Use the following data to determine the length L of a wire when a fore of 30N is applied, assuming Hooke’s law is obeyed
\(\begin{array}{c|c}\text{Force applied/N} & 0 & 5 & 10 & 30\\ \hline \text{length of wire/mm} & 500.0 & 500.0 & 501.0 & L \end{array}\)
- 3.0mm
- 3.5mm
-
503.00mm
- 503.5mm
- 506.0mm
The correct answer is: C
Explanation
F \(\alpha\) eF = ke
K = \(\frac{F}{e} = \frac{5}{0.5} = 10\)
when F = 30, e = \(\frac{F}{K} = \frac{30}{10} = 3\)
L = 500 + 3 = 503.00