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ICSE Class 10 Physics 2019 Paper

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Question : 49 of 74
Marks: +1, -0
A pendulum has a frequency of 4 vibrations per second. An observer starts the pendulum and fires a gun simultaneously. He hears the echo from the cliff after 6 vibrations of the pendulum. If the velocity of sound in air is 340m/s340 \mathrm{m/s}, find the distance between the cliff and the observer.
Solution:  
Time taken for 4 vibrations of the pendulum =1sec=1 \mathrm{sec}
Time taken for 1 vibrations of the pendulum =  14sec=\;\frac{1}{4} \mathrm{sec}
∴\therefore Time taken for 6 vibrations of the pendulum =  14×6=1.5sec=\;\frac{1}{4} \times 6 = 1.5 \mathrm{sec}
We know,
v=  2dtv = \;\frac{2d}{t}
or d=  vt2d = \;\frac{v t}{2}
∴    d=  340×1.52=255m\therefore \;\; d = \;\frac{340 \times 1.5}{2} = 255 \mathrm{m}
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