MHT CET · Physics · Mechanical Properties of Fluids
If a capillary tube of radius \(1 \mathrm{~mm}\) is immersed in water, the mass of water rising in the capillary tube is \(\mathrm{m}\). If the radius of the capillary tube is doubled, then the mass of water that rises in the same capillary tube will be
- A \(3 \mathrm{~m}\)
- B \(\mathrm{m} / 2\)
- C \(\mathrm{m}\)
- D \(2 \mathrm{~m}\)
Answer & Solution
Correct Answer
(D) \(2 \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
For capillary rise, \(h_1 r_1=h_2 r_2\)
\(\therefore \frac{h_2}{h_1}=\frac{r_1}{r_2}=\frac{1}{2}\)
And \(m=\pi r^2 h_1 \rho\) and \(M=p(2 r)^2 \times h_2 \times \rho\)
\(\begin{aligned} & \therefore \frac{M^{\prime}}{M}=\left(\frac{2 r}{r}\right)^2 \times\left(\frac{1}{2}\right)=4 \times \frac{1}{2}=2 \\ & \therefore M=2 \mathrm{~m}\end{aligned}\)
\(\therefore \frac{h_2}{h_1}=\frac{r_1}{r_2}=\frac{1}{2}\)
And \(m=\pi r^2 h_1 \rho\) and \(M=p(2 r)^2 \times h_2 \times \rho\)
\(\begin{aligned} & \therefore \frac{M^{\prime}}{M}=\left(\frac{2 r}{r}\right)^2 \times\left(\frac{1}{2}\right)=4 \times \frac{1}{2}=2 \\ & \therefore M=2 \mathrm{~m}\end{aligned}\)
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