Model Paper 1 — Question 11

Figure given below shows that two cliffs and the position of the person.

(a) First echo is heard from the nearest cliff.

Let d₁, be the distance of the nearest cliff 1 from the person.

Total distance travelled by the sound in going and then coming back = 2d₁ = 2 x 640 m = 1280 m

Time taken (t) = 4 s

$\text {Speed of sound (v)} = \dfrac{\text {Tot. dist. sound travels} }{\text {Time taken}} \\[1em] {\text {Time taken}}=\dfrac{2d_1}{\text t} \\[1em] =\dfrac{1280}{4}=320\text{ ms}^{-1}$

(b) The second echo is heard from the farther cliff 2. If d₂ is the distance of the farther cliff 2 from the person, then total distance travelled by the sound in going and then coming back = 2d₂.

Time taken (t) = 4 + 3 = 7 s

Now,

$\text {Speed of sound (v)} = \dfrac{\text {Tot. dist. sound travels} }{\text {Time taken}} \\[1em] \Rightarrow \text v = \dfrac{2d_2}{\text t} \\[1em] \Rightarrow \text d_2 = \dfrac{\text{vt}}{2} \\[1em] \Rightarrow \text d_2 = \dfrac{320\times7}{2} \\[1em] \Rightarrow \text d_2 = 1120 \text{ m}$

Distance between the two cliffs = d₁ + d₂ = 640 m + 1120 m = 1760 m

Hence, speed of sound in air is 320 ms^-1 and distance between the cliffs is 1760 m.