Question

If $Q$ inversely varies as square of $P$ and if $Q=8$ when $P=2$, find $Q$ when $P=2\sqrt{3}$

Answer 1

Since $Q\propto \frac{1}{P^2}$, we get
$Q{p}^2=k,k$ a constant
Putting $Q=8$ and $P=2$, we obtain $k=8\times 2^2=32$. Hence
$Q{P}^2=32$.
Putting $P=2\sqrt{3}$ in this relation we obtain
$Q=\frac{32}{(\sqrt{3}{)}^2}=\frac{32}{12}=\frac{8}{3}$.