If a simple pendulum is taken to place where g decreases by 2% , then time period

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13.Oscillations

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If a simple pendulum is taken to place where g decreases by $2\%$, then the time period

A

Decreases by $1\%$

B

Decreases by $2\%$

C

Increases by $2\%$

D

Increases by $1\%$

Solution

(d) $T = 2\pi \sqrt {\frac{l}{g}} $ ==> $T \propto \frac{l}{{\sqrt g }}$

$ \Rightarrow \frac{{\Delta T}}{T} \times 100 = – \frac{1}{2}\left( {\frac{{\Delta g}}{g}} \right) \times 100$$ = – \frac{1}{2}( – \,2\% ) = 1\% .$

Standard 11

Physics

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