SSC
PHYSICS
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Question
[CLICK ON ANY CHOICE TO KNOW THE RIGHT ANSWER]
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A pendulum oscillating near the surface of the Earth swings with a time period T. What is the time period of the same pendulum near the surface of the planet Mercury where the gravitational field strength is 0.4g?
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0.2 T
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0.6 T
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1.2 T
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1.6 T
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Explanation:
Detailed explanation-1: -The period of oscillation of a simple pendulum of length l is given by T=2√l/g. The length l is about 10 cm and is known to 1 mm accuracy. The period of oscillation is about 0.5 s. The time of 100 oscillations has been measured with a stop watch of 1 s resolution.
Detailed explanation-2: -Updated On: 27-06-2022 T1=2s, T2(moon) =2√l/g2, T2=T1√g(g/6)=2×√6=4.89 second . Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams. The acceleration due to gravity on the surface of the moon is 1.7ms^(-2) .
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