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प्रश्न
A seconds pendulum is taken to a place where acceleration due to the gravity falls to one-fourth. How is the time period of the pendulum affected, if at all? Give reasons. What will be its new time period?
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उत्तर
As we know,
`T = 2pisqrt(l/g)`
We observe that time period is inversely related to the square root of gravity's acceleration.
As a result, as 'g' decreases by one-fourth, the time period increases.
When the acceleration due to gravity is reduced by one-fourth, we see that
`T = 2pisqrt((l/g)/4)`
`T = 2pisqrt((4l)/g)`
`T = 2 xx 2pisqrt(l/g)`
As a result, we can conclude that when gravity's acceleration is lowered to one-fourth, the time period of a basic pendulum doubles.
As, the given pendulum is a seconds' pendulum so T = 2s
∴ New T = 2 x 2 = 4s
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