Verified Solution • 2025–26 Session moderate

QJEEM20Phb1dd: JEE Main 2020 Physics - Class 11 JEEM20Phb1dd Solution: A force N acts at a point m. Then the magnitude of torque about …

A force F→=(i^+2j^+3k^)\overrightarrow F = \left( {\widehat i + 2\widehat j + 3\widehat k} \right) N acts at a point
(4i^+3j^−k^)\left( {4\widehat i + 3\widehat j - \widehat k} \right) m. Then the magnitude of torque
about the point (i^+2j^+k^)\left( {\widehat i + 2\widehat j + \widehat k} \right) m will be x\sqrt x N m.
The value of x is _______.

Step-by-Step Verified Solution

τ→=r→×F=(3i^+j^−2k^)×(i^+2j^+3k^)\overrightarrow \tau = \overrightarrow r \times F = (3\widehat i + \widehat j - 2\widehat k) \times (\widehat i + 2\widehat j + 3\widehat k)

=|ijk31−2123|

=i^(3+4)−j^(9+2)+k^(6−1)= \widehat i(3 + 4) - \widehat j(9 + 2) + \widehat k(6 - 1)

τ→=7j^−11j^+5k^\overrightarrow \tau = 7\widehat j - 11\widehat j + 5\widehat k

∣τ→∣=49+121+25=195\left| {\overrightarrow \tau } \right| = \sqrt {49 + 121 + 25} = \sqrt {195}

∴\therefore x=195x = 195

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