starting from rest it travels a distance $$a,$$ and in next $$\tau s$$ it travels $$2a,$$ in same direction, then:

A

amplitude of motion is $$3a$$

B

time period of oscillations is $$8\tau $$

C

amplitude of motion is $$4a$$

D

time period of oscillations is $$6\tau $$

In simple harmonic motion, starting from rest,

At $$t=0,$$ $$x=A$$

$$x = A\cos \omega t\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$

When $$t = \tau ,\,\,x = A - a$$

When $$t = 2\,\tau ,\,x = A - 3a$$

From equation $$(i)$$

$$A - a = A\cos \omega \,\tau \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$

$$A - 3a = A\cos 2\omega \,\tau \,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {iii} \right)$$

As $$\cos 2\omega \,\tau = 2{\cos ^2}\omega \tau - 1...\left( {iv} \right)$$

From equation $$(ii),$$ $$(iii)$$ and $$(iv)$$

$${{A - 3A} \over A} = 2{\left( {{{A - a} \over A}} \right)^2} - 1$$

$$ \Rightarrow {{A - 3a} \over A} = {{2{A^2} + 2{a^2} - 4Aa - {A^2}} \over {{A^2}}}$$

$$ \Rightarrow {A^2} - 3aA = {A^2} + 2{a^2} - 4Aa$$

$$ \Rightarrow 2{a^2} = aA \Rightarrow \,\,\,\,\,\,\,A = 2a$$

$$ \Rightarrow {a \over A} = {1 \over 2}$$

Now, $$A-a=A$$ $$\cos \omega \tau $$

$$ \Rightarrow \cos \omega \tau = {{A - a} \over A} \Rightarrow \,\,\cos \omega \tau = {1 \over 2}$$

or, $${{2\pi } \over T}\tau = {\pi \over 3} \Rightarrow \,\,\,T - 6\,\tau $$

At $$t=0,$$ $$x=A$$

$$x = A\cos \omega t\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$

When $$t = \tau ,\,\,x = A - a$$

When $$t = 2\,\tau ,\,x = A - 3a$$

From equation $$(i)$$

$$A - a = A\cos \omega \,\tau \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$

$$A - 3a = A\cos 2\omega \,\tau \,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {iii} \right)$$

As $$\cos 2\omega \,\tau = 2{\cos ^2}\omega \tau - 1...\left( {iv} \right)$$

From equation $$(ii),$$ $$(iii)$$ and $$(iv)$$

$${{A - 3A} \over A} = 2{\left( {{{A - a} \over A}} \right)^2} - 1$$

$$ \Rightarrow {{A - 3a} \over A} = {{2{A^2} + 2{a^2} - 4Aa - {A^2}} \over {{A^2}}}$$

$$ \Rightarrow {A^2} - 3aA = {A^2} + 2{a^2} - 4Aa$$

$$ \Rightarrow 2{a^2} = aA \Rightarrow \,\,\,\,\,\,\,A = 2a$$

$$ \Rightarrow {a \over A} = {1 \over 2}$$

Now, $$A-a=A$$ $$\cos \omega \tau $$

$$ \Rightarrow \cos \omega \tau = {{A - a} \over A} \Rightarrow \,\,\cos \omega \tau = {1 \over 2}$$

or, $${{2\pi } \over T}\tau = {\pi \over 3} \Rightarrow \,\,\,T - 6\,\tau $$

3

MCQ (Single Correct Answer)

An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass $$M.$$ The piston and the cylinder have equal cross sectional area $$A$$. When the piston is in equilibrium, the volume of the gas is $${V_0}$$ and its pressure is $${P_0}.$$ The piston is slightly displaced from the equilibrium position and released,. Assuming that the system is completely isolated from its surrounding, the piston executes a simple harmonic motion with frquency

