A force of 200N is applied to the rim of a pulley wheel of diameter 200mm. The torque is:a.10Nmb.2Nmc.20Nmd.20kNm
Question
A force of 200N is applied to the rim of a pulley wheel of diameter 200mm. The torque is:a.10Nmb.2Nmc.20Nmd.20kNm
Solution
The torque (τ) is given by the formula:
τ = F * r
where: F is the force applied, r is the radius of the pulley wheel.
Given: F = 200 N, The diameter of the pulley wheel = 200 mm = 0.2 m (since 1 m = 1000 mm), So, the radius r = diameter/2 = 0.2 m/2 = 0.1 m.
Substituting these values into the formula, we get:
τ = 200 N * 0.1 m τ = 20 Nm
So, the torque is 20 Nm. Therefore, the correct answer is c. 20 Nm.
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