 JEE Mains Previous Years Questions with Solutions

4.5
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1

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

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

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

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

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 ${1 \over 2}m{\omega ^2}{a^2}.$
3

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

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

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

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

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

$= 10\%$
4

AIEEE 2003

Two particles $A$ and $B$ of equal masses are suspended from two massless springs of spring of spring constant ${k_1}$ and ${k_2}$, respectively. If the maximum velocities, during oscillation, are equal, the ratio of amplitude of $A$ and $B$ is
A
$\sqrt {{{{k_1}} \over {{k_2}}}}$
B
${{{{k_2}} \over {{k_1}}}}$
C
$\sqrt {{{{k_2}} \over {{k_1}}}}$
D
${{{{k_1}} \over {{k_2}}}}$

Explanation

Maximum velocity during $SHM$ $= A\omega = A\sqrt {{k \over m}}$

$\left[ {\,\,} \right.$ $\therefore$ $\omega = \sqrt {{k \over m}}$ $\left. {\,\,} \right]$

Here the maximum velocity is same and $m$ is also same

$\therefore$ ${A_1}\sqrt {{k_1}} = {A_2}\sqrt {{k_2}}$

$\therefore$ ${{{A_1}} \over {{A_2}}} = \sqrt {{{{k_2}} \over {{k_1}}}}$