WAEC - Physics (2016 - No. 32)
An aluminium cable of diameter 4 x 10-3m and resistivity 3.0 x 10-8\(\omega\)m has a resistance of 21\(\Omega\). Calculate the length of the cable. (\(\pi = 3.14\))
8.0 x 102m
8.8 x 103m
8.0 x 104m
8.8 x 104m
Explanation
R = \(\frac{4eL}{\pid^2}\)
L = \(\frac{R\pid^2}{4e}\)
= \(\frac{21 \times 3.14(4 \times 10^{-3})^2}{4 \times 3.0 x 10^{-8}}\)
= 8.8 x 103m
L = \(\frac{R\pid^2}{4e}\)
= \(\frac{21 \times 3.14(4 \times 10^{-3})^2}{4 \times 3.0 x 10^{-8}}\)
= 8.8 x 103m
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