Answer
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Hint: In order to solve this question, we should know that surface area means the area of the outside surface of any object and here we will use the general formula of pressure inside a soap bubble for both cases and then will compare both equations in order to find the relation between surface area of soap bubbles for both cases.
Complete answer:
According to the question, we have given that pressure inside of two soap bubbles as;
For soap bubble one, $P = 1.02atm$ and as we know that, the relation between tension force inside the soap bubble and radius of spherical bubble is,
$\Delta P = \dfrac{{4T}}{R}$
where $T$ is surface tension force acting on the spherical soap bubble of radius $R$.
For first soap bubble we have the error change in given pressure is,
$\Delta P = 0.02\,atm$
$\Rightarrow 0.02 = \dfrac{{4T}}{{{R_1}}} \to (i)$
For soap bubble second having pressure $P = 1.05\,atm$
again using same formula keeping the surface tension same we have,
$\Delta P = \dfrac{{4T}}{R}$ and for this soap bubble $\Delta P = 0.05\,atm$
$\Rightarrow 0.05 = \dfrac{{4T}}{{{R_2}}} \to (ii)$
where ${R_2}$ denotes the radius of spherical bubble soap.
Divide the equation (ii) by equation (i) we have,
$\dfrac{5}{2} = \dfrac{{{R_1}}}{{{R_2}}}$
and now since surface area of sphere is given by,
$S = 4\pi {R^2}$
where $R$ is radius and surface area is directly proportional to square of radius so we have, squaring the equation $\dfrac{5}{2} = \dfrac{{{R_1}}}{{{R_2}}}$ we get,
$\Rightarrow \dfrac{{25}}{4} = \dfrac{{4\pi {{({R_1})}^2}}}{{4\pi {{({R_2})}^2}}}$
$\therefore \dfrac{{{S_1}}}{{{S_2}}} = \dfrac{{25}}{4}$
where, $S$ denotes the surface area of the soap bubble.
Hence, the correct option is B.
Note: It should be remembered that, we have given the pressure inside the soap bubble with two digits after decimal and we use the change in pressure from inside and outside in the formula $\Delta P = \dfrac{{4T}}{R}$ so, change in pressure $\Delta P$ is just the difference between given pressure and atmospheric pressure which has a value of $1atm$.
Complete answer:
According to the question, we have given that pressure inside of two soap bubbles as;
For soap bubble one, $P = 1.02atm$ and as we know that, the relation between tension force inside the soap bubble and radius of spherical bubble is,
$\Delta P = \dfrac{{4T}}{R}$
where $T$ is surface tension force acting on the spherical soap bubble of radius $R$.
For first soap bubble we have the error change in given pressure is,
$\Delta P = 0.02\,atm$
$\Rightarrow 0.02 = \dfrac{{4T}}{{{R_1}}} \to (i)$
For soap bubble second having pressure $P = 1.05\,atm$
again using same formula keeping the surface tension same we have,
$\Delta P = \dfrac{{4T}}{R}$ and for this soap bubble $\Delta P = 0.05\,atm$
$\Rightarrow 0.05 = \dfrac{{4T}}{{{R_2}}} \to (ii)$
where ${R_2}$ denotes the radius of spherical bubble soap.
Divide the equation (ii) by equation (i) we have,
$\dfrac{5}{2} = \dfrac{{{R_1}}}{{{R_2}}}$
and now since surface area of sphere is given by,
$S = 4\pi {R^2}$
where $R$ is radius and surface area is directly proportional to square of radius so we have, squaring the equation $\dfrac{5}{2} = \dfrac{{{R_1}}}{{{R_2}}}$ we get,
$\Rightarrow \dfrac{{25}}{4} = \dfrac{{4\pi {{({R_1})}^2}}}{{4\pi {{({R_2})}^2}}}$
$\therefore \dfrac{{{S_1}}}{{{S_2}}} = \dfrac{{25}}{4}$
where, $S$ denotes the surface area of the soap bubble.
Hence, the correct option is B.
Note: It should be remembered that, we have given the pressure inside the soap bubble with two digits after decimal and we use the change in pressure from inside and outside in the formula $\Delta P = \dfrac{{4T}}{R}$ so, change in pressure $\Delta P$ is just the difference between given pressure and atmospheric pressure which has a value of $1atm$.
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