Physics Elasticity Question 152
Question: A copper wire of length 4.0m and area of cross-section $ 1.2,cm^{2} $ is stretched with a force of $ 4.8\times 10^{3} $ N. If Young’s modulus for copper is $ 1.2\times 10^{11},N/m^{2}, $ the increase in the length of the wire will be
[MP PET 2001]
Options:
A) 1.33 mm
B)1.33 cm
C) 2.66 mm
D)2.66 cm
Show Answer
Answer:
Correct Answer: A
Solution:
$ l=\frac{FL}{AY}=\frac{4.8\times 10^{3}\times 4}{1.2\times {{10}^{-4}}\times 1.2\times 10^{11}}=1.33\ mm $
BETA
