 JEE Mains Previous Years Questions with Solutions

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1

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

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

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

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

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

$\Rightarrow {m \over M} = {{16} \over 9}$
2

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.
3

AIEEE 2002

In a simple harmonic oscillator, at the mean position
A
kinetic energy is minimum, potential energy is maximum
B
both kinetic and potential energies are maximum
C
kinetic energy is maximum, potential energy is minimum
D
both kinetic and potential energies are minimum.

Explanation

$K.E = {1 \over 2}k\left( {{A^2} - {x^2}} \right);\,\,\,U = {1 \over 2}k{x^2}$
At the mean position $x=0$
$\therefore$ $K.E. = {1 \over 2}k{A^2} =$ Maximum and $U=0$
4

AIEEE 2002

If a spring has time period $T,$ and is cut into $n$ equal parts, then the time period of each part will be
A
$T\sqrt n$
B
$T/\sqrt n$
C
$nT$
D
$T$

Explanation

Let the spring constant of the original spring be $k.$

Then its time period $T = 2\pi \sqrt {{m \over k}}$ where $m$ is the mass of oscillating body.
When the spring is cut into $n$ equal parts, the spring constant of one part becomes $nk.$ Therefore the new time period,

$T' = 2\pi \sqrt {{m \over {nk}}} = {T \over {\sqrt n }}$