A man standing in front of a hill beats a drum at regular intervals. The rate of drumming is generally…
A man standing in front of a hill beats a drum at regular intervals. The rate of drumming is generally increased and he finds that the echo is not heard distinctly when the rate becomes 40 per minute. Then, he moves closer to hill by $90\text{ m}$ and he again finds that echo is again not heard when the drumming rate becomes 60 per minute.
The distance between the mountain and initial position of the man is [BHU 2012]
180 m
205 m
270 m
300 m
Solution
For not hearing the echo, the time interval between the beats of drum must be equal to that of time of echo.
$\Rightarrow t_1 = \frac{2d}{v} = \frac{60}{40} = \frac{3}{2} \Rightarrow 2d = \frac{3}{2}v$ ...(i)
Also, $t_2 = \frac{2(d - 90)}{v} = \frac{60}{60} = 1$
$\Rightarrow 2d - 180 = v$ ...(ii)
From Eqs. (i) and (ii), we get
$\frac{3}{2}v - 180 = v \Rightarrow v = 360\text{ ms}^{-1}$
$\therefore \frac{2d}{360} = \frac{3}{2} \Rightarrow d = \frac{3}{2} \times \frac{360}{2} = 270\text{ m}$