Physics Elasticity Question 83
Question: In CGS system, the Young’s modulus of a steel wire is $ 2\times 10^{12} $ . To double the length of a wire of unit cross-section area, the force required is
[MP PMT 1989]
Options:
A) $ 4\times 10^{6} $ dynes
B) $ 2\times 10^{12} $ dynes
C) $ 2\times 10^{12} $ newtons
D) $ 2\times 10^{8} $ dynes
Show Answer
Answer:
Correct Answer: B
Solution:
To double the length of wire, Stress = Young’s modulus \ $ \frac{F}{A}=2\times 10^{12}\frac{dyne}{cm^{2}}. $ If A = 1 then F = 2 × 1012 dyne
BETA
