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

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


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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$
$F= 20 \sec 45^{\circ}=20 \sqrt 2 \;N$
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.
answered Oct 8, 2013 by meena.p
edited Mar 21, 2014 by balaji.thirumalai

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