Physics Elasticity Question 115
Question: Young’s modulus of rubber is $ 10^{4},N/m^{2} $ and area of cross-section is $ 2,cm^{2} $ . If force of $ 2\times 10^{5} $ dynes is applied along its length, then its initial length lbecomes
Options:
A) 3L
B) 4L
C) 2L
D) None of the above
Show Answer
Answer:
Correct Answer: C
Solution:
$ Y=10^{4}N/m^{2},A=2\times {{10}^{-4}}m^{2},F=2\times 10^{5}dyne=2N $
$ l=\frac{FL}{AY}=\frac{2\times L}{2\times {{10}^{-4}}\times 10^{4}}=L $ Final length = initial length + increment = 2L
BETA
