Verified Solution • 2025–26 Session moderate

QJEEM21Phc7e5: JEE Main 2021 Physics - Class 11 JEEM21Phc7e5 Solution: A force = is applied on an intersection point of x = 2 plane…

A force F→\overrightarrow F = 4i^+3j^+4k^{4\widehat i + 3\widehat j + 4\widehat 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)

Step-by-Step Verified Solution

τ→=r→×F→\overrightarrow \tau = \overrightarrow r \times \overrightarrow F

=[(2−2)i^+(0−3)j^+(0−4)k^]×(4i^+3j^+4k^)= \left[ {(2 - 2)\widehat i + (0 - 3)\widehat j + (0 - 4)\widehat k} \right] \times (4\widehat i + 3\widehat j + 4\widehat k)

=(−3j^−4k^)×(4i^+3j^+4k^)= ( - 3\widehat j - 4\widehat k) \times (4\widehat i + 3\widehat j + 4\widehat k)

=−16j^+12k^= - 16\widehat j + 12\widehat k

∣τ→∣ =20|\overrightarrow \tau |\, = 20 units

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