MHT CET · Physics · Mechanical Properties of Fluids
Water rises in a capillary tube of radius ' \(r\) ' upto height ' \(h\) ' The mass of water in the capillary is ' \(m\) ' The mass of water will rise in a capillary of radius will \(\frac{r}{4}\) be
- A \(4 \mathrm{~m}\)
- B \(\frac{\mathrm{m}}{4}\)
- C \(\mathrm{m}\)
- D \(\frac{4}{\mathrm{~m}}\)
Answer & Solution
Correct Answer
(B) \(\frac{\mathrm{m}}{4}\)
Step-by-step Solution
Detailed explanation
The height of water column in a capillary tube is given by
\(\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{perg}}\)
\(\Rightarrow h \propto \frac{1}{r}\quad\ldots(1)\)
also, mass of liquid in a column of radius \(r\) and height \(h\), is
\(m=\left(\pi r^2 h\right) \rho\)
\(\Rightarrow m \propto r^2 h\quad\ldots(2)\)
From eqs. (i) and (ii), we get
\(\begin{aligned} & m \propto r^2 \times \frac{1}{r} \\ & m \propto r \\ & \Rightarrow \frac{m^{\prime}}{m}=\frac{\frac{r}{4}}{r} \Rightarrow m^{\prime}=\frac{m}{4}\end{aligned}\)
\(\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{perg}}\)
\(\Rightarrow h \propto \frac{1}{r}\quad\ldots(1)\)
also, mass of liquid in a column of radius \(r\) and height \(h\), is
\(m=\left(\pi r^2 h\right) \rho\)
\(\Rightarrow m \propto r^2 h\quad\ldots(2)\)
From eqs. (i) and (ii), we get
\(\begin{aligned} & m \propto r^2 \times \frac{1}{r} \\ & m \propto r \\ & \Rightarrow \frac{m^{\prime}}{m}=\frac{\frac{r}{4}}{r} \Rightarrow m^{\prime}=\frac{m}{4}\end{aligned}\)
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