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A light spiral spring supports 200g weig...

A light spiral spring supports `200g` weight at its lower end oscillates with a period of `1s`. The weight that must be removed from the lower end to reduce the period to `0.5s` is

A

100 g

B

50 g

C

150 g

D

200 g

Text Solution

Verified by Experts

The correct Answer is:
C
`(T_(2))/(T_(1)) = sqrt((m_(2))/(m_(1))) rArr (0.5)/(1) = sqrt((200 - m)/(200))`
`rArr m = 150 g` must be removed
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NARAYNA-OSCILLATIONS-EXERCISE - I (H.W)
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