Kerala KEAM 5 June 2024 Solved Paper

When the bullet is fired, two echoes are heard:- The first echo after $6 \, s$ is from the shorter hillock.- The second echo after $12 \, s$ is from the taller hillock.The distance traveled by sound in time $t$ is given by:$$d = v \times t$$where:- $v = 330 \, m/s$ is the velocity of sound in air,- $t$ is the time taken for the sound to travel to the object and back.For the shorter hillock:- The total time for the sound to travel to the shorter hillock and return is $t_1 = 6 \, s$.- The distance traveled by sound is $d_1 = v \times t_1 = 330 \times 6 = 1980 \, m$.- The distance to the shorter hillock is half of this distance: $d_{short} = \frac{1980}{2} = 990 \, m$.For the taller hillock:- The total time for the sound to travel to the taller hillock and return is $t_2 = 12 \, s$.- The distance traveled by sound is $d_2 = v \times t_2 = 330 \times 12 = 3960 \, m$.- The distance to the taller hillock is half of this distance: $d_{tall} = \frac{3960}{2} = 1980 \, m$.Thus, the distance between the two hillocks is:$$Distance between hillocks = d_{tall} - d_{short} = 1980 - 990 = 990 \, m$$Therefore, the distance between the hillocks is $990 \, m$.$$\boxed{990 \, m}$$ Quick Tip: The distance to the hillocks can be calculated by using the time for the sound to travel to the hillock and back. Since the sound travels both to and from the hillock, divide the total distance by 2.