Question:

A uniform rod of length $1$ m and mass of $2$ kg is attached to a side support at $O$ as shown in the figure. The rod is at equilibrium due to upward force $T$ acting at $P$. Assume the acceleration due to gravity as $10$ m/s$^2$. The value of $T$ is

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For rods, weight acts at center; always take torque about pivot to simplify.
Updated On: May 1, 2026
  • $0$
  • $2$ N
  • $5$ N
  • $10$ N
  • $20$ N
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The Correct Option is D

Solution and Explanation


Concept:
For rotational equilibrium: \[ \sum \tau = 0 \]

Step 1:
Weight of rod.
\[ W = mg = 2 \times 10 = 20 \text{ N} \] Acts at center (0.5 m from O).

Step 2:
Take moments about O.
\[ T \times 1 = 20 \times 0.5 \]

Step 3:
Solve.
\[ T = 10 \text{ N} \]
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