Question

The metallic bob o fa simple pendulum has the relative density $\rho$. The time period of this pendulum is T. If the metallic bob is immersed in water, then the new time period is given by.

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

Answer 1

Answer: (D)

$*$ Relative density of Bob $=\rho$
$* $ Time period in water $=$ ?
$Mg _{\text {eff }}=\sigma vg -\rho_{ w } \cdot vg =\left(\sigma-\rho_{ w }\right) \cdot vg$
$\sigma \cdot v \cdot g _{ \text{eff }}=\left(\sigma-\rho_{ w }\right) \cdot vg$
$\Rightarrow g _{\text {eff }}=\left(\frac{\sigma-\rho_{ w }}{\sigma}\right) \cdot g =\left(\frac{\rho-1}{\rho}\right) \cdot g$
Time period $T '=2 \pi \cdot \sqrt{\frac{1}{ g }}$
$T '=2 \pi \cdot \sqrt{\frac{1}{ g } \cdot\left(\frac{ p }{ p -1}\right)}$
$= T \cdot \sqrt{\left(\frac{ p }{ p -1}\right)} T '$
$= T \cdot \sqrt{\frac{ p }{ p -1}}$