Pressure inside two soap bubbles are 1 . 01 and 1 . 02 atmosphere,...

Excess pressure inside a soap bubble is $\dfrac{4T}{R}$. Therefore,

${P}_{1} - {P}_{0} = \dfrac{4T}{{R}_{1}} = 0 . 01atm$

${P}_{2} - {P}_{0} = \dfrac{4T}{{R}_{2}} = 0 . 02atm$

Dividing,

$\dfrac{1}{2} = \dfrac{{R}_{2}}{{R}_{1}}$

$⇒ {R}_{1} = 2{R}_{2}$

$⇒ \dfrac{{V}_{1}}{{V}_{2}} = {R}_{1}^{3}{R}_{2}^{3} = 8{R}_{2}^{3}{R}_{2}^{3} = \dfrac{8}{1}$