MHT CET · Physics · Mechanical Properties of Fluids
What should be the radius of water drop so that excess pressure inside it is \(72 \mathrm{Nm}^{-2}\) ? (The surface tension of water \(7.2 \times 10^{-2} \mathrm{Nm}^{-1}\) )
- A \(1 \mathrm{~mm}\)
- B \(2 \mathrm{~mm}\)
- C \(8\mathrm{~mm}\)
- D \(4 \mathrm{~mm}\)
Answer & Solution
Correct Answer
(B) \(2 \mathrm{~mm}\)
Step-by-step Solution
Detailed explanation
Excess pressure in a water drop \(=\frac{2 T}{R}\)
\(
\begin{aligned}
& \mathrm{T}=7.2 \times 10^{-2} \mathrm{Nm}^{-2} \\
& \therefore 72=\frac{2 \times 7.2 \times 10^{-2}}{\mathrm{R}} \\
& \therefore \mathrm{R}=\frac{2 \times 7.2 \times 10^{-2}}{72}=2 \times 10^{-3} \mathrm{~m}=2 \mathrm{~mm}
\end{aligned}
\)
\(
\begin{aligned}
& \mathrm{T}=7.2 \times 10^{-2} \mathrm{Nm}^{-2} \\
& \therefore 72=\frac{2 \times 7.2 \times 10^{-2}}{\mathrm{R}} \\
& \therefore \mathrm{R}=\frac{2 \times 7.2 \times 10^{-2}}{72}=2 \times 10^{-3} \mathrm{~m}=2 \mathrm{~mm}
\end{aligned}
\)
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