1

### JEE Main 2018 (Online) 16th April Morning Slot

Two sitar strings, A and B, playing the note 'Dha' are slightly out of tune and produce beats of frequency 5 Hz. The tension of the string B s slightly increased and the beat frequency is found to decrease by 3Hz. If the frequency of A is 425 Hz, the original frequency of B is :
A
430 Hz
B
420 Hz
C
428 Hz
D
422 Hz

## Explanation

Frequency of B, fB = 425 $\pm$ 5 = 420 or 430 Hz

As tension of string B is increased

So, frequency of B, fB should also increase [as f $\propto$ $\sqrt T$]

If initially fB = 430 Hz then when fB increases by increasing the tension then fB $-$ fA increases that means beat frequency increase.

So, fB can't be 430 Hz

When fB = 420 then when fB increases fA $-$ fB decreases means beat frequency decreases. So, correct fB = 420 Hz.
2

### JEE Main 2018 (Online) 16th April Morning Slot

The end correction of a resonance column is 1 cm. If the shortest length resonating with the tunning fork is 10 cm, the next resonating length should be :
A
28 cm
B
32 cm
C
36 cm
D
40 c

## Explanation

Given, End correction (e) = 1 cm

For first resonance,

${\lambda \over 4} = {l_1} + e$ = 10 + 1 = 11 cm

For second resonance,

${3\lambda \over 4} = {l_2} + e$

$\Rightarrow$ ${l_2}$ = 3 $\times$ 11 - 1 = 32 cm
3

### JEE Main 2019 (Online) 9th January Morning Slot

Two coherent sources produce waves of different intensities which interfere. After interference, the ratio of the maximum intensity to the minimum intensity is 16. The intensity of the waves are in the ratio :
A
16 : 9
B
25 : 9
C
4 : 1
D
5 : 3

## Explanation

Given that,

${{{{\rm I}_{\max }}} \over {{{\mathop{\rm I}\nolimits} _{min}}}} = {{16} \over 1}$

We know,

Imax $=$ ${\left( {\sqrt {{{\rm I}_1}} + \sqrt {{{\rm I}_2}} } \right)^2}$

and Imin $= {\left( {\sqrt {{{\rm I}_1}} - \sqrt {{{\rm I}_2}} } \right)^2}$

$\therefore$   ${{{{\left( {\sqrt {{{\rm I}_1}} + \sqrt {{{\rm I}_2}} } \right)}^2}} \over {{{\left( {\sqrt {{{\rm I}_1}} - \sqrt {{{\rm I}_2}} } \right)}^2}}} = {{16} \over 1}$

$\Rightarrow$   ${{\sqrt {{{\rm I}_1}} + \sqrt {{{\rm I}_2}} } \over {\sqrt {{{\rm I}_1}} - \sqrt {{{\rm I}_2}} }} = {4 \over 1}$

$\Rightarrow$   $4\sqrt {{{\rm I}_1}} - 4\sqrt {{{\rm I}_2}} = \sqrt {{{\rm I}_1}} + \sqrt {{{\rm I}_2}}$

$\Rightarrow$   $3\sqrt {{{\rm I}_1}} = 5\sqrt {{{\rm I}_2}}$

$\Rightarrow$   ${{\sqrt {{{\rm I}_1}} } \over {\sqrt {{{\rm I}_2}} }} = {5 \over 3}$

$\Rightarrow$   ${{{{\rm I}_1}} \over {{{\rm I}_2}}} = {{25} \over 9}$
4

### JEE Main 2019 (Online) 9th January Evening Slot

A rod of mass 'M' and length '2L' is suspended at its middle by a wire. It exhibits torsional oscillations; If two masses each of 'm' are attached at distance 'L/2' from its centre on both sides, it reduces the oscillation frequency by 20%. The value of radio m/M is close to :
A
0.77
B
0.57
C
0.37
D
0.17

## Explanation

Initially : After putting 2 masses of each 'm' at a distance ${L \over 2}$ from center : We know,

Time period (T) = 2$\pi$ $\sqrt {{{\rm I} \over C}}$

$\therefore$  T $\propto$ $\sqrt {\rm I}$

$\therefore$   Frequency (f) $\propto$ $\sqrt {{1 \over {\rm I}}}$

$\therefore$   ${{{f_1}} \over {{f_2}}}$ = $\sqrt {{{{{\rm I}_2}} \over {{{\rm I}_1}}}}$

Also given that,
After putting two masses 'm' at both end new frequency becomes 80% of initial frequency.

$\therefore$   f2 = 0.8f1

$\therefore$   ${{{f_1}} \over {0.8{f_1}}}$ = $\sqrt {{{{{\rm I}_2}} \over {{{\rm I}_1}}}}$

$\therefore$   ${{{{{\rm I}_1}} \over {{{\rm I}_2}}}}$ = 0.64

Initial moment of inertia of the system,

${{{\rm I}_1}}$ = ${{M{{\left( {2L} \right)}^2}} \over {12}}$

Final moment of inertia of the system,

I2 = ${{M{{\left( {2L} \right)}^2}} \over {12}}$ + 2$\left( {m{{\left( {{L \over 2}} \right)}^2}} \right)$

$\therefore$   ${{M{{\left( {2L} \right)}^2}} \over {12}}$ = 0.64 $\left[ {{{M{L^2}} \over 3} + {{m{L^2}} \over 2}} \right]$

$\Rightarrow$   ${{M{L^2}} \over {3 \times 0.64}}$ = ${{M{L^2}} \over 3}$ + ${{M{L^2}} \over 2}$

$\Rightarrow$   ${M \over {1.92}} - {M \over 3} = {m \over 2}$

$\Rightarrow$   ${{1.08M} \over {3 \times 1.92}}$ = ${m \over 2}$

$\Rightarrow$   ${m \over M}$ = ${{1.08 \times 2} \over {3 \times 1.92}}$ = 0.37