QJEEM18Ph6c5a: JEE Main 2018 Physics - Class 11 JEEM18Ph6c5a Solution: When an air bubble of radius r rises from the bottom to the…

Given, bubble of radius r rises from bottom of lake. The radius of bubble at top becomes 5r/4. Therefore, pressure in bubble at bottom is ${P_1} = {P_0} + \rho gh + {{4T} \over r}$

Pressure in bubble at top is ${P_2} = {P_0} + {{4T} \over {5r/4}}$

Now, we know ${P_1}{V_1} = {P_2}{V_2}$ , therefore,

$\left( {{P_0} + \rho gh + {{4T} \over 4}} \right){{4\pi } \over 3}{r^3} = \left( {{P_0} + {{4T} \over {5r/4}}} \right){{4\pi } \over 3}{\left( {{{5r} \over 4}} \right)^3}$

${P_0} + \rho hg + {{4T} \over r} = \left( {{P_0} + {{4T \times 4} \over {5r}}} \right){{125} \over {64}}$

Now given P₀ = 10 $\rho$ g, therefore,

$10\rho g + \rho gh + {{4T} \over r} = {{125} \over {64}} \times 10\rho g + {{16T} \over {5r}} \times {{125} \over {64}}$

Neglecting effect of temperature, we get

$10\rho g + \rho gh = {{125} \over {64}} \times 10\rho g$

$\Rightarrow 10 + h = {{125} \over {64}} \times 10 \Rightarrow h = {{1250} \over {64}} - 10 = 9.53125\,m$

$\Rightarrow$ h $\sim$ 9.5 m