Q116 / 150 · MHT CET 2026 (17 April Shift 2) · Physics
When a capillary tube of radius 'r' is immersed in water, rise of water is upto height 'h'. The mass of water in capillary tube is 'm'. When another capillary tube of radius 'xr' is immersed in water, the mass of water that will rise in this tube is
- Correct answerA$x \cdot m$
- B$\dfrac{m}{x}$
- C$x^2 m$
- D$(x + 1)m$
Solution
The height of water rise in a capillary tube is given by Jurin's law:
$h = \dfrac{2T \cos \theta}{r \rho g}$
The mass of water in the capillary tube is:
$m = \text{Volume} \times \text{density} = \pi r^2 h \rho$
Substituting the expression for $h$ into the mass equation:
$m = \pi r^2 \left( \dfrac{2T \cos \theta}{r \rho g} \right) \rho = \dfrac{2 \pi r T \cos \theta}{g}$
This shows that the mass of water rising in the capillary tube is directly proportional to its radius:
$m \propto r$
For the first capillary tube of radius $r$, the mass is $m$. For the second capillary tube of radius $xr$, the new mass $m'$ will be:
$m' = x \cdot m$
Answer: $x \cdot m$
$h = \dfrac{2T \cos \theta}{r \rho g}$
The mass of water in the capillary tube is:
$m = \text{Volume} \times \text{density} = \pi r^2 h \rho$
Substituting the expression for $h$ into the mass equation:
$m = \pi r^2 \left( \dfrac{2T \cos \theta}{r \rho g} \right) \rho = \dfrac{2 \pi r T \cos \theta}{g}$
This shows that the mass of water rising in the capillary tube is directly proportional to its radius:
$m \propto r$
For the first capillary tube of radius $r$, the mass is $m$. For the second capillary tube of radius $xr$, the new mass $m'$ will be:
$m' = x \cdot m$
Answer: $x \cdot m$
Hence, the correct answer is option: 'A'.