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Answer (Detailed Solution Below)

Option 2 : 8.66 N·m / 8.66 N·m

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Detailed Solution

सही उत्तर 8.66 N·m / 8.66 N·m है।

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Question:

A force of $20$ N is applied at a point $0.5$ m from the pivot of a rod, at $60°$ to the rod. The torque about the pivot is / छड़ के आलंब से $0.5$ m दूर बिंदु पर $20$ N का बल छड़ से $60°$ पर लगाया गया है। आलंब के परितः बल-आघूर्ण है

1.
10.00 N·m / 10.00 N·m
2.
8.66 N·m / 8.66 N·m
3.
5.00 N·m / 5.00 N·m
4.
17.32 N·m / 17.32 N·m

Correct Answer: 8.66 N·m / 8.66 N·m

$\tau = rF\sin\theta = 0.5 \times 20 \times \sin 60° = 10 \times 0.866 = 8.66$ N·m. / $\tau = rF\sin\theta = 0.5 \times 20 \times \sin 60° = 10 \times 0.866 = 8.66$ N·m।

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