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BITSAT 2016
Exam
BITSAT
Year
2016
Subject
Physics
Chapter
Mechanical Properties of Fluids

BITSAT 2016 · Physics · Mechanical Properties of Fluids

A capillary tube of radius r is immersed vertically in a liquid...

A capillary tube of radius r\mathrm{r} is immersed vertically in a liquid such that liquid rises in it to height h\mathrm{h} (less than the length of the tube). Mass of liquid in the capillary tube is m\mathrm{m}. If radius of the capillary tube is increased by 50%50 \%, then mass of liquid that will rise in the tube, is

Options

  1. A23m\frac{2}{3} \mathrm{ m}
  2. Bm\mathrm{m}
  3. C32m\frac{3}{2} \mathrm{ m}
  4. D94m\frac{9}{4} \mathrm{ m}

Correct answer

C. 32m\frac{3}{2} \mathrm{ m}

Step-by-step solution

\mathrm{h}=\frac{2 \mathrm{ T} \cos \theta}{\mathrm{r} \rho \mathrm{g}} \Rightarrow \mathrm{h} \propto \frac{1}{\mathrm{r}} \Rightarrow \frac{\mathrm{h}_{2}}{\mathrm{ h}_{1}}=\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}=\frac{2}{3} (∵r1=r,r2=r+50% of r=32r) New mass m2=πr22h2ρ=π(32r1)2(23h1)ρ=32(πr12h1)ρ=32m\begin{aligned}\left(\because \mathrm{r}_{1}=\mathrm{r}, \quad \mathrm{r}_{2}\right.&=\mathrm{r}+50 \% \text{ of } \left.\mathrm{r}=\frac{3}{2} \mathrm{r}\right) \\ \text{ New mass } \mathrm{m}_{2}=\pi \mathrm{r}_{2}^{2} \mathrm{ h}_{2} \rho&=\pi\left(\frac{3}{2} \mathrm{r}_{1}\right)^{2}\left(\frac{2}{3} \mathrm{ h}_{1}\right) \rho\\ &=\frac{3}{2}\left(\pi \mathrm{r}_{1}^{2} \mathrm{ h}_{1}\right) \rho=\frac{3}{2} \mathrm{ m} \end{aligned}$$
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