First slide

Pressure in a fluid-mechanical properties of fluids

Question

A uniform rod of density $ρ$ is placed in a wide tank containing a liquid of density $ρ_{0} (ρ_{0} > ρ)$ . The depth of liquid in the tank is half the length of the rod. The rod is in equilibrium, with its lower end resting on the bottom of the tank. In this position the rod makes an angle $θ$ with the horizontal

Moderate

A

$\sin θ = \frac{1}{2} \sqrt{ρ_{0} / ρ}$
B

$\sin θ = \frac{1}{2} . \frac{ρ_{0}}{ρ}$
C

$\sin θ = \sqrt{ρ / ρ_{0}}$
D

$\sin θ = ρ_{0} / ρ$

Solution

Let L = PQ = length of rod
$∴ SP = SQ = \frac{L}{2}$
Weight of rod, $W = Alρg$ , acting
At point S
And force of buoyancy,
$F_{B} = {Alρ}_{0} g$ , [l = PR]
which acts at mid-point of PR.
For rotational equilibrium,
$\begin{array}{l} {Alρ}_{0} g \times \frac{l}{2} \cos θ = ALρg \times \frac{L}{2} \cos θ \\ \Rightarrow \frac{l^{2}}{L^{2}} = \frac{ρ}{ρ_{0}} \Rightarrow \frac{l}{L} = \sqrt{\frac{ρ}{ρ_{0}}} \end{array}$
From figure, $sinθ = \frac{h}{l} = \frac{L}{2 l} = \frac{1}{2} \sqrt{\frac{ρ_{0}}{ρ}}$

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