JEE PHYSICS MCQ: SYSTEM OF PARTICLES

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Application +4 | -1

The handle of a door is at a distance 40 cm from axis of rotation. If a force 5 N is applied on the handle in a direction

{30^0}

with plane of door, then the torque is

View answer and explanation

Correct option: B

1 Nm

Option B (1 Nm) is correct because the torque is calculated using the formula

\tau = rF\sin \theta

, where

\theta

is the angle between the force and the radial vector.

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\mathrm{AB}

\mathrm{L}

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x

\mathrm{A}

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\left(\frac{7}{12}\right) \mathrm{L}

\mathrm{A}

\mathrm{b}

1.0 \mathrm{~g}

8 \mathrm{~cm}

\omega = {t^4} + 2{t^3}

t = 1\,sec

\left( {rad\,{s^{ - 2}}} \right)