1

### 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
2

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

${{{f_1}} \over {{f_2}}} = {{340 - 17} \over {340 - 34}} = {{323} \over {306}} \Rightarrow {{{f_1}} \over {{f_2}}} = {{19} \over {18}}$
3

### 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
4

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

A string of length 1 m and mass 5 g is fixed at both ends. The tension in the string is 8.0 N. The string is set into vibration using an external vibrator of frequency 100 Hz. The separation between successive nodes on the string is close to -
A
16.6 cm
B
10.0 cm
C
20.0 cm
D
33.3 cm

## Explanation

Velocity of wave on string

$V = \sqrt {{T \over \mu }} = \sqrt {{8 \over 5} \times 1000} = 40m/s$

Now, wavelength of wave

$\lambda = {v \over n} = {{40} \over {100}}m$

Separation b/w successive nodes,

${\lambda \over 2} = {{20} \over {100}}\,m$ $=$ 20 cm