Physics Elasticity Question 97
Question: An area of cross-section of rubber string is $ 2,cm^{2} $ . Its length is doubled when stretched with a linear force of $ 2\times 10^{5} $ dynes. The Young’s modulus of the rubber in $ dyne/cm^{2} $ will be
[MP PET 1985]
Options:
A) $ 4\times 10^{5} $
B) $ 1\times 10^{5} $
C) $ 2\times 10^{5} $
D) $ 1\times 10^{4} $
Show Answer
Answer:
Correct Answer: B
Solution:
If length of the wire is doubled then strain = 1 Y = $ r=\frac{r}{r} $ = $ \frac{2\times 10^{5}}{2}=10^{5}\frac{dyne}{cm^{2}} $
BETA
