If a capillary tube of radius $1 \mathrm{~mm}$ is immersed in water, the mass of water rising in the…

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
  1. $3 \mathrm{~m}$
  2. $\mathrm{m} / 2$
  3. $\mathrm{m}$
  4. $2 \mathrm{~m}$

Solution

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 $M=p(2 r)^2 \times h_2 \times \rho$ $\begin{aligned} & \therefore \frac{M^{\prime}}{M}=\left(\frac{2 r}{r}\right)^2 \times\left(\frac{1}{2}\right)=4 \times \frac{1}{2}=2 \\ & \therefore M=2 \mathrm{~m}\end{aligned}$

Asked in: MHT CET 2022 (05 Aug Shift 2)

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