A person standing between two parallel cliffs fires a gun and hears two echoes, first echo after 1 second and the second echo after 3 second. The distance between the two cliffs is (Velocity of sound $=340 \mathrm{~m} / \mathrm{s}$ )

Distance between cliffs using echo timing

Given sound velocity $v = 340 \text{ m/s}$, first echo time $t_1 = 1 \text{ s}$, and second echo time $t_2 = 3 \text{ s}$.

Since each echo corresponds to sound traveling to a cliff and back, the distance to the first cliff is $d_1 = v t_1 / 2 = 340 \text{ m/s} \times 1 \text{ s} / 2 = 170 \text{ m}$.

Similarly, distance to the second cliff is $d_2 = v t_2 / 2 = 340 \text{ m/s} \times 3 \text{ s} / 2 = 510 \text{ m}$.

The total distance between cliffs is $d_1 + d_2 = 170 \text{ m} + 510 \text{ m} = 680 \text{ m}$.

Therefore, the distance is $680 \text{ m}$ (Option B).