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Standard XII

Physics

Variation of G Due to Height and Depth

If the length...

Question

If the length of a simple pendulum is increased by $2$ %, then the time period

Increases by

2

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Decreases by

2

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Increases by

1

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Decreases by

1

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Solution

The correct option is B Increases by $1$ %
Given that $l_{2} = 1.02 l_{1}$
We know that $T = 2 π \sqrt{(\frac{l}{g})}$
or $T \propto \sqrt{l}$
$∴ \frac{T_{2}}{T_{1}} = \sqrt{(\frac{l_{2}}{l_{1}})} = \sqrt{\frac{(1.02 l_{1})}{l_{1}}} = 1.01$
Thus time period increases by $1$ %

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If the length of a simple pendulum is increased by 2%, then the time period

The correct option is B Increases by 1%Given that l2=1.02l1We know that T=2π√(lg) or T∝√l∴T2T1=√(l2l1)=√(1.02l1)l1=1.01Thus time period increases by 1%

If the length of a simple pendulum is increased by $2$ %, then the time period

The correct option is B Increases by $1$ %
Given that $l_{2} = 1.02 l_{1}$
We know that $T = 2 π \sqrt{(\frac{l}{g})}$
or $T \propto \sqrt{l}$
$∴ \frac{T_{2}}{T_{1}} = \sqrt{(\frac{l_{2}}{l_{1}})} = \sqrt{\frac{(1.02 l_{1})}{l_{1}}} = 1.01$
Thus time period increases by $1$ %