MHT CET · Physics · Waves and Sound
A person standing between two parallel cliffs fires a gun and hears two echoes, first echo after 1 second and the second echo after 3 second. The distance between the two cliffs is (Velocity of sound \(=340 \mathrm{~m} / \mathrm{s}\) )
- A 340 m
- B 680 m
- C 1020 m
- D 1360 m
Answer & Solution
Correct Answer
(B) 680 m
Step-by-step Solution
Detailed explanation
\(d_1 = \frac{v t_1}{2} = \frac{340 \mathrm{~m/s} \times 1 \mathrm{~s}}{2} = 170 \mathrm{~m}\) \(d_2 = \frac{v t_2}{2} = \frac{340 \mathrm{~m/s} \times 3 \mathrm{~s}}{2} = 510 \mathrm{~m}\)
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