$(a)$ Given, drag force $F_{d}$ depends on air density $\sigma$, velocity of ball $v$ and area of cross-section of ball $A$.

$\therefore \quad F_{d} \propto \sigma^{\alpha}\propto v^{b} A^{c}$

Substituting dimensions of different physical quantities, we have

$\left[ MLT ^{-2}\right]=\left[ ML ^{-3}\right]^{\alpha}\left[ LT ^{-1}\right]^{b}\left[ L ^{2}\right]^{c}$

Equating dimensions of fundamental quantities, we have

$a=1 \quad \dots(i)$

$-3 a+b+2 c=1\quad \dots(ii)$

$-b=-2 \Rightarrow b=2\quad \dots(iii)$

Putting the values of $a$ and $b$ in Eq $(ii)$, we get

$\Rightarrow \quad-3 \times 1+2+2 c=1 \Rightarrow c=1$

$\because a=1, b=2, c=1$

So, drag force on football is

$F_{d}=k \cdot \sigma \cdot v^{2} \cdot A$

When football reaches terminal speed, its weight is balanced by drag force.

$\Rightarrow m g=F_{d}$

$\Rightarrow m g=k \sigma v_{T}^{2} A$

where, $u_{T}=$ terminal speed of football.

$\Rightarrow \quad m g=k \sigma v_{T}^{2} \pi R^{2}$

$\Rightarrow \quad m g=k \sigma u_{T}^{2} \pi\left(\frac{m}{\frac{4}{3} \pi \rho}\right)^{\frac{2}{3}}$

[from $m=\frac{4}{3} \pi R^{3} \rho$,

where $\rho=$ density of football]

$\Rightarrow \quad u_{T} \propto m^{1 / 6}$

$\Rightarrow \quad \frac{v_{1}}{v_{2}}=\left(\frac{m_{1}}{m_{2}}\right)^{\frac{1}{6}}=\left(\frac{250}{125}\right)^{\frac{1}{6}}=2^{1 / 6}$