An observer moves towards a staionary source of sound, with a velocity one-fifth of the velocity of sound. What is the percentage increase in the apparent frequency Option 1) 5% Option 2) 20% Option 3) 0% Option 4) 0.5%

V Vakul

$n' = \left [ \frac{v+v_{0}}{v} \right ]n\Rightarrow \left [ \frac{v+v/5}{v} \right ]n = 6/5n$

$\Rightarrow 1.2n$

Increament in frequency = 0.2

50% increament $\frac{0.2n}{n}\times 100= 20 \%$

Frequency when observer is stationary and source is moving away from observer -

$\nu {}'= \nu _{0}.\frac{C}{C+V_{s}}$

- wherein

$C=$ speed of sound

$V_{s}=$ speed of source

$\nu _{0}=$ original frequency

$\nu {}'=$ apparent frequency

Option 1)

5%

This is incorrect

Option 2)

20%

This is correct

Option 3)

0%

This is incorrect

Option 4)

0.5%

This is incorrect

V Vakul

$n' = \left [ \frac{v+v_{0}}{v} \right ]n\Rightarrow \left [ \frac{v+v/5}{v} \right ]n = 6/5n$

$\Rightarrow 1.2n$

Increament in frequency = 0.2

50% increament $\frac{0.2n}{n}\times 100= 20 \%$

Frequency when observer is stationary and source is moving away from observer -

$\nu {}'= \nu _{0}.\frac{C}{C+V_{s}}$

- wherein

$C=$ speed of sound

$V_{s}=$ speed of source

$\nu _{0}=$ original frequency

$\nu {}'=$ apparent frequency

Option 1)

5%

This is incorrect

Option 2)

20%

This is correct

Option 3)

0%

This is incorrect

Option 4)

0.5%

This is incorrect

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