A force of 3i+2j-k N acts on a particle with position vector i+j-k m. The magnitude of torque of given force is (a) √(5) N m (b) √(8) N m (c) √(6) N m (d) √(10) N m

Answer: (c). τ = r with r = (1,1,-1), F = (3,2,-1). τx = (1)(-1)-(-1)(2) = 1, τy = -[(1)(-1)-(-1)(3)] = -2, τz = (1)(2)-(1)(3) = -1 |τ| = √(1+4+1) = √(6) N m So the answer is (c) √(6) N m.

A force of 3i+2j-k N acts on a particle with position vector i+j-k m… | MHT CET 2026 (19 Apr, Shift 2)

Q1MHT CET 2026 · 19 APR · SHIFT 2Easy

Torque, Couple & Equilibrium of Rigid Bodies

A force of $3\hat{i}+2\hat{j}-\hat{k}$ N acts on a particle with position vector $\hat{i}+\hat{j}-\hat{k}$ m. The magnitude of torque of given force is

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Answer: (c)

$\vec{\tau} = \vec{r}\times\vec{F}$ with $\vec{r} = (1,1,-1)$ , $\vec{F} = (3,2,-1)$ .
$\tau_x = (1)(-1)-(-1)(2) = 1$ , $\tau_y = -[(1)(-1)-(-1)(3)] = -2$ , $\tau_z = (1)(2)-(1)(3) = -1$
$|\vec{\tau}| = \sqrt{1+4+1} = \sqrt{6}$ N m
So the answer is (c) $\sqrt{6}$ N m.

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