The bob of a simple pendulum is displaced from its equilibrium position O to a position Q which is at height h above O and the bob is then released. Assuming the mass of the bob to be m and time period of oscillations to be 2.0 sec, the tension in the string when the bob passes through O is

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m (g+π2g h)

m g+π22g h

m g+π23g h

answer is A.

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Detailed Solution

Tension in the string when bob passes through lowest point T=mg+mv2r=mg+mvω (∵v=rω)putting v=2gh and ω==2πT=2π2=πwe get T=m (g+π2gh)

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