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?
- 1.12cm
- 2.14cm
-
2.24cm
- 2.52cm
The correct answer is: C
Explanation
\(\frac{F_1}{e_1}\) = \(\frac{F_2}{e_2}\)\(\frac{5}{0.56}\) = \(\frac{20}{e_2}\)
e2 = 2.24cm