A person standing between two parallel cliffs fires a gun and hears two echoes, first echo after 1 second and second echo after 3 second. The distance between the two cliffs is (The velocity of sound $= 330$ m/s)

Let the distance of the person from the first cliff be $d_1$ and from the second cliff be $d_2$.

The time taken to hear the first echo is $t_1 = 1$ s. The sound travels to the first cliff and back, so the distance covered is $2d_1$.

$2d_1 = v \times t_1$

$2d_1 = 330 \times 1 = 330$ m

$d_1 = 165$ m

The time taken to hear the second echo is $t_2 = 3$ s. The sound travels to the second cliff and back, so the distance covered is $2d_2$.

$2d_2 = v \times t_2$

$2d_2 = 330 \times 3 = 990$ m

$d_2 = 495$ m

The total distance between the two cliffs is $d = d_1 + d_2$.

$d = 165 + 495 = 660$ m

Answer: $660$ m