### JEE Mains Previous Years Questions with Solutions

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1

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

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

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

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

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

$= 10\%$
2

### 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

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

$\therefore$ ${{{A_1}} \over {{A_2}}} = \sqrt {{{{k_2}} \over {{k_1}}}}$
3

### AIEEE 2003

A mass $M$ is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes $SHM$ of time period $T.$ If the mass is increased by $m.$ the time period becomes ${{5T} \over 3}$. Then the ratio of ${{m} \over M}$ is
A
${3 \over 5}$
B
${25 \over 9}$
C
${16 \over 9}$
D
${5 \over 3}$

## Explanation

$T = 2\pi \sqrt {{M \over k}}$

$T' = 2\pi \sqrt {{{M + m} \over k}} = {{5T} \over 3}$

$\therefore$ $2\pi \sqrt {{{M + m} \over k}}$

$= {5 \over 3} \times 2\pi \sqrt {{M \over k}}$

$\Rightarrow {m \over M} = {{16} \over 9}$
4

### AIEEE 2002

A child swinging on a swing in sitting position, stands up, then the time period of the swing will
A
increase
B
decrease
C
remains same
D
increases of the child is long and decreases if the child is short

## Explanation

KEY CONCEPT : The time period $T = 2\pi \sqrt {{\ell \over g}}$ where

$\ell$ $=$ distance between the point of suspension and the center of mass of the child. This distance decreases when the child stands

$\therefore$ $T' < T$ i.e., the period decreases.