Physics Elasticity Question 67
Question: A wire of length 50 cm and cross sectional area of 1 sq. mm is extended by 1 mm. The required work will be $ (Y=2\times 10^{10},N{{m}^{-2}}) $
[RPET 1999]
Options:
A) $ 6\times {{10}^{-2}},J $
B) $ 4\times {{10}^{-2}},J $
C) $ 2\times {{10}^{-2}},J $
D) $ 1\times {{10}^{-2}},J $
Show Answer
Answer:
Correct Answer: C
Solution:
$ W=\frac{YAl^{2}}{2L}=\frac{2\times 10^{10}\times {{10}^{-6}}\times {{({{10}^{-3}})}^{2}}}{2\times 50\times {{10}^{-2}}} $
$ =2\times {{10}^{-2}}\ J $
BETA
