Question

A uniform rod of length 2.0 m, specific gravity 0.5 and mass 2 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 m, as shown in figure. Taking, the case $\theta \ne 0$, the force exrted by hinge on the rod is : $(g=10m/s^2)$

• A. 4.3 N
• B. 6.3 N
• C. 8.3 N
• D. 10.3 N

Answer 1

The correct option is C 8.3 N

Length of rod inside water = 1sec$\theta$

Thus volume of rod inside water = $\frac{2sec\theta }{2\times 500}$

Thus $F_B=Volumeofliquiddisplaced\times {\rho }_{water}\times g=20Sec\theta$

Let's write the equation of torque about ) balance for rotational equilibrium,

$\frac{F_{B}sec\theta sin\theta }{2}$=$W_{rod}\times Sin\theta$ which gives $\theta ={45}^0$ and F=20$\sqrt{2}$N

Now writing equation for force equilibrium in vertical direction

Force at hinge = $20\sqrt{2}-20=8.28N$