JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2004

The bob of a simple pendulum executes simple harmonic motion in water with a period $t,$ while the period of oscillation of the bob is ${t_0}$ in air. Neglecting frictional force of water and given that the density of the bob is $\left( {4/3} \right) \times 1000\,\,kg/{m^3}.$ What relationship between $t$ and ${t_0}$ is true
A
$t = 2{t_0}$
B
$t = {t_0}/2$
C
$t = {t_0}$
D
$t = 4{t_0}$

Explanation

$t = 2\pi \sqrt {{\ell \over {{g_{eff}}}}} ;\,{t_o}\,\, = 2\pi \sqrt {{\ell \over g}}$

$m{g_{eff}} = mg - B = my - V \times 100 \times g$

$\therefore$ ${g_{eff}} = g - {{100} \over {\left( {m/v} \right)}}g$

$= g - {{1000} \over {{4 \over 3} \times 1000}}g = {g \over 4}$

$\therefore$ $t = 2\pi \sqrt {{\ell \over {g/4}}} \,\,\,\,\,\,\,\,\,\,\,t = 2{t_0}$
2

AIEEE 2003

The displacement of particle varies according to the relation
$x=4$$\left( {\cos \,\pi t + \sin \,\pi t} \right).$ The amplitude of the particle is
A
$-4$
B
$4$
C
$4\sqrt 2$
D
$8$

Explanation

$x = 4\left( {\cos \pi t + \sin \pi t} \right)$

$= \sqrt 2 \times 4\left( {{{\sin \pi t} \over {\sqrt 2 }} + {{\cos \pi t} \over {\sqrt 2 }}} \right)$

$x = 4\sqrt 2 \sin \left( {\pi t + {{45}^ \circ }} \right)$
3

AIEEE 2003

A body executes simple harmonic motion. The potential energy $(P.E),$ the kinetic energy $(K.E)$ and total energy $(T.E)$ are measured as a function of displacement $x.$ Which of the following statements is true ?
A
$K.E$ is maximum when $x=0$
B
$T.E$ is zero when $x=0$
C
$K.E$ is maximum when $x$ is maximum
D
$P.E$ is maximum when $x=0$

Explanation

$K.E. = {1 \over 2}m{\omega ^2}\left( {{a^2} - {x^2}} \right)$

When $x=0,$ $K.E$ is maximum and is equal to ${1 \over 2}m{\omega ^2}{a^2}.$
4

AIEEE 2003

The length of a simple pendulum executing simple harmonic motion is increased by $21\%$. The percentage increase in the time period of the pendulum of increased length is
A
$11\%$
B
$21\%$
C
$42\%$
D
$10\%$

Explanation

$T = 2\pi \sqrt {{\ell \over g}}$ and $T' = 2\pi \sqrt {{{1.21\ell } \over g}}$

$\left( \, \right.$ as $\ell ' = \ell + 21\%$ of $\left. \ell \right)$

$\%$ increase $= {{T' - T} \over T} \times 100$

$= {{\sqrt {1.21\ell } - \sqrt \ell } \over {\sqrt \ell }} \times 100$

= ${{\sqrt {{{121} \over {100}}l} - \sqrt l } \over {\sqrt l }} \times 100$

= ${{{{11} \over {10}}\sqrt l - \sqrt l } \over {\sqrt l }} \times 100$

= ${{{{11} \over {10}} - 1} \over 1} \times 10$

$= 10\%$