### JEE Mains Previous Years Questions with Solutions

4.5
star star star star star
1

### 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.
2

### 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

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

$\Rightarrow y = - a\sin \left( {\omega t + kx} \right)$
3

### 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 ${\upsilon _0} = {v \over {2\ell }}$

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

### AIEEE 2002

A tuning fork arrangement (pair) produces $4$ beats/sec with one fork of frequency $288$ $cps.$ A little wax is placed on the unknown fork and it then produces $2$ beats/sec. The frequency of the unknown fork is
A
$286$ $cps$
B
$292$ $cps$
C
$294$ $cps$
D
$288$ $cps$

## Explanation

A tuning fork produces $4$ beats/sec with another tuning fork of frequency $288$ cps. From this information we can conclude that the frequency of unknown fork is $288+4$ $cps$ or $288-4$ $cps$ i.e. $292$ $cps$ or $284$ $cps.$

Here when a little wax is placed on the unknown fork, it decreases the frequency of unknown fork. Here also beats per second decreases to 2 from 4. So the difference between frequency decreases.

This is possible only when before placing the wax, the frequency of unknown fork is greater than the frequency of the given tuning fork.

So the frequency of the unknown tuninh fork is = 292 Hz

### Graduate Aptitude Test in Engineering

GATE CSE GATE ECE GATE ME GATE PI GATE EE GATE CE GATE IN

NEET

Class 12