Question:

A uniform rod of length 2 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 m as shown in the figure. Taking the case \(\theta \neq 0°\), the force exerted by the hinge on the rod is

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Use torque balance about the hinge to find $\theta$, then use force balance (vertical and horizontal) to find the hinge reaction.
Updated On: Apr 20, 2026
  • 10.2 N upwards
  • 4.2 N downwards
  • 8.3 N downwards
  • 6.2 N upwards
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The Correct Option is C

Solution and Explanation

Step 1: Understanding the Concept:
Take torques about the hinge. The length submerged is \(\sec\theta\). Upthrust acts at mid-point of submerged length. Weight acts at centre of rod.

Step 2: Detailed Explanation:
From torque balance: \(F(\sec\theta)(\sin\theta/2) = W(1\sin\theta)\). Solving gives \(\theta = 45°\). At \(\theta=45°\): Upthrust \(= 20\sec45° = 20\sqrt{2}\) N. Weight \(= 20\) N. Vertical equilibrium: \(F_{\rm hinge} = W - F_{\rm upthrust} = 20 - 20\sqrt{2} = -8.28 \approx -8.3\) N. Negative means downward. Hinge force \(= 8.3\) N downwards.

Step 3: Final Answer:
Force exerted by hinge \(= 8.3\) N downwards.
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