1

### 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
2

### 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
3

### 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 $\left[ {{{340} \over {340 - 17}}} \right]$

${{{f_1}} \over {{f_2}}} = {{340 - 17} \over {340 - 34}} = {{323} \over {306}} \Rightarrow {{{f_1}} \over {{f_2}}} = {{19} \over {18}}$
4

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

In an electron microscope, the resolution that can be achieved is of the order of the wavelength of electrons used. To resolve a width of 7.5 × 10–12 m, the minimum electron energy required is close to -
A
25 keV
B
500 keV
C
100 keV
D
1 keV

## Explanation

$\lambda$ = ${h \over p}$          {$\lambda$ = 7.5 $\times$ 10$-$12}

P = ${h \over \lambda }$

KE = ${{{P^2}} \over {2m}} = {{{{\left( {h/\lambda } \right)}^2}} \over {2m}}$

$= {{\left\{ {{{6.6 \times {{10}^{ - 34}}} \over {7.5 \times {{10}^{ - 12}}}}} \right\}} \over {2 \times 9.1 \times {{10}^{ - 31}}}}$ J

KE = 25 Kev