A

$${1 \over {2\pi }}\,{{A\gamma {P_0}} \over {{V_0}M}}$$

B

$${1 \over {2\pi }}\,{{{V_0}M{P_0}} \over {{A^2}\gamma }}$$

C

$${1 \over {2\pi }}\,\sqrt {{{A\gamma {P_0}} \over {{V_0}M}}} $$

D

$${1 \over {2\pi }}\,\sqrt {{{M{V_0}} \over {A\gamma {P_0}}}} $$

$${{Mg} \over A} = {P_0}$$

$$Mg = {P_0}A\,\,\,\,...\left( 1 \right)$$

$${P_0}V_0^\gamma = P{V^\gamma }$$

$$P = {{{P_0}x_0^\gamma } \over {{{\left( {{x_0} - x} \right)}^y}}}$$

Let piston is displaced by distance $$x$$

$$Mg - \left( {{{{P_0}x_0^\gamma } \over {{{\left( {{x_0} - x} \right)}^\gamma }}}} \right)A = {F_{restoring}}$$

$${P_0}A\left( {1 - {{x_0^\gamma } \over {{{\left( {{x_0} - x} \right)}^\gamma }}}} \right) = {F_{restoring}}$$

$$\left[ {{x_0} - x \approx {x_0}} \right]$$

$$F = - {{\gamma {P_0}Ax} \over {{x_0}}}$$

$$\therefore$$ Frequency with which piston executes $$SHM.$$

$$f = {1 \over {2\pi }}\sqrt {{{\gamma {P_0}A} \over {{x_0}M}}} = {1 \over {2\pi }}\sqrt {{{\gamma {P_0}{A^2}} \over {M{V_0}}}} $$

$$Mg = {P_0}A\,\,\,\,...\left( 1 \right)$$

$${P_0}V_0^\gamma = P{V^\gamma }$$

$$P = {{{P_0}x_0^\gamma } \over {{{\left( {{x_0} - x} \right)}^y}}}$$

Let piston is displaced by distance $$x$$

$$Mg - \left( {{{{P_0}x_0^\gamma } \over {{{\left( {{x_0} - x} \right)}^\gamma }}}} \right)A = {F_{restoring}}$$

$${P_0}A\left( {1 - {{x_0^\gamma } \over {{{\left( {{x_0} - x} \right)}^\gamma }}}} \right) = {F_{restoring}}$$

$$\left[ {{x_0} - x \approx {x_0}} \right]$$

$$F = - {{\gamma {P_0}Ax} \over {{x_0}}}$$

$$\therefore$$ Frequency with which piston executes $$SHM.$$

$$f = {1 \over {2\pi }}\sqrt {{{\gamma {P_0}A} \over {{x_0}M}}} = {1 \over {2\pi }}\sqrt {{{\gamma {P_0}{A^2}} \over {M{V_0}}}} $$

4

MCQ (Single Correct Answer)

The amplitude of a damped oscillator decreases to $$0.9$$ times its original magnitude in $$5s$$. In another $$10s$$ it will decrease to $$\alpha $$ times its original magnitude, where $$\alpha $$ equals

A

$$0.7$$

B

$$0.81$$

C

$$0.729$$

D

$$0.6$$

as $$\,\,A = {A_0}{e^{{{bt} \over {2m}}}}$$ (where, $${A_0} = $$ maximum amplitude)

According to the questions, after $$5$$ second,

$$0.9{A_0} = {A_0}{e^{ - {{b\left( 5 \right)} \over {2m}}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$

After $$10$$ more second,

$$A = {A_0}{e^{ - {{b\left( {15} \right)} \over {2m}}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$

From eqns $$(i)$$ and $$(ii)$$

$$A = 0.729\,{A_0}$$

$$\therefore$$ $$\alpha = 0.729$$

According to the questions, after $$5$$ second,

$$0.9{A_0} = {A_0}{e^{ - {{b\left( 5 \right)} \over {2m}}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$

After $$10$$ more second,

$$A = {A_0}{e^{ - {{b\left( {15} \right)} \over {2m}}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$

From eqns $$(i)$$ and $$(ii)$$

$$A = 0.729\,{A_0}$$

$$\therefore$$ $$\alpha = 0.729$$

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