Physics Elasticity Question 93
Question: To double the length of a iron wire having $ 0.5,cm^{2} $ area of cross-section, the required force will be $ (Y=10^{12},dyne/cm^{2}) $
[MP PMT 1987]
Options:
A) $ 1.0\times {{10}^{-7}}N $
B) $ 1.0\times 10^{7}N $
C) $ 0.5\times {{10}^{-7}}N $
D) $ 0.5\times 10^{12} $ dyne
Show Answer
Answer:
Correct Answer: D
Solution:
If length of wire doubled then strain = 1 $ Y=r $
Therefore $ F=Y\times A $
$ =10^{12}\times 0.5 $
$ =0.5\times 10^{12}dyne $
BETA
