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A uniform rod of length 2.0 \mathrm{~m} specific gravity 0.5 and mass 2 \mathrm{~kg} is hinged at one end to the bottom of a tank of water (specific gravity \mathrm{=1.0} ) filled upto a height of 1.0 \mathrm{~m} as shown in figure. Taking the \operatorname{case} \theta \neq 0^{\circ} the force exerted by the hinge on the rodis :\mathrm{ (\mathrm{g}=10 \mathrm{~m} / \mathrm{s} 2} )

Option: 1

\mathrm{10.2 \mathrm{~N} \, \, upwards}
 


Option: 2

\mathrm{4.2 \mathrm{~N}\, \, downwards}


Option: 3

\mathrm{\text { 8.3 } \mathrm{N} \text { downwards }}


Option: 4

\mathrm{6.2 \mathrm{~N} \text { upwards }}


Answers (1)

best_answer

\mathrm{\begin{aligned} & F_b=\frac{m g}{2} \times \sec \theta \times \frac{1}{0.5} \\ & F_b= F_g \times 1=F_b \times \frac{\sec \theta}{2} \\ & \Rightarrow F_b=F_g \sqrt{2} \\ & F_b-F_g=(\sqrt{2}-1) m g \end{aligned}}

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Rishi

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