Oscillations Ques 56
56. If the length of a simple pendulum is increased by $2 \%$, then the time period
[1997]
(a) increases by $2 \%$
(b) decrease by $2 \%$
(c) increases by $1 \%$
(d) decrease by $1 \%$
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Answer:
Correct Answer: 56.(c)
Solution:
- (c) We know that $T=2 \pi \sqrt{\frac{l}{g}}$
$\frac{\Delta T}{T} \times 100=\frac{1}{2} \frac{\Delta l}{l} \times 100$
If length is increased by $2 \%$., time period increases by $1 \%$.
(i) If $g$ is constant and length varies by $n \%$. Then $\%$ change in time period $\frac{\Delta T}{T} \times 100=\frac{n}{2} \%$
(ii) If $l$ is constant and $g$ varies by $n \%$. Then $\%$ change in time period $\frac{\Delta T}{T} \times 100=\frac{n}{2} \%$
Valid only for percentage less than $10 \%$.
BETA
