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Q.

A force F→=i^+2j^+3k^N acts at a point 4i^+3j^−k^m. Then the magnitude of torque about the point i^+2j^+k^m will be x N−m. The value of x is _____________

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answer is 195.

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Detailed Solution

Relative Position =r→=4−1i^+3−2j^+−1−1k^⇒r→=3i^+j^−2k^Also , F→=i^+2j^+3j^ Torque=τ→=r→×F→=i^j^k^31−2123 =7i^−11j^+5k^Magnitude of torque =72+112+52=195

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