A load of 5N gives an extension of 0.56cm in a wire which obeys Hooke's law. What is the extension caused by a load of 20N?
A
1.12cm
B
2.14cm
C
2.24cm
D
2.52cm
Verified Correct Answer: C.
Physical Principles & Numerical Solution
\(\frac{F_1}{e_1}\) = \(\frac{F_2}{e_2}\)
\(\frac{5}{0.56}\) = \(\frac{20}{e_2}\)
e2 = 2.24cm
Source: Myschool Past Questions Archive (Record #28409)