 JEE Mains Previous Years Questions with Solutions

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1

JEE Main 2015 (Offline)

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 :
($g=$ $gravitational$ $acceleration$)
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
$\left[ {{{\left( {{{{T_M}} \over T}} \right)}^2} - 1} \right]{{Mg} \over A}$

Explanation

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

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

JEE Main 2014 (Offline)

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 $6\tau$

Explanation

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

JEE Main 2013 (Offline)

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

Explanation

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

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

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

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

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 $SHM.$

$f = {1 \over {2\pi }}\sqrt {{{\gamma {P_0}A} \over {{x_0}M}}} = {1 \over {2\pi }}\sqrt {{{\gamma {P_0}{A^2}} \over {M{V_0}}}}$
4

JEE Main 2013 (Offline)

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
$0.6$

Explanation

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 $(ii)$

$A = 0.729\,{A_0}$

$\therefore$ $\alpha = 0.729$