JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
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\(........\)
- A \(200\)
- B \(195\)
- C \(150\)
- D \(175\)
Answer & Solution
Correct Answer
(B) \(195\)
Step-by-step Solution
Detailed explanation
\(\vec{\tau}=\left(\overrightarrow{ r}_{2}-\overrightarrow{ r }_{1}\right) \times \overrightarrow{ F }\) \(=[(4 \hat{ i }+3 \hat{ j }-\hat{ k })-(\hat{ i }+2 \hat{ j }+\hat{ k })] \times \overrightarrow{ F }\)…
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