### JEE Mains Previous Years Questions with Solutions

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1

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

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

### 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
3

### AIEEE 2002

length of a string tied to two rigid supports is $40$ $cm$. Maximum length (wavelength in $cm$) of a stationary wave produced on it is
A
$20$
B
$80$
C
$40$
D
$120$

## Explanation

This will happen for fundamental mode of vibration as shown in the figure. ${S_1}$ and ${S_2}$ are rigid support

Here ${\lambda \over 2} = 40\,\,\,\,\,\,$ $\therefore$ $\lambda = 80\,cm$