Physics Elasticity Question 112
Question: The length of a wire is 1.0 m and the area of cross-section is $ 1.0\times {{10}^{-2}},cm^{2} $ . If the work done for increase in length by 0.2 cm is 0.4 joule, then Young’s modulus of the material of the wire is
Options:
A) $ 2.0\times 10^{10},N/m^{2} $
B) $ 4\times 10^{10},N/m^{2} $
C) $ 2.0\times 10^{11},N/m^{2} $
D) $ 2\times 10^{10},N/m^{2} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ W=\frac{1}{2}\frac{YAl^{2}}{L} $
Therefore $ 0.4=\frac{1}{2}\times \frac{Y\times {{1}^{-6}}\times {{(0.2\times {{10}^{-2}})}^{2}}}{1} $ Y$ =2\times 10^{11}N/m^{2} $
BETA
