Question

Bullets are fired at regular intervals of 14 s from an armoured car A moving towards car B with a velocity of 180 kmhr${}^{-1}$. The intervals at which an officer seated in B and dashing with a speed of 72 kmhr${}^{-1}$ towards A hears the bullet shots is (velocity of sound = 330 ms${}^{-1}$)

• A. 10 s
• B. 11.2 s
• C. 14.2 s
• D. 16 s

Answer 1

Answer: (B)

11.2 s

Here velocity of A is $V_A=V_s=50m/s$ and the velocity of B is

$V_B=V_o=20m/s$. Here $f=\frac{1}{T}$

$f=\frac{1}{14}=0.07$

Now the officer seated in B hears the bullet shots with

a frequency:

$\overline{n}=n\frac{V+V_o}{V-V_s}=0.07\frac{330+20}{330-50}=0.09Hz$. So

the time will be $\overline{T}=1/\overline{n}=11.2s$