Rotational Motion

JEE Main level, single correct, medium

112 of 280

Question 112 of 280

The torque of a force 5i^+3j^−7k^5\hat{i}+3\hat{j}-7\hat{k} about the origin is τ\tau. If the force acts on a particle whose position vector is 2i^+2j^+k^2\hat{i}+2\hat{j}+\hat{k}, then the value of τ\tau will be

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Answer: (C) −17i^+19j^−4k^-17\hat{i}+19\hat{j}-4\hat{k}

Using the determinant rule to find the torque
τ⃗=r⃗×F⃗=∣i^j^k^22153−7∣\vec{\tau }=\vec{r}\times \vec{F}=\left|\begin{matrix}\hat{i} & \hat{j} & \hat{k} \\ 2 & 2 & 1 \\ 5 & 3 & -7\end{matrix}\right|
=i^(−14−3)−j^(−14−5)+k^(6−10)=\hat{i}\left(-14-3\right)-\hat{j}\left(-14-5\right)+\hat{k}\left(6-10\right)
Thus, the value of τ⃗=−17i^+19j^−4k^\vec{\tau }=-17\hat{i}+19\hat{j}-4\hat{k}

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