Q133 / 160 · TS EAMCET 2024 (09 May Shift 2) · Physics
The excess pressure inside a soap bubble of radius 0.5 cm is balanced by the pressure due to an oil column of height 4 mm . If the density of the oil is $900 \mathrm{~kg} \mathrm{~m}^{-3}$, then the surface tension of the soap solution is (Acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )
- A$9 \times 10^{-2} \mathrm{Nm}^{-1}$
- B$2.25 \times 10^{-2} \mathrm{Nm}^{-1}$
- Correct answerC$4.5 \times 10^{-2} \mathrm{Nm}^{-1}$
- D$7 \times 10^{-2} \mathrm{Nm}^{-1}$
Solution
Given,
Radius of soap bubble, $\mathrm{R}=0.5 \mathrm{~cm}$ Height of oil column, $\mathrm{h}=4 \times 10^{-3} \mathrm{~m}$
Density of the oil, $\rho=900 \mathrm{~kg} \mathrm{~m}^{-3}$
Excess pressure inside soap bubble, $\mathrm{P}=\frac{4 \mathrm{~S}}{\mathrm{R}}$
Pressure due to oil column, $\mathrm{P}^{\prime}=\rho g h$
According to question
$\begin{aligned}
& \rho g h=\frac{4 \mathrm{~S}}{\mathrm{R}} \\
& \Rightarrow 900 \times 10 \times 4 \times 10^{-3}=\frac{4 \mathrm{~S}}{0.5 \times 10^{-2}} \\
& \Rightarrow 36000 \times 10^{-3}=8 \mathrm{~S} \times 102 \\
& \Rightarrow \mathrm{~S}=4.5 \times 10^{-2} \mathrm{Nm}^{-1}
\end{aligned}$
Hence, the correct answer is option: 'C'.