Solved 2024 Specimen Paper ICSE Class 10 Physics — Question 10

(a) Let the distance of the cliff from the initial position of the person be x m.

Time after which echo is heard = 3 sec

So, the distance travelled by sound in 3 sec = 2x m

Speed of sound in air = 336 m/s

Distance = speed x time

Substituting the values we get,

$2x = 336 \times 3 \\[1em] \Rightarrow x = \dfrac{336 \times 3}{2} \\[1em] \Rightarrow x = 168 \times 3 \\[1em] \Rightarrow x = 504 \text{ m}$

∴ The initial distance of the person from the cliff is 504 m.

(b) As time taken to hear the echo is increased by 1.5 sec, hence, the person has moved away from the cliff.

Let the distance moved by the person from his initial position be y m.

Time after which echo is heard = 1.5 + 3 = 4.5 sec

New distance = 504 + y

So, the distance travelled by sound in 4.5 sec = 2(504 + y) m

Speed of sound in air = 336 m/s

Distance = speed x time

Substituting the values we get,

$2(504 + y) = 336 \times 4.5 \\[1em] \Rightarrow 504 + y = \dfrac{336 \times 4.5}{2} \\[1em] \Rightarrow y = (168 \times 4.5) - 504 \\[1em] \Rightarrow y = 756 - 504 \\[1em] \Rightarrow y = 252 \text{ m}$

∴ Distance moved by the person = 252 m