A thin cylindrical rod PQ of length L and density d1 is pivoted at its lowest point textP , inside a stationary homogeneous and non-viscous liquid of density d2 . The rod is always fully submerged inside the liquid and is free to rotate in a vertical plane about a horizontal axis passing through P . If d1

Question

A thin cylindrical rod PQ of length L and density d1 is pivoted at its lowest point textP , inside a stationary homogeneous and non-viscous liquid of density d2 . The rod is always fully submerged inside the liquid and is free to rotate in a vertical plane about a horizontal axis passing through P . If d1 < d2 , then the time period of small angular oscillations of the rod about its vertical equilibrium position will be (A) T=2π √(2 L/3 g) ((d1/d2 - d1)) (B) T=2π √(L/3 g) ((d2/d2 - d1)) (C) T=2π √(2 L/g) ((d1/d2 - d1)) (D) T=2π √(2 L/3 g) ((d2 - d1/d1))

Tardigrade Admin · Accepted Answer

Consider the diagram in a displaced position.
Solution

The weight and upthrust

F_{B}

, both pass through the centre of gravity

G

W = (π r^{2} L) d_{1} g

F_{B} = (π r^{2} L) d_{2} g

The restoring torque about the pivot is

τ = - (π r^{2} Lg (d_{2} - d_{1}) \frac{L}{2}) θ

α = - \frac{[ π r ^{2} Lg ( d _{2} - d _{1} ) \frac{L}{2} ] θ}{( ( π r ^{2} L d _{1} ) L ^{2} ) /3}

α = - [\frac{3 g}{2 L} (\frac{d _{2} - d _{1}}{d _{1}})] θ

T = 2 π \frac{2 L}{3 g} (\frac{d _{1}}{d _{2} - d _{1}})

$T = 2 π \frac{2 L}{3 g} (\frac{d _{1}}{d _{2} - d _{1}})$

$T = 2 π \frac{L}{3 g} (\frac{d _{2}}{d _{2} - d _{1}})$

$T = 2 π \frac{2 L}{g} (\frac{d _{1}}{d _{2} - d _{1}})$

$T = 2 π \frac{2 L}{3 g} (\frac{d _{2} - d _{1}}{d _{1}})$

T=2π3g2L​(d2​−d1​d1​​)​

T=2π3gL​(d2​−d1​d2​​)​

T=2πg2L​(d2​−d1​d1​​)​

T=2π3g2L​(d1​d2​−d1​​)​

$T = 2 π \frac{2 L}{3 g} (\frac{d _{1}}{d _{2} - d _{1}})$

$T = 2 π \frac{L}{3 g} (\frac{d _{2}}{d _{2} - d _{1}})$

$T = 2 π \frac{2 L}{g} (\frac{d _{1}}{d _{2} - d _{1}})$

$T = 2 π \frac{2 L}{3 g} (\frac{d _{2} - d _{1}}{d _{1}})$