Question

Given, 16 c.c. of water flows per second through a capillary tube of radius a cm and of length l cm when connected to a pressure head h cm of water. If a tube of same length and radius a/2 cm is connected to the same pressure head, the quantity of water flowing through the tube per second will be
A. 16c.c
B. 4c.c
C. 1c.c
D. 8c.c

Answer 1

At first we need to write all the values that are given in the question all the possible values that we can get, then we have to form the equation for volumes for both the capillary tube and then divide the equations of volumes for both the capillary tube. From this we will get the relation between the volume of the first tube and the volume of the second tube replaced with the values that are given in the question to get the required result.

Formula Used:
${{V}_{1}}=\dfrac{\pi {{p}_{1}}r_{1}^{4}}{8\pi {{l}_{1}}}$
${{V}_{2}}=\dfrac{\pi {{p}_{2}}r_{2}^{4}}{8\pi {{l}_{2}}}$

Complete answer:
According to the question,
For the first tube,
${{V}_{1}}=16c{{m}^{3}}/\sec$ of water is flowing,
Radius of the capillary tube is ${{r}_{1}}=a\text{ cm}$
And the length of the tube is ${{l}_{1}}=l\text{ cm}$ .
Pressure =${{P}_{1}}=\rho gh$
Pressure head of the water is h cm.
For the second tube,
${{V}_{2}}=?$ ${{P}_{2}}=\rho gh$ ${{l}_{2}}=l$
Radius of the second tube is ${{r}_{2}}=\dfrac{a}{2}$ cm.
So, for the first case and the second case the volume of water flowing in the 1st capillary and the second capillary respectively would be,
${{V}_{1}}=\dfrac{\pi {{p}_{1}}r_{1}^{4}}{8\pi {{l}_{1}}}\ \;\text{and}\ \;{{V}_{2}}=\dfrac{\pi {{p}_{2}}r_{2}^{4}}{8\pi {{l}_{2}}}$, respectively.
Now, if we divide both the volumes, we get
$\therefore \dfrac{{{V}_{2}}}{{{V}_{1}}}=\dfrac{{{p}_{2}}}{{{P}_{1}}}\times \dfrac{r_{2}^{4}}{r_{1}^{4}}\times \dfrac{{{l}_{1}}}{{{l}_{2}}}$
On further solving,
$\dfrac{{{V}_{2}}}{{{V}_{1}}}=\dfrac{{{(a/2)}^{4}}}{{{a}^{4}}}\times \dfrac{l}{l}$
Now, placing the values in the equation
$\dfrac{{{V}_{2}}}{{{V}_{1}}}={{(\dfrac{1}{2})}^{4}}=\dfrac{1}{16}$
Now, we know that ${{V}_{2}}=\dfrac{{{V}_{1}}}{16}$
${{V}_{2}}=\dfrac{16}{16}=1c{{m}^{3}}/\sec$, which means $60c{{m}^{3}}/\min$

So, now according to the given explanation option C, that is 1cc is the correct answer.

Note:
In the equation ${{V}_{1}}=\dfrac{\pi {{p}_{1}}r_{1}^{4}}{8\pi {{l}_{1}}}$, r is the radius of the capillary tube, l is the length of the tube, P is the pressure in the tube. We have to find a ratio between the volume of the first and the second capillary tube in that way only we can solve the question.