Physics Elasticity Question 143
Question: A steel wire of lm long and $ 1,mm^{2} $ cross section area is hang from rigid end. When weight of 1kg is hung from it then change in length will be (given $ Y=2\times 10^{11}N/m^{2}) $
[RPMT 2000]
Options:
A) 0.5 mm
B)0.25 mm
C) 0.05 mm
D)5 mm
Show Answer
Answer:
Correct Answer: C
Solution:
$ l=\frac{MgL}{YA}=\frac{1\times 10\times 1}{2\times 10^{11}\times {{10}^{-6}}}=0.05\ mm $
BETA
