A seconds' pendulum is taken to a place where acceleration due to gravity falls to one-fourth. How is the time period of the pendulum affected, if at all?

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Solution

We know, $T = 2 π \sqrt{\frac{l}{g}} = 2$ ...(i)
$T^{'} = 2 π \sqrt{\frac{l}{\frac{g}{4}}} = 2 \times 2 π \sqrt{\frac{l}{g}}$ ...(ii)
From (i) and (ii) we have, T’ =2T
Thus, the time period will become 4s.

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A seconds' pendulum is taken to a place where acceleration due to gravity falls to one-fourth. How is the time period of the pendulum affected, if at all?

We know, T=2π√lg=2 ...(i) T′=2π√lg4=2×2π√lg ...(ii) From (i) and (ii) we have, T’ =2T Thus, the time period will become 4s.

We know, $T = 2 π \sqrt{\frac{l}{g}} = 2$ ...(i)
$T^{'} = 2 π \sqrt{\frac{l}{\frac{g}{4}}} = 2 \times 2 π \sqrt{\frac{l}{g}}$ ...(ii)
From (i) and (ii) we have, T’ =2T
Thus, the time period will become 4s.