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The length of second’s pendulum on the surface of earth is nearly 1 m. Its length on the surface of moon should be [Given: acceleration due to gravity (g) on moon is 1/6 th of that on the earth’s surface](A) 1/6 m (B) 6 m (C) 1/36 m(D) 1/\(\sqrt{6}\) |
Answer» Correct option is: (A) 1/6 m |
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