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Q.

A uniform rod of length L pivoted at the bottom of a pond of depth $\frac{L}{2}$ stays in stable equilibrium as shown in figure. Find the angle $θ$ if the density of the material of the rod is half of the density of water

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a

$45^{0}$

b

$37^{0}$

c

$60^{0}$

d

$30^{0}$

answer is C.

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Detailed Solution

$W = L A ρ g$
$O T = \frac{L}{2 \sin θ}$
$F_{B} = (\frac{L}{2 \sin θ}) A ρ_{w} g$
$F_{B}$ acts at P, where $O P = \frac{O T}{2} = \frac{L}{4 \sin θ}$
Question Image
Balancing torque about O,
$W (O Q \cos θ) = F_{B} (O P \cos θ) L A ρ g [\frac{L}{2}] = (\frac{L}{2 \sin θ} A ρ_{w} g) (\frac{L}{4 \sin θ})$ $\sin^{2} θ = \frac{ρ_{_{w}}}{4 ρ} \Rightarrow \sin^{2} θ = \frac{1}{2} \Rightarrow θ = 45^{0}$

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