Question

Two capillary tubes of the diameter 5mm and 4mm axe are held vertically inside water one by one. How much high the water will rise in each tube (Given g = 10$m/{{s}^{2}}$ and the surface tension water =$7\times {{10}^{-2}}N/{{m}^{2}}$)
A) 5mm and 7mm
B) 5.6mm and 7mm
C) 12mm and 5mm
D) 10mm and 5.6mm

Answer 1

Hint: A very narrow tube of the glass with a fine bore and open at both the ends are called capillary tubes when in a capillary tube is dipped in a liquid the liquid in the tube will raise or fall, in order to measure height of the capillary tube, we have to use the formula height for capillary tube.
Formula used:
$h=\dfrac{2T\cos \theta }{\rho rg}....\left( 1 \right)$

Complete answer:
Given data:-
$\begin{align}
  & g=10m/{{s}^{2}} \\
 & T=7.0\times {{10}^{2}}N/m \\
\end{align}$
Now in order to know total rise in the water we have to use the below formula
$h=\dfrac{2T\cos \theta }{\rho rg}....\left( 1 \right)$
Where, h = height by liquid in the capillary tube
T = surface tension
$\rho $= density of the given liquid.
r = radius of the capillary tube.
g = gravitational acceleration.
Θ = angle of the contact.
Now for the tube with the diameter 5.0 mm:-
d = 5.0 mm
Therefore radius,
$r=\dfrac{5}{2}mm=2.5mm$
Converting unit from the mm to m
$\therefore r=2.5\times {{10}^{-3}}m$
Now substitute values in the equation (1)
${{h}_{1}}=\dfrac{2\times 7\times {{10}^{-2}}\times \cos \left( {{0}^{o}} \right)}{1\times {{10}^{3}}\times 25\times {{10}^{-3}}\times 10}$
For the water angle of the contact is approximately ${{0}^{o}}\left( \therefore \theta ={{0}^{o}} \right)$ and the value of the density is $1\times {{10}^{3}}kg/{{m}^{3}}\left( \rho ={{10}^{3}}kg/{{m}^{3}} \right)$
$\begin{align}
  & \Rightarrow {{h}_{1}}=\dfrac{14\times {{10}^{-2}}\times 1}{25} \\
 & \therefore {{h}_{1}}=0.56\times {{10}^{-2}}m \\
\end{align}$
Converting unit from m to mm
$\therefore {{h}_{1}}=5.6mm$
Now the height for the capillary tube with the diameter d = 4.0mm
Therefore the radius is,
r = 2mm
Now converting unit from mm to m
$\therefore r=2\times {{10}^{-3}}m$
Now substitute value in the equation (1)
$\begin{align}
  & {{h}_{2}}=\dfrac{2\times 7\times {{10}^{-2}}\times \cos \left( {{0}^{o}} \right)}{1\times {{10}^{3}}\times 25\times {{10}^{-3}}\times 10} \\
 & {{h}_{2}}=\dfrac{7}{10}\times {{10}^{-2}}m \\
\end{align}$
Now converting units from m to mm
$\therefore {{h}_{2}}=7mm$

Therefore correct option is (b) 5.6mm and 7mm.

Note:
We often get confused that first we convert units of the radius of the capillary tube from mm to m but in the final answer we again convert units from the m to mm the reason behind that is the surface tension of the water and the acceleration of the gravity is given in the meter so if we do not convert units of the given radius then the answer will be wrong.