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Physics Question on simple harmonic motion

The metallic bob o fa simple pendulum has the relative density ρ\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(ρ−1ρ)T\left(\frac{\rho-1}{\rho}\right)

B

T(ρρ−1)T\left(\frac{\rho}{\rho-1}\right)

C

Tρ−1ρT\sqrt\frac{\rho-1}{\rho}

D

(Tρρ−1)(T\sqrt\frac{\rho}{\rho-1})

Answer

(Tρρ−1)(T\sqrt\frac{\rho}{\rho-1})

Explanation

Solution

∗* Relative density of Bob =ρ=\rho ∗* Time period in water == ? Mgeff =σvg−ρw⋅vg=(σ−ρw)⋅vgMg _{\text {eff }}=\sigma vg -\rho_{ w } \cdot vg =\left(\sigma-\rho_{ w }\right) \cdot vg σ⋅v⋅geff =(σ−ρw)⋅vg\sigma \cdot v \cdot g _{ \text{eff }}=\left(\sigma-\rho_{ w }\right) \cdot vg ⇒geff =(σ−ρwσ)⋅g=(ρ−1ρ)⋅g\Rightarrow g _{\text {eff }}=\left(\frac{\sigma-\rho_{ w }}{\sigma}\right) \cdot g =\left(\frac{\rho-1}{\rho}\right) \cdot g Time period T′=2π⋅1gT '=2 \pi \cdot \sqrt{\frac{1}{ g }} T′=2π⋅1g⋅(pp−1)T '=2 \pi \cdot \sqrt{\frac{1}{ g } \cdot\left(\frac{ p }{ p -1}\right)} =T⋅(pp−1)T′= T \cdot \sqrt{\left(\frac{ p }{ p -1}\right)} T ' =T⋅pp−1= T \cdot \sqrt{\frac{ p }{ p -1}}