JEE MAIN - Physics (2021 - 16th March Evening Shift - No. 25)
A force $$\overrightarrow F $$ = $${4\widehat i + 3\widehat j + 4\widehat k}$$ 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)
Answer
20
Explanation
$$\overrightarrow \tau = \overrightarrow r \times \overrightarrow F $$
$$ = \left[ {(2 - 2)\widehat i + (0 - 3)\widehat j + (0 - 4)\widehat k} \right] \times (4\widehat i + 3\widehat j + 4\widehat k)$$
$$ = ( - 3\widehat j - 4\widehat k) \times (4\widehat i + 3\widehat j + 4\widehat k)$$
$$ = - 16\widehat j + 12\widehat k$$
$$|\overrightarrow \tau |\, = 20$$ units
$$ = \left[ {(2 - 2)\widehat i + (0 - 3)\widehat j + (0 - 4)\widehat k} \right] \times (4\widehat i + 3\widehat j + 4\widehat k)$$
$$ = ( - 3\widehat j - 4\widehat k) \times (4\widehat i + 3\widehat j + 4\widehat k)$$
$$ = - 16\widehat j + 12\widehat k$$
$$|\overrightarrow \tau |\, = 20$$ units
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