Let surface charge density of both the spheres are $$\sigma$$ and their charges are $$\mathrm{q}_{1}$$
and $$\mathrm{q}_{2}$$
$$\therefore
q_{1}=\sigma \cdot A_{1}=\sigma \cdot 4 \pi R^{2}$$
$$q_{2}=\sigma
\cdot A_{2}=\sigma \cdot 4 \pi(2 R)^{2}=\sigma \cdot 4 \pi R^{2} \cdot 4=4
q_{1}$$
Both charged spheres are kept in contact, so charge flows between them and their potential becomes
equal, let the charges on them now become $$q_{1}$$ 'and $$q_{2}$$ ‘.
$$\mathrm{So},
\mathrm{V}_{1}=\mathrm{V}_{2}\left(\because V=\dfrac{k q}{r}\right)$$
$$\mathrm{So}
\dfrac{\mathrm{kq}_{1}}{R}=\dfrac{\mathrm{kq}_{2}}{(2
\mathrm{R})}\left[\because k=\dfrac{1}{4 \pi \varepsilon_{0}}\right]$$
Where $$q_{1}$$ 'and $$q_{2}$$ ' are the charges on spheres after redistribution of charges
$$\dfrac{q_{1}}{R}=\dfrac{q_{2}}{2
R}$$
$$\therefore
q_{2}^{\prime}-2 q_{1}^{\prime} \dots I.$$
By law of conservation of charges
$$\mathrm{q}_{1}+\mathrm{q}_{2}=q_{1}+q_{2}$$
$$q_{1}+4
q_{1}=q_{1}+2 q_{1}(\text { from } D)$$
$$5 q_{1}=3
q_{1}$$
$$q_{1}’=\dfrac{20}{3}
\pi R^{2} \sigma$$
$$\sigma_{1}=\dfrac{q_{1}’}{A_{1}}=\dfrac{\dfrac{20}{3}
\pi R^{2} \sigma}{4 \pi R^{2}}=\dfrac{5}{3} \sigma$$
$$\sigma_{2}=\dfrac{q_{2}’}{A_{2}}=\dfrac{2
. q_{1}’}{4 . \pi(2 R)^{2}}=\dfrac{2 . \dfrac{20}{3} \pi R^{2} \sigma}{4 \pi .4
R^{2}}=\dfrac{5}{6} \sigma$$
Hence, $$\sigma_{1}=\dfrac{5}{3}
\sigma$$ and $$\sigma_{2}=\dfrac{5}{6} \sigma$$