JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2020
- Medium
- Numerical
A force \(\overrightarrow{ F }=(\hat{ i }+2 \hat{ j }+3 \hat{ k }) N\) acts at a point \((4 \hat{ i }+3 \hat{ j }-\hat{ k }) m \cdot\) Then the magnitude of torque about the point \((\hat{i}+2 \hat{j}+\hat{k}) m\) will be \(\sqrt{ x } N - m .\) The value of \(x\) is\(........\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(195\)
Step-by-step Solution
Detailed explanation
\(\vec{\tau}=\left(\overrightarrow{ r}_{2}-\overrightarrow{ r }_{1}\right) \times \overrightarrow{ F }\)
\(=[(4 \hat{ i }+3 \hat{ j }-\hat{ k })-(\hat{ i }+2 \hat{ j }+\hat{ k })] \times \overrightarrow{ F }\)
\(=(3 \hat{i}+\hat{j}-2 \hat{k}) \times(\hat{i}+2 \hat{j}+3 \hat{k})\)
\(\tau=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & -2 \\ 1 & 2 & 3\end{array}\right|\)
\(=7 \hat{ i }-11 \hat{ j }+5 \hat{ k }\)
\(|\vec{\tau}|=\sqrt{195}\)
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