JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Rotational Mechanics

If force

F = 3 \hat{i} + 4 \hat{j} - 2 \hat{k}

acts on a particle having position vector

2 \hat{i} + \hat{j} + 2 \hat{k}

then, the torque about the origin will be :-

(A) $3 \hat{i} + 4 \hat{j} - 2 \hat{k}$
(B) $- 10 \hat{i} + 10 \hat{j} + 5 \hat{k}$ ✓ Correct
(C) $10 \hat{i} + 5 \hat{j} - 10 \hat{k}$
(D) $10 \hat{i} + \hat{j} - 5 \hat{k}$

Answer: (B)

Solution

Torque is given by

τ = r \times F

. Computing the cross product:

τ = \hat{i} 23 \hat{j} 14 \hat{k} 2 - 2

= \hat{i} (1 \cdot (- 2) - 2 \cdot 4) - \hat{j} (2 \cdot (- 2) - 2 \cdot 3) + \hat{k} (2 \cdot 4 - 1 \cdot 3)

= \hat{i} (- 2 - 8) - \hat{j} (- 4 - 6) + \hat{k} (8 - 3) = - 10 \hat{i} + 10 \hat{j} + 5 \hat{k}

Answer: (B)

JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Single correct option

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