Given that \(P\propto \frac{1}{\sqrt{r}}\) and p = 3 when r = 16, find the value of r when p =-5
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Correct Answer: Option B
Explanation:
\(P \propto \frac{1}{\sqrt{r}} P=3\hspace{1mm}r=16\\
P = \frac{k}{\sqrt{r}}\Rightarrow 3 = \frac{k}{\sqrt{16}}\Rightarrow \frac{3}{1} = \frac{k}{4} \\ \Rightarrow K = 12 \Rightarrow K = 12; P\sqrt{r} = k \Rightarrow \sqrt{r} = \frac{k}{p}\\
r = \frac{(k)^2}{P}=12^2 \div \frac{3}{2}=\left(\frac{12}{1}\times \frac{2}{3}\right)=8^2\\
r = 64\)
\(P \propto \frac{1}{\sqrt{r}} P=3\hspace{1mm}r=16\\
P = \frac{k}{\sqrt{r}}\Rightarrow 3 = \frac{k}{\sqrt{16}}\Rightarrow \frac{3}{1} = \frac{k}{4} \\ \Rightarrow K = 12 \Rightarrow K = 12; P\sqrt{r} = k \Rightarrow \sqrt{r} = \frac{k}{p}\\
r = \frac{(k)^2}{P}=12^2 \div \frac{3}{2}=\left(\frac{12}{1}\times \frac{2}{3}\right)=8^2\\
r = 64\)