Physics Elasticity Question 103
Question:A weight of 200 kg is suspended by vertical wire of length 600.5 cm. The area of cross-section of wire is $ 1,mm^{2} $ . When the load is removed, the wire contracts by 0.5 cm. The Young’s modulus of the material of wire will be
Options:
A) $ 2.35\times 10^{12},N/m^{2} $
B) $ 1.35\times 10^{10},N/m^{2} $
C) $ 13.5\times 10^{11},N/m^{2} $
D) $ 23.5\times 10^{9},N/m^{2} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ F=2000N,\ L=6m,\ l=0.5\ cm,A={{10}^{-6}}m^{2} $
$ Y=\frac{FL}{Al}=\frac{2000\times 6}{{{10}^{-6}}\times 0.5\times {{10}^{-2}}}=2.35\times 10^{12}\ N/m^{2} $
BETA
