 JEE Mains Previous Years Questions with Solutions

4.5
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1

AIEEE 2003

A metal wire of linear mass density of $9.8$ $g/m$ is stretched with a tension of $10$ $kg$-$wt$ between two rigid supports $1$ metre apart. The wire passes at its middle point between the poles of a permanent magnet, and it vibrates in resonance when carrying an alternating current of frequency $n.$ The frequency $n$ of the alternating source is
A
$50$ $Hz$
B
$100$ $Hz$
C
$200$ $Hz$
D
$25$ $Hz$

Explanation

KEY CONCEPT : For a string vibrating between two rigid support, the fundamental frequency is given by

$n = {1 \over {2\ell }}\sqrt {{T \over \mu }} = {1 \over {2 \times }}\sqrt {{{10 \times 9.8} \over {9.8 \times {{10}^{ - 3}}}}} = 50Hz$

As the string is vibrating in resonance to a.c of frequency $n,$ therefore both the frequencies are same.
2

AIEEE 2002

When temperature increases, the frequency of a tuning fork
A
increases
B
decreases
C
remains same
D
increases or decreases depending on the material

Explanation

KEY CONCEPT : The frequency of a tuning fork is given by the expression

$f = {{{m^2}k} \over {4\sqrt 3 \pi {\ell ^2}}}\sqrt {{Y \over \rho }}$

As temperature increases, $\ell$ increases and therefore $f$ decreases.
3

AIEEE 2002

A wave $y=a$ $\sin \left( {\omega t - kx} \right)$ on a string meets with another wave producing a node at $x=0.$ Then the equation of the unknown wave is
A
$y = \alpha \,\sin \,\left( {\omega t + kx} \right)$
B
$y = - \alpha \,\sin \,\left( {\omega t + kx} \right)$
C
$y = \alpha \,\sin \,\left( {\omega t - kx} \right)$
D
$y = - \alpha \,\sin \,\left( {\omega t - kx} \right)$

Explanation

To form a node there should be superposition of this wave with the reflected wave. The reflected wave should travel in opposite direction with a phase change of $\pi$. The equation of the reflected wave will be

$y = a\sin \left( {\omega t + kx + \pi } \right)$

$\Rightarrow y = - a\sin \left( {\omega t + kx} \right)$
4

AIEEE 2002

Tube $A$ has bolt ends open while tube $B$ has one end closed, otherwise they are identical. The ratio of fundamental frequency of tube $A$ and $B$ is
A
$1:2$
B
$1:4$
C
$2:1$
D
$4:1$

Explanation

KEY CONCEPT : The fundamental frequency for closed organ pipe is given by ${\upsilon _c} = {v \over {4\ell }}$ and

For open organ pipe is given by ${\upsilon _0} = {v \over {2\ell }}$

$\therefore$ ${{{\upsilon _0}} \over {{\upsilon _c}}} = {v \over {2\ell }} \times {{4\ell } \over v} = {2 \over 1}$