Physics Elasticity Question 158
Question: The area of cross section of a steel wire $ (Y=2.0\times 10^{11}N/m^{2}) $ is $ 0.1\ cm^{2} $ . The force required to double its length will be
[MP PET 2002]
Options:
A) $ 2\times 10^{12}N $
B)$ 2\times 10^{11}N $
C) $ 2\times 10^{10}N $
D)$ 2\times 10^{6}N $
Show Answer
Answer:
Correct Answer: D
Solution:
When the length of wire is doubled then $ l=L $ and strain = 1 \ Y = strain =$ \frac{F}{A} $ Force = Y × A $ =2\times 10^{11}\times 0.1\times {{10}^{-4}} $
$ =2\times 10^{6}N $
BETA
