Question
Physics Question on mechanical properties of solids
If the work done in stretching a wire by 1mm is 2J, the work necessary for stretching another wire of same material but with double radius of cross-section and half the length by 1mm is
A
16J
B
8J
C
4J
D
41 J
Answer
16J
Explanation
Solution
Stretching force, F= LYπr2ΔL where the symbols have their usual meanings. Both the wires are of same material, so Y will be equal, extension in both the wires is same, so ΔL will be equal. ∴ F ∝ Lr2 ∴ FF′ =(L/2)(2r)2 ×r2L=8 or F′ =8F …(i) Work done in stretching a wire, W=21× F×ΔL For same extension W ∝ F ∴ WW′ =FF′=8 [Using(i)] W′ =8W=8×2J =16J