A simple pendulum has a hollow sphere containing mercury suspended by means of a wire. If a little mercury is drained off, the period of the pendulum will

increase

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decrease

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remains unchanged

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become erratic

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Solution

The correct option is D increase
$a = - ω^{2} x$
$ω = \sqrt{\frac{a}{x}}$
$\frac{2 π}{T} = \sqrt{\frac{a}{x}}$
$\Rightarrow T \propto \sqrt{x}$
$∴$ When mercury is drained, the CM of system goes down thus increasing effective length of pendulum, and hence increasing the time period.

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A simple pendulum has a hollow sphere containing mercury suspended by means of a wire. If a little mercury is drained off, the period of the pendulum will

The correct option is D increasea=−ω2xω=√ax2πT=√ax⇒T∝√x∴ When mercury is drained, the CM of system goes down thus increasing effective length of pendulum, and hence increasing the time period.

The correct option is D increase
$a = - ω^{2} x$
$ω = \sqrt{\frac{a}{x}}$
$\frac{2 π}{T} = \sqrt{\frac{a}{x}}$
$\Rightarrow T \propto \sqrt{x}$
$∴$ When mercury is drained, the CM of system goes down thus increasing effective length of pendulum, and hence increasing the time period.