TGPSC HMWSSB Manager 2015 Paper 2 · Question 104
Question 104 of the TSPSC Manager (Engineering), HMWSSB, 2015: Paper 2 (Civil Engineering) (1 November 2015). Try it first, then open the answer and explanation.
A weight W falls freely on a body from a height of h. If the instantaneous deformation of the body in the direction of weight is δ, the work done by the force is ______.
- ½ W (h + δ)
- W (h + δ)
- ½ W h + W δ
- W h + ½ W δ
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Answer: (2) W (h + δ)
The weight falls through h and then moves a further δ while the body deforms, so the work done by the constant force W is W(h + δ), which equals the strain energy stored at maximum deformation. The factor ½ applies to a gradually applied load, not to a falling weight.