JEE Main 202229 JulyEvening Shift
- Exam
- JEE Main
- Year
- 2022
- Date
- 29 July
- Shift
- Evening Shift
- Subject
- Physics
- Chapter
- Rotational Motion
JEE Main 2022 · 29 July · Evening Shift · Physics · Rotational Motion
The torque of a force 5 ^ i + 3 ^ j - 7 ^ k about the origin is τ ....
The torque of a force 5\overset{^}{i} + 3\overset{^}{j} - 7\overset{^}{k} about the origin is . If the force acts on a particle whose position vector is 2\overset{^}{i} + 2\overset{^}{j} + \overset{^}{k}, then the value of will be
Options
- A11\overset{^}{i} + 19\overset{^}{j} - 4\overset{^}{k}
- B- 11\overset{^}{i} + 9\overset{^}{j} - 16\overset{^}{k}
- C- 17\overset{^}{i} + 19\overset{^}{j} - 4\overset{^}{k}
- D17\overset{^}{i} + 9\overset{^}{j} + 16\overset{^}{k}
Correct answer
C. - 17\overset{^}{i} + 19\overset{^}{j} - 4\overset{^}{k}
Step-by-step solution
Using the determinant rule to find the torque
\overset{\to}{τ} = \overset{\to}{r} \times \overset{\to}{F} = \left|\begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ 2 & 2 & 1 \\ 5 & 3 & - 7\end{cases}\right|
= \overset{^}{i}\left(- 14 - 3\right) - \overset{^}{j}\left(- 14 - 5\right) + \overset{^}{k}\left(6 - 10\right)
Thus, the value of \overset{\to}{τ} = - 17\overset{^}{i} + 19\overset{^}{j} - 4\overset{^}{k}