Physics Elasticity Question 64
Question: When a force is applied on a wire of uniform cross-sectional area $ 3\times {{10}^{-6}},m^{2} $ and length 4m, the increase in length is 1 mm. Energy stored in it will be $ (Y=2\times 10^{11},N/m^{2}) $
[MP PET 1995; Pb. PET 2002]
Options:
A) 6250 J
B) 0.177 J
C) 0.075 J
D) 0.150 J
Show Answer
Answer:
Correct Answer: C
Solution:
$ U=\frac{1}{2}\times \frac{YAl^{2}}{L} $
$ =\frac{1}{2}\times \frac{2\times 10^{11}\times 3\times {{10}^{-6}}\times {{(1\times {{10}^{-3}})}^{2}}}{4} $
$ =0.075\ J $
BETA
