Question

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

• A. $10.2\text{N}$ upwards
• B. $4.2\text{N}$ downwards
• C. $8.3\text{N}$ downwards
• D. $6.2\text{N}$ upwards

Answer 1

The correct option is C $8.3\text{N}$ downwards

Let ${}^{\prime }{L}^{\prime }$ be the height upto which water is filled and $l^{\prime }$ be the total length of the rod.
Length of rod inside the water is given by $\left(l\right)=L\sec \theta \dots \dots \left(1\right)$
Upthrust $\left(F\right)=\frac{\lambda l}{\sigma }g$
where, $\sigma \to$ specific gravity
$\lambda \to$ linear mass density of the rod.

From the data given in the question,

$F=\frac{1\times L\sec \theta \times 10}{0.5}=20L\sec \theta \dots \dots \left(2\right)$

Weight of the rod $\left(W\right)=mg=2\times 10=20\text{N}\dots \dots \left(3\right)$

For rotational equilibrium of rod, net torque about ${}^{\prime }{O}^{\prime }$ should be zero.
$\therefore F\frac{l}{2}\sin \theta =W\frac{l^{\prime }}{2}\sin \theta$

Using $\left(2\right)$ in the above equation,
$\frac{20\times L\sec \theta \times l\sin \theta }{2}=W\frac{l^{\prime }}{2}\sin \theta$

Using $\left(1\right)$ and $\left(3\right)$ in the above equation

$10L^2{\sec }^2\theta =20$

Given,
Height upto which water is filled is $L=1\text{m}$
$\therefore {\sec }^2\theta =2\Rightarrow \theta ={45}^{\circ }$

Substituting this in equation $\left(2\right)$ we get,

$F=20\sec {45}^{\circ }=20\sqrt{2}\text{N}$

For vertical equilibrium of rod,
$W+F_{hinge}=F$
or $20+F_{hinge}=20\sqrt{2}$
i.e Force exerted by the hinge on the rod will be $(20\sqrt{2}-20)\text{N}$ or $8.3\text{N}$ downwards.

Thus, option (c) is the correct answer.