Question

The metallic bob of a simple pendulum has the relative density p. The time period of this pendulum is T. If the metallic bob is immersed in water, then the new time period is

• A. $T\left(\frac{p-1}{p}\right)$
• B. $T\left(\frac{p}{p-1}\right)$
• C. $T\sqrt{\frac{p-1}{p}}$
• D. $T\sqrt{\frac{p}{p-1}}$

Answer 1

Answer: (D)

$T\sqrt{\frac{p}{p-1}}$

$\begin{array}{c}\ast \text{Relative density of}Bob=\rho \\ \ast \text{Time period in water}=?\end{array}$

$\begin{array}{c}M{g}_{eff}=\sigma vg-{\rho }_w\cdot vg=(\sigma -{\rho }_w)\cdot vg\\ \sigma \cdot v\cdot g_{eff}=(\sigma -{\rho }_w)\cdot vg\\ \Rightarrow g_{eff}=\left(\frac{\sigma -{\rho }_w}{\sigma }\right)\cdot g=\left(\frac{\rho -1}{\rho }\right)\cdot g\end{array}$

$\text{Time period}{T}^{\prime }=2\pi \cdot \sqrt{\frac{l}{g}}$

$T^{\prime }=2\pi \cdot \sqrt{\frac{l}{g}\cdot \left(\frac{p}{p-1}\right)}=T\cdot \sqrt{\left(\frac{p}{p-1}\right)}$ $T^{\prime }=T\cdot \sqrt{\frac{p}{p-1}}$