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Question

A uniform rod of length 2m, specific gravity 0.5 and mass 2kg is hinged at one end to the bottom of a tank of water (specific gravity =1.0) filled upto a height of 1m as shown in the figure. Taking the case θ0o, the force exerted by the hinge on the rod is :
670037_701ac470a23942ca9778c5d3a918836d.jpg

A
10.2N upwards
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B
4.2N downwards
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C
8.3N downwards
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D
6.2N upwards
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Solution

The correct option is C 8.3N downwards
Length of rod inside the water =1secθ=secθ
Upthrust, F=(22)(secθ)(1500)(1000)(10)
F=20secθ
Weight of rod, w=20×10=20N
For rotational equilibrium of rod, net torque about O should be zero.
F(secθ2)(sinθ)=w=(1sinθ)
202sec2θ=20
θ=45o
F=20sec45o
F=202N
For vertical equilibrium of rod, force exerted by the hinge on the rod will be (20220)N downwards i.e. 8.3N downwards.

707415_670037_ans_01cbbcb2d85041018a438f7a3f777891.jpg

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