JEE MAIN - Physics (2020 - 5th September Morning Slot - No. 14)
A force $$\overrightarrow F = \left( {\widehat i + 2\widehat j + 3\widehat k} \right)$$ N acts at a point
$$\left( {4\widehat i + 3\widehat j - \widehat k} \right)$$ m. Then the magnitude of torque
about the point $$\left( {\widehat i + 2\widehat j + \widehat k} \right)$$ m will be $$\sqrt x $$ N m.
The value of x is _______.
$$\left( {4\widehat i + 3\widehat j - \widehat k} \right)$$ m. Then the magnitude of torque
about the point $$\left( {\widehat i + 2\widehat j + \widehat k} \right)$$ m will be $$\sqrt x $$ N m.
The value of x is _______.
Answer
195
Explanation
$$\overrightarrow \tau = \overrightarrow r \times F = (3\widehat i + \widehat j - 2\widehat k) \times (\widehat i + 2\widehat j + 3\widehat k)$$
$$ = \left| {\matrix{ i & j & k \cr 3 & 1 & { - 2} \cr 1 & 2 & 3 \cr } } \right|$$
$$ = \widehat i(3 + 4) - \widehat j(9 + 2) + \widehat k(6 - 1)$$
$$\overrightarrow \tau = 7\widehat j - 11\widehat j + 5\widehat k$$
$$\left| {\overrightarrow \tau } \right| = \sqrt {49 + 121 + 25} = \sqrt {195} $$
$$ \therefore $$ $$x = 195$$
$$ = \left| {\matrix{ i & j & k \cr 3 & 1 & { - 2} \cr 1 & 2 & 3 \cr } } \right|$$
$$ = \widehat i(3 + 4) - \widehat j(9 + 2) + \widehat k(6 - 1)$$
$$\overrightarrow \tau = 7\widehat j - 11\widehat j + 5\widehat k$$
$$\left| {\overrightarrow \tau } \right| = \sqrt {49 + 121 + 25} = \sqrt {195} $$
$$ \therefore $$ $$x = 195$$
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