Physics Elasticity Question 141
Question: An iron rod of length 2m and cross section area of $ 50,mm^{2} $ , stretched by 0.5 mm, when a mass of 250 kg is hung from its lower end. Young’s modulus of the iron rod is
[AFMC 1999]
Options:
A) $ 19.6\times 10^{10},N/m^{2} $
B) $ 19.6\times 10^{15},N/m^{2} $
C) $ 19.6\times 10^{18},N/m^{2} $
D) $ 19.6\times 10^{20},N/m^{2} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ Y=\frac{MgL}{Al}=\frac{250\times 9.8\times 2}{50\times {{10}^{-6}}\times 0.5\times {{10}^{-3}}} $
$ =19.6\times 10^{10}N/m^{2} $
BETA
