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  • The time period of a simple pendulum is T. The pendulum is oscillated with its bob immersed in a liquid of density <img alt="\mathrm{\sigma}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5

The time period of a simple pendulum is T. The pendulum is oscillated with its bob immersed in a liquid of density \mathrm{\sigma}. If the density of the bob is \mathrm{\rho} and viscous effect is neglected, the time period of the pendulum will be

Option: 1

\mathrm{\left(\frac{\rho}{\rho-\sigma}\right)^{1 / 2} T}


Option: 2

\mathrm{\left(\frac{\sigma}{\rho-\sigma}\right)^{1 / 2} T}


Option: 3

\mathrm{\left(\frac{\rho}{\sigma}\right)^{1 / 2} T}


Option: 4

\mathrm{\left(\frac{\sigma}{\rho}\right)^{1 / 2} T}


Answers (1)

best_answer

The net downward force acting on the bob is
\mathrm{ \begin{aligned} F= & (\rho-\sigma) V g=\left(\frac{\rho-\sigma}{\rho}\right) m g \\ & (\because m=\rho V) \end{aligned} }
\mathrm{ \text { or } \quad m g^{\prime}=\left(\frac{\rho-\sigma}{\rho}\right) m g }
or
\mathrm{ g^{\prime}=\left(\frac{\rho-\sigma}{\rho}\right) g }
This is the effective acceleration due to gravity.
\mathrm{ \begin{aligned} \therefore \quad T^{\prime} & =2 \pi \sqrt{\frac{l}{g^{\prime}}}=2 \pi \sqrt{\frac{\rho l}{(\rho-\sigma) g}} \\ & =\left(\frac{\rho}{\rho-\sigma}\right)^{1 / 2} T \quad\left(\because T=2 \pi \sqrt{\frac{l}{g}}\right) \end{aligned} }

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Divya Prakash Singh

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