A force of 3 i ^ + 2 j ^ − k ^ N acts on a particle with position vector i ^ + j ^ − k ^ m. The magnitude of torque of given force is

Given force F = 3 i ^ + 2 j ^ − k ^ N Position vector r = i ^ + j ^ − k ^ m Torque τ = r × F τ = i ^ j ^ k ^ 1 1 − 1 3 2 − 1 τ = i ^ ( − 1 − ( − 2 )) − j ^ ( − 1 − ( − 3 )) + k ^ ( 2 − 3 ) τ = i ^ − 2 j ^ − k ^ Magnitude of torque ∣ τ ∣ = 1 2 + ( − 2 ) 2 + ( − 1 ) 2 = 1 + 4 + 1 = 6 N m

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Physics - Rotation Question with Solution | TestHub

PhysicsRotationTorque / angular accelerationMedium2 minPYQ_2026

PhysicsMediumsingle choice

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:

Solution:

Given force $F = 3 \hat{i} + 2 \hat{j} - \hat{k}$ N

Position vector $r = \hat{i} + \hat{j} - \hat{k}$ m

Torque $τ = r \times F$

$τ = \hat{i} \hat{j} \hat{k} 1 1 - 13 2 - 1$

$τ = \hat{i} (- 1 - (- 2)) - \hat{j} (- 1 - (- 3)) + \hat{k} (2 - 3)$

$τ = \hat{i} - 2 \hat{j} - \hat{k}$

Magnitude of torque $∣ τ ∣ = 1^{2} + (- 2)^{2} + (- 1)^{2} = 1 + 4 + 1 = 6$ N m

Stream:MHT_CETSubject:PhysicsTopic:RotationSubtopic:Torque / angular acceleration

⏱ 2mℹ️ Source: PYQ_2026

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