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