1

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

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

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

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

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

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

$\Rightarrow$   ${{{{\rm I}_1}} \over {{{\rm I}_2}}} = {{25} \over 9}$
2

### 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
3

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

A musician using an open flute of length 50 cm producess second harmonic sound waves. A person runs towards the musician from another end of hall at a speed of 10 km/h. If the wave speed is 330 m/s, the frequency heard by the running person shall be close to :
A
666 Hz
B
753 Hz
C
500 Hz
D
333 Hz

## Explanation

Frequency of sound wave produce by flute

= ${{2{V_S}} \over {2\ell }}$

= ${{2 \times 330} \over {2 \times 50 \times {{10}^{ - 2}}}}$

= 660 Hz

Speed of the observer

= 10km/hr

= 10 $\times$ ${5 \over {18}}$ m/s

= ${{25} \over 9}$ m/s

Frequency heard by the observer,

f' = $\left( {{{{V_S} + {V_o}} \over {{V_S}}}} \right)f$

= $\left( {{{330 + {{25} \over 9}} \over {330}}} \right) \times 660$

= 666 Hz
4

### JEE Main 2019 (Online) 10th January Morning Slot

A train moves towards a stationary observer with speed 34 m/s. The train sounds a whistle and its frequency registered by the observer is ƒ1. If the speed of the train is reduced to 17 m/s, the frequency registered is ƒ2. If speed of sound is 340 m/s, then the ratio ƒ12 is -
A
19/18
B
20/19
C
21/20
D
18/17

## Explanation

fapp = f0 $\left[ {{{{v_2} \pm {v_0}} \over {{v_2} \pm {v_s}}}} \right]$

f1 = f0 $\left[ {{{340} \over {340 - 34}}} \right]$

f2 = f0 $\left[ {{{340} \over {340 - 17}}} \right]$

${{{f_1}} \over {{f_2}}} = {{340 - 17} \over {340 - 34}} = {{323} \over {306}} \Rightarrow {{{f_1}} \over {{f_2}}} = {{19} \over {18}}$