= Solution
By <dimensional analysis>, <power> has dimensions $\mathsf{M L^2 T^{-3}}$, <mass density> has dimensions $\mathsf{M L^{-3}}$, and <angular velocity> has dimensions $\mathsf{T^{-1}}$; <radians> are dimensionless. Thus
$$
\mathsf{M L^2 T^{-3}}=\mathsf{M^\alpha L^{\gamma-3\alpha}T^{-\beta}}.
$$
Matching the three independent dimensions gives $\alpha=1$, $\beta=3$ and $\gamma-3\alpha=2$, hence
$$
\boxed{\alpha=1,\quad\beta=3,\quad\gamma=5,\qquad P\propto\rho\omega^3D^5.}
$$
With unchanged <mass density>, <angular velocity> and the same performance coefficient (a <dimensionless quantity>), the <pump power affinity law> gives
$$
\frac{P_{\rm new}}{P_{\rm old}}=\left(\frac{D_{\rm new}}{D_{\rm old}}\right)^5=1.1^5=1.61051.
$$
The required <power> therefore \b[increases by $61.1\%$]. This uses the stipulated dependence on only these three quantities, with comparable pump geometry and performance; <dimensional analysis> alone does not guarantee that unrelated pumps have the same performance coefficient.
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