What percentage increase in the radius of a sphere will cause its volume to increase by 45%?
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Correct Answer: Option n
Explanation:
Let the original volume be V with radius r.
\(V = \frac{4}{3}\pi r^{3}\)
45% increased volume = 145%V.
Let the %age increase in radius = m%r
\(\frac{145}{100}V = \frac{4}{3}\pi (\frac{mr}{100})^{3}\)
\(1.45V = (\frac{4}{3}\pi r^{3})(\frac{m}{100})^{3}\)
\(1.45V = V(\frac{m}{100})^{3}\)
\(\implies 1.45 \times 10^{6} = m^{3}\)
\(m = \sqrt[3]{1.45 \times 10^{6}} = 113.2%\)
\(\therefore \text{%age increase =} 113.2 - 100 = 13.2%\)
\(\approxeq 13%\)
Let the original volume be V with radius r.
\(V = \frac{4}{3}\pi r^{3}\)
45% increased volume = 145%V.
Let the %age increase in radius = m%r
\(\frac{145}{100}V = \frac{4}{3}\pi (\frac{mr}{100})^{3}\)
\(1.45V = (\frac{4}{3}\pi r^{3})(\frac{m}{100})^{3}\)
\(1.45V = V(\frac{m}{100})^{3}\)
\(\implies 1.45 \times 10^{6} = m^{3}\)
\(m = \sqrt[3]{1.45 \times 10^{6}} = 113.2%\)
\(\therefore \text{%age increase =} 113.2 - 100 = 13.2%\)
\(\approxeq 13%\)