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JEE Mains · Physics · STD 11 - System of particles and rotational motion
  • JEE Mains 2021
  • Hard
  • Numerical

A force \(\overrightarrow{ F }=4 \hat{ i }+3 \hat{ j }+4 \hat{ k }\) is applied on an intersection point of \(x =2\) plane and \(x\) -axis. The magnitude of torque of this force about a point \((2,3,4)\) is.......... . (Round off to the Nearest Integer)

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Answer & Solution

Correct Answer

(B) \(20\)

Step-by-step Solution

Detailed explanation

\(\vec{\tau}=\overrightarrow{ r } \times \overrightarrow{ F }\)

\(\overrightarrow{ r }=(2 \hat{ i })-(2 \hat{ i }+3 \hat{ j }+4 \hat{ k })=-3 \hat{ j }-4 \hat{ k }\)

\( \overrightarrow{ F }=4 \hat{ i }+3 \hat{ j }+4 \hat{ k }\)

\(\vec{\tau}=\overrightarrow{ r } \times \overrightarrow{ F }=\left|\begin{array}{ccc}\hat{i} & \hat{ j } & \hat{ k } \\ 0 & -3 & -4 \\ 4 & 3 & 4\end{array}\right|\)

\(\quad=\hat{ i }(-12+12)-\hat{ j }(0+16)+\hat{ k }(0+12)\)

\(=-16 \hat{ i }+12 \hat{ k }\)

\(\therefore \quad|\vec{\tau}|=\sqrt{16^{2}+12^{2}}=20\)

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