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?
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Correct Answer: Option C
Explanation:
\(\frac{F_1}{e_1}\) = \(\frac{F_2}{e_2}\)
\(\frac{5}{0.56}\) = \(\frac{20}{e_2}\)
e2 = 2.24cm
\(\frac{F_1}{e_1}\) = \(\frac{F_2}{e_2}\)
\(\frac{5}{0.56}\) = \(\frac{20}{e_2}\)
e2 = 2.24cm