JAMB - Mathematics (2011 - No. 15)
T varies inversely as the cube of R. When R = 3, T = \(\frac{2}{81}\), find T when R = 2
\(\frac{1}{18}\)
\(\frac{1}{12}\)
\(\frac{1}{24}\)
\(\frac{1}{6}\)
Explanation
T \(\alpha \frac{1}{R^3}\)
T = \(\frac{k}{R^3}\)
k = TR3
= \(\frac{2}{81}\) x 33
= \(\frac{2}{81}\) x 27
dividing 81 by 27
k = \(\frac{2}{2}\)
therefore, T = \(\frac{2}{3}\) x \(\frac{1}{R^3}\)
When R = 2
T = \(\frac{2}{3}\) x \(\frac{1}{2^3}\) = \(\frac{2}{3}\) x \(\frac{1}{8}\)
= \(\frac{1}{12}\)
T = \(\frac{k}{R^3}\)
k = TR3
= \(\frac{2}{81}\) x 33
= \(\frac{2}{81}\) x 27
dividing 81 by 27
k = \(\frac{2}{2}\)
therefore, T = \(\frac{2}{3}\) x \(\frac{1}{R^3}\)
When R = 2
T = \(\frac{2}{3}\) x \(\frac{1}{2^3}\) = \(\frac{2}{3}\) x \(\frac{1}{8}\)
= \(\frac{1}{12}\)
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