A man standing in front of a mountain 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. He then moves nearer to the mountain by $90 m$ and finds that echo is again not heard when the drumming rate becomes $60$ per minute. The distance between the mountain and the initial position of the man is .... $m$
Diffcult
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(d) For not hearing the echo the time interval between the beats of drum must be equal to time of echo.
==> ${t_1} = \frac{{2d}}{v} = \frac{{60}}{{40}} = \frac{3}{2}$......$(i)$
and ${t_2} = \frac{{2(d - 90)}}{v} = \frac{{60}}{{60}} = 1$
==> $2d - 180 = v$......$(ii)$
Form $(i),$ we get $2d = \frac{3}{2}v$. Substituting in $(ii),$ we get
==> $\frac{3}{2}v - 180 = v$ ==> $180 = \frac{v}{2}$ ==> $v = 360m{s^{ - 1}}$
==> $\frac{{2(d)}}{{360}} = \frac{3}{2}$ ==> $d = 270m$.
==> ${t_1} = \frac{{2d}}{v} = \frac{{60}}{{40}} = \frac{3}{2}$......$(i)$
and ${t_2} = \frac{{2(d - 90)}}{v} = \frac{{60}}{{60}} = 1$
==> $2d - 180 = v$......$(ii)$
Form $(i),$ we get $2d = \frac{3}{2}v$. Substituting in $(ii),$ we get
==> $\frac{3}{2}v - 180 = v$ ==> $180 = \frac{v}{2}$ ==> $v = 360m{s^{ - 1}}$
==> $\frac{{2(d)}}{{360}} = \frac{3}{2}$ ==> $d = 270m$.
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