Physics Elasticity Question 87
Question: The diameter of a brass rod is 4 mm and Young’s modulus of brass is $ 9\times 10^{10},N/m^{2} $ . The force required to stretch by 0.1% of its length is
[MP PET 1991; BVP 2003]
Options:
A) $ 360,\pi N $
B) 36 N
C) $ 144\pi \times 10^{3}N $
D) $ 36\pi \times 10^{5}N $
Show Answer
Answer:
Correct Answer: A
Solution:
$ F=\frac{YAl}{L}=\frac{9\times 10^{10}\times \pi \times 4\times {{10}^{-6}}\times 0.1}{100}=360\ \pi \ N $
BETA
