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