JEE Main 20205 SeptemberMorning Shift
- Exam
- JEE Main
- Year
- 2020
- Date
- 5 September
- Shift
- Morning Shift
- Subject
- Physics
- Chapter
- Rotational Motion
JEE Main 2020 · 5 September · Morning Shift · Physics · Rotational Motion
A force → F = ( ^ i + 2 ^ j + 3 ^ k ) N acts at a point ( 4 ^ i + 3...
A force \overset{\to}{F} = ( \overset{^}{i} + 2\overset{^}{j} + 3\overset{^}{k} ) N acts at a point ( 4\overset{^}{i} + 3\overset{^}{j} - \overset{^}{k} ) m. Then the magnitude of torque about the point ( \overset{^}{i} + 2\overset{^}{j} + \overset{^}{k} ) m will be The value of is..........
Correct answer
0
Step-by-step solution
\overset{\to}{r} = \left(4 - 1\right)\overset{^}{i} + \left(3 - 2\right)\overset{^}{j} + \left(- 1 - 1\right)\overset{^}{k}
= 3\overset{^}{i} + \overset{^}{j} - 2\overset{^}{k}
τ = \overset{\to}{r} \times \overset{\to}{F} = \left|\begin{cases}\overset{^}{i} & j & \overset{^}{k} \\ 3 & 1 & - 2 \\ 1 & 2 & 3\end{cases}\right|
= \overset{^}{i}\left(7\right) - \overset{^}{j}\left(11\right) + \overset{^}{k}\left(5\right) = 7\overset{^}{i} - 11\overset{^}{j} + 5\overset{^}{k}