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

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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.

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