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ICSE Class 10 Physics Question 7 of 27

Solved 2024 Question Paper ICSE Class 10 Physics — Question 22

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Question

Question 2(vii)

Rohan conducted experiments on echo in different media. He observed that a minimum distance of ‘x’ meters is required for the echo to be heard in oxygen and ‘y’ meters in benzene. Compare ‘x’ and ‘y’. Justify your answer.

Speed of sound in oxygen: 340 ms-1

Speed of sound in benzene: 200 ms-1

Answer

Given,

Speed of sound in oxygen = 340 ms-1

Speed of sound in benzene = 200 ms-1

Minimum distance required to hear an echo = distance sound travels in 0.1 s (since the human ear can distinguish sound if the reflected sound reaches after 0.1 s)

So,

Minimum distance for echo=Speed of sound×Minimum time to hear an echo2=v×tmin2\text {Minimum distance for echo} = \dfrac{\text {Speed of sound}\times \text {Minimum time to hear an echo}}{2}=\dfrac{\text v \times \text t_\text{min}}{2}

and

tmin = 0.1 sec

For medium containing oxygen :

x=340×0.12=342=17 m\text x = \dfrac{340\times 0.1}{2}=\dfrac{34}{2} = 17\ \text m

For medium containing benzene :

y=200×0.12=202=10 m\text y = \dfrac{200\times 0.1}{2}=\dfrac{20}{2} = 10\ \text m

So, from above calculations x > y.