# 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 $(g=10 \;m/s^2)$

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

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$
$\theta=45^{\circ}$
$F= 20 \sec 45^{\circ}=20 \sqrt 2 \;N$
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.
edited Mar 21, 2014