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Question:

The magnitude of torque on a particle of mass 1kg is 2.5 Nm about the origin. If the force acting on it is 1 N, and the distance of the particle from the origin is 5 m, the angle between the force and the position vector is?

π/3

π/8

π/6

π/4

Solution:

Torque (τ) is given by the formula τ = rFsinθ, where r is the distance from the origin, F is the force, and θ is the angle between the force and the position vector.

We are given:
τ = 2.5 Nm
F = 1 N
r = 5 m

Substituting these values into the formula, we get:
2.5 = 5 × 1 × sinθ
2.5 = 5sinθ
sinθ = 2.5/5
sinθ = 0.5
θ = sin⁻¹(0.5)
θ = π/6