Physics Elasticity Question 115
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 l becomes
Options:
3L
4L
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=2\times 10^{-1}N $
$ l=\frac{FL}{AY}=\frac{2\times L}{2\times {{10}^{-4}}\times 10^{4}}=L $ Final length = initial length + increment = L
BETA
