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 q¹ 0ŗ the force exerted by the hinge on the rod is : (g = 10 m/s²) –

• A. 10.2 N upwards
• B. 4.2 N downwards
• C. 8.3 N downwards
• D. 6.2 N upwards

Answer 1

Answer: (C)

8.3 N downwards

Answer 2

Length of rod inside the water = 1.0 secq = secq

Upthrust F = $\left( \frac{2}{2} \right)$ (sec q) $\left( \frac{1}{500} \right)$ (1000) (10)

or F = 20 sec q

Weight of rod W = 2 × 10 = 20 N

For rotational equilibrium of rod net torque about O should be

zero.

\ F $\left( \frac{\sec\theta}{2} \right)$ (sin q) = W = (1.0 sin q)

or $\frac{20}{2}$ sec²q = 20

or q = 45⁰

\ F = 20 sec 45⁰

= 20$\sqrt{2}$ N