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Q.

If the period of oscillation of a simple pendulum is increased by a% then the percentage increase in its length is

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a

$\frac{a}{2}$

b

a

c

$\frac{1}{2}$

d

2a

answer is C.

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Detailed Solution

$T = 2 π \sqrt{\frac{l}{g}} \Rightarrow \frac{Δ T}{T} \times 100 = \frac{1}{2} (\frac{Δ l}{l} \times 100)$

$\Rightarrow a = \frac{1}{2} (\frac{Δ l}{l} \times 100) ∴ \frac{Δ l}{l} \times 100 = 2 a$

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