JEE MAIN - Physics (2022 - 29th July Evening Shift - No. 15)
The torque of a force $$5 \hat{i}+3 \hat{j}-7 \hat{k}$$ about the origin is $$\tau$$. If the force acts on a particle whose position vector is $$2 i+2 j+k$$, then the value of $$\tau$$ will be
$$11 \hat{i}+19 \hat{j}-4 \hat{k}$$
$$-11 \hat{i}+9 \hat{j}-16 \hat{k}$$
$$-17 \hat{i}+19 \hat{j}-4 \hat{k}$$
$$17 \hat{i}+9 \hat{j}+16 \hat{k}$$
Explanation
$\vec{\tau}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 2 & 1 \\ 5 & 3 & -7\end{array}\right|$
$$ \begin{aligned} &=\hat{i}(-14-3)+\hat{j}(5+14)+\hat{k}(6-10) \\\\ &=-17 \hat{i}+19 \hat{j}-4 \hat{k} \end{aligned} $$
$$ \begin{aligned} &=\hat{i}(-14-3)+\hat{j}(5+14)+\hat{k}(6-10) \\\\ &=-17 \hat{i}+19 \hat{j}-4 \hat{k} \end{aligned} $$
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