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A uniform rod of length $2.0 m$ specific $0.5$ and mass$2 \;kg$ is hinged at one end to bottom of tank of water (Specific gravity =1.0) filled upto a height of $1m$. as shown when $\theta \neq 0$. the force exerted by hinge on the rod is $(g=10 \;m/s^2)$

\[\begin {array} {1 1} (a)\;10.2\;N\;upward \\ (b)\;4.2\;N\; downward \\ (c)\;8.3\;N\;downward \\ (d)\;6.2\;N\;upward \end {array}\]


1 Answer

We consider the portion of rod inside water and the upthrust on this portion (F) and the weight of the rod (W) acting down wards.
Length of rod inside water $= 1.0\; \sec \theta $
$\qquad= \sec \theta$
upthrust (F)=Volume of water displaced $\times $ density of water $\times$ g
Volume of rod inside water:
$\qquad= \large\frac{(mass/unit \;length) \times length}{density}$
$\qquad= \large\frac{\Large\frac{2}{2} \times \sec \theta}{500}$
[Specific gravity=0.5]
$F= \large\frac{2}{2}$$ \sec \theta \times \large\frac{1}{500}$$ \times 1000 \times 10$
$F= 20\;sec \theta$
Weight of rod $(W)=2 \times 10=20\;N$
For rotational equilibrium net torque about 0 should be zero.
$F \large\frac{\sec \theta}{2}$$ \sin \theta$$=W(1.0 \sin \theta)$
$\large\frac{20}{2}$$ \sec^2 \theta=20$
$F= 20 \sec 45^{\circ}=20 \sqrt 2 \;N$
For vertical equilibrium of rod force exerted by the hinge will be
$(20 \sqrt {2} -20)=8.28\;N\; down ward$
Hence c is the correct answer.
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