MHT CET · Physics · Mechanical Properties of Fluids
Water rises in a capillary tube of radius ' \(r\) ' to a height ' \(h\) '. The mass of water in a capillary is ' \(m\) '. The mass of water will rise in a capillary tube of radius \(\frac{r}{3}\) will be
- A \(3m\)
- B \(\frac{\mathrm{m}}{3}\)
- C \(m\)
- D \(\frac{2 \mathrm{~m}}{3}\)
Answer & Solution
Correct Answer
(B) \(\frac{\mathrm{m}}{3}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{h} \propto \frac{1}{\mathrm{r}} \quad \therefore \frac{\mathrm{h}_2}{\mathrm{~h}_1}=\frac{\mathrm{r}_1}{\mathrm{r}_2}=3 \\ & \therefore \mathrm{h}_2=3 \mathrm{~h}_1 \\ & \mathrm{~m}_1=\mathrm{h}_1 \pi \mathrm{r}_1^2 \rho, \mathrm{m}_2=\mathrm{h}_2 \pi \mathrm{r}_2^2 \rho \\ & \therefore \frac{\mathrm{m}_2}{\mathrm{~m}_1}=\frac{\mathrm{h}_2}{\mathrm{~h}_1}\left(\frac{\mathrm{r}_2}{\mathrm{r}_1}\right)^2=3 \times\left(\frac{1}{3}\right)^2=\frac{1}{3} \\ & \therefore \mathrm{m}_2=\frac{\mathrm{m}_1}{3}\end{aligned}\)
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