JEE MAIN - Physics (2021 - 17th March Morning Shift - No. 6)
A triangular plate is shown. A force $$\overrightarrow F $$ = 4$$\widehat i$$ $$-$$ 3$$\widehat j$$ is applied at point P. The torque at point P with respect to point 'O' and 'Q' are :
_17th_March_Morning_Shift_en_6_1.png)
$$-$$ 15 + 20$$\sqrt 3 $$, 15 + 20$$\sqrt 3 $$
15 $$-$$ 20$$\sqrt 3 $$, 15 + 20$$\sqrt 3 $$
15 + 20$$\sqrt 3 $$, 15 $$-$$ 20$$\sqrt 3 $$
$$-$$ 15 $$-$$ 20$$\sqrt 3 $$, 15 $$-$$ 20$$\sqrt 3 $$
Explanation
$$\overrightarrow F = 4\widehat i - 3\widehat j$$
$${\overrightarrow r _1} = 5\widehat i + 5\sqrt 3 \widehat j$$ & $${\overrightarrow r _2} = - 5\widehat i + 5\sqrt 3 \widehat j$$
Torque about 'O'
$${\overrightarrow \tau _O} = {\overrightarrow r _1} \times \overrightarrow F = \left( { - 15 - 20\sqrt 3 } \right)\widehat k = \left( {15 + 20\sqrt 3 } \right)\left( { - \widehat k} \right)$$
Torque about 'Q'
$${\overrightarrow \tau _Q} = {\overrightarrow r _2} \times \overrightarrow F = \left( { - 15 + 20\sqrt 3 } \right)\widehat k = \left( {15 - 20\sqrt 3 } \right)\left( { - \widehat k} \right)$$
$${\overrightarrow r _1} = 5\widehat i + 5\sqrt 3 \widehat j$$ & $${\overrightarrow r _2} = - 5\widehat i + 5\sqrt 3 \widehat j$$
Torque about 'O'
$${\overrightarrow \tau _O} = {\overrightarrow r _1} \times \overrightarrow F = \left( { - 15 - 20\sqrt 3 } \right)\widehat k = \left( {15 + 20\sqrt 3 } \right)\left( { - \widehat k} \right)$$
Torque about 'Q'
$${\overrightarrow \tau _Q} = {\overrightarrow r _2} \times \overrightarrow F = \left( { - 15 + 20\sqrt 3 } \right)\widehat k = \left( {15 - 20\sqrt 3 } \right)\left( { - \widehat k} \right)$$
Comments (0)
