Question
A man is standing between two parallel cliffs and fires a gun. If he hears first and second echoes after $1.5 \,s$ and $3.5\,s$ respectively, the distance between the cliffs is .... $ m$ (Velocity of sound in air $= 340 ms^{-1}$)
==> $2({d_1} + {d_2}) = v({t_1} + {t_2})$
${d_1} + {d_2} = \frac{{v({t_1} + {t_2})}}{2}$
$=\frac{{340 \times (1.5 + 3.5)}}{2} = 850\, m.$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
