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JEE Main 202225 JuneMorning Shift
Exam
JEE Main
Year
2022
Date
25 June
Shift
Morning Shift
Subject
Physics
Chapter
Rotational Motion

JEE Main 2022 · 25 June · Morning Shift · Physics · Rotational Motion

If force → F = 3 ^ i + 4 ^ j - 2 ^ k acts on a particle having...

If force \overset{\to}{F} = 3\overset{^}{i} + 4\overset{^}{j} - 2\overset{^}{k} acts on a particle having position vector 2\overset{^}{i} + \overset{^}{j} + 2\overset{^}{k} then, the torque about the origin will be:

Options

  1. A- 10\overset{^}{i} + 10\overset{^}{j} + 5\overset{^}{k}
  2. B3\overset{^}{i} + 4\overset{^}{j} - 2\overset{^}{k}
  3. C10\overset{^}{i} + 5\overset{^}{j} - 10\overset{^}{k}
  4. D10\overset{^}{i} + \overset{^}{j} - 5\overset{^}{k}

Correct answer

A. - 10\overset{^}{i} + 10\overset{^}{j} + 5\overset{^}{k}

Step-by-step solution

Torque in vector form is expressed as vector multiplication of position vector and force.

τ→=r→×F→\overset{\to}{τ} = \overset{\to}{r} \times \overset{\to}{F}

⇒ \overset{\to}{τ} = \left|\begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ 2 & 1 & 2 \\ 3 & 4 & - 2\end{cases}\right|

⇒ \overset{\to}{τ} = \left(- 2 - 8\right)\overset{^}{i} - \left(- 4 - 6\right)\overset{^}{j} + \left(8 - 3\right)\overset{^}{k}

⇒ \overset{\to}{τ} = \left(- 10\overset{^}{i} + 10\overset{^}{j} + 5\overset{^}{k}\right)Nm

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