MCQ
The mass of water rises in capillary tube of radius $R$ is $M.$ The mass of water that rises in tube of radius $2R$ is:
- A$M$
- ✓$\frac{\text{M}}{2}$
- C$2M$
- D$4M$
✓
Answer
Correct option: B.
$\frac{\text{M}}{2}$
$\text{h}=\frac{2\text{S}\cos\theta}{\text{r}\rho\text{g}}$
i.e.,$\text{ h}\propto\frac{1}{\text{r}}$
$\therefore\frac{\text{h}_1}{\text{h}}=\frac{\text{R}}{\text{2R}}=\frac12$ or $\text{h}_1=\frac{\text{h}}{2}$
$\text{M}=\text{Ah}\rho$
and $\text{M}_1=\text{Ah}_1\rho=\text{A}\Big(\frac{\text{h}}{2}\Big)\rho=\frac{\text{M}}{2}$
i.e.,$\text{ h}\propto\frac{1}{\text{r}}$
$\therefore\frac{\text{h}_1}{\text{h}}=\frac{\text{R}}{\text{2R}}=\frac12$ or $\text{h}_1=\frac{\text{h}}{2}$
$\text{M}=\text{Ah}\rho$
and $\text{M}_1=\text{Ah}_1\rho=\text{A}\Big(\frac{\text{h}}{2}\Big)\rho=\frac{\text{M}}{2}$
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