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# A policeman on duty detects a drop of 10% in the frequency of a motor car as it crosses him. If the velocity of sound is 330m/s, the speed of car is a) 20m/sb) 36.6m/sc) 25m/sd) None of these

Hint: The frequency of the sound of the motor car heard by the policeman drops. Hence the car moves away from him. Since there is relative motion between the policeman and the car the actual frequency of the motor car will be different from that heard by the policeman. Therefore we will use Doppler’s effect to determine the velocity of the car.

The apparent frequency of sound heard by the observer when the source is moving away from the observer is given by ${{\text{V}}_{\text{A}}}\text{=}\dfrac{v}{v-{{v}_{S}}}\text{V}....\text{(1)}$ where ${{\text{V}}_{\text{A}}}$ is the apparent frequency heard by the observer, V is the actual frequency of the source $v$ is the speed of sound in air and ${{v}_{S}}$ is the speed of the source.
In the above case the policeman detects a drop in the frequency by 10 percent of the actual frequency. Hence the frequency of sound heard by him is 90 percent of the actual frequency i.e. ${{\text{V}}_{\text{A}}}$= 0.9V where V is the actual frequency of the motor car and ${{\text{V}}_{\text{A}}}$ is the frequency of sound heard by the policeman.
Now using equation 1 let us determine the speed of the source ${{v}_{S}}$ i.e. of the car.
\begin{align} & {{\text{V}}_{\text{A}}}\text{=}\dfrac{v}{v-{{v}_{S}}}\text{V} \\ & \text{0}\text{.9V=}\dfrac{v}{v-{{v}_{S}}}\text{V} \\ & \text{0}\text{.9(}v-{{v}_{S}})=v \\ \end{align}
\begin{align} & \text{0}\text{.9(}v-{{v}_{S}})=v \\ & 0.9{{v}_{S}}=v-0.9v \\ & {{v}_{S}}=\dfrac{0.1}{0.9}v=\dfrac{1}{9}\text{v m/s} \\ \end{align}
It is given to us that the speed of sound in air i.e. v= 330m/s. Hence substituting in the above equation we get ${{v}_{S}}$ i.e. speed of the car as,
${{v}_{S}}=\dfrac{1}{9}\text{v m/s}=\dfrac{330}{9}m/s=36.6m/s$.