TGPSC HMWSSB Manager 2015 Paper 2 · Question 106

Question 106 of the TSPSC Manager (Engineering), HMWSSB, 2015: Paper 2 (Civil Engineering) (1 November 2015). Try it first, then open the answer and explanation.

1. Manager 2015 Paper 2 (Civil) · Q106 · Strength of Materials & Structural Analysis · Strength of Materials

   A force of magnitude 5 N moves through a distance of 4mm in a direction, inclined at 60° to the direction of force. The magnitude of the work done by the force is ______.

   1. 10√3 N.mm
   2. 0 N.mm
   3. 10 N.mm
   4. 20 N.mm
   Show answer

   Answer: (3) 10 N.mm

   Work = F × d × cos θ = 5 × 4 × cos 60° = 5 × 4 × 0.5 = 10 N.mm. Only the component of displacement along the force does work.

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