Physics Elasticity Question 81
Question: On increasing the length by 0.5 mm in a steel wire of length 2 m and area of cross-section $ 2,mm^{2} $ , the force required is
[Y for steel$ =2.2\times 10^{11},N/m^{2}] $ ][MP PET/PMT 1988]
Options:
A) $ 1.1\times 10^{5},N $
B) $ 1.1\times 10^{4},N $
C) $ 1.1\times 10^{3},N $
D) $ 1.1\times 10^{2},N $
Show Answer
Answer:
Correct Answer: D
Solution:
$ F=\frac{YAl}{L}=\frac{2.2\times 10^{11}\times 2\times {{10}^{-6}}\times 5,\times {{10}^{-4}}}{2} $
$ =1.1\times 10^{2}N $
BETA
