`16cm^(3)` of water flows per second through a capillary tube of radius a cm and of length l cm when connected to a pressure head of h cm of water if a tube of the 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-

Options

A. `16cm^(3)`
B. `4cm^(3)`
C. `1cm^(3)`
D. `8cm^(3)`

Answer

To solve the problem, we will use the principles of fluid flow through a capillary tube, specifically applying Poiseuille's law. The flow rate \( q \) through a capillary tube is given by the formula:

\[ q = \frac{\pi \Delta P r^4}{8 \eta l} \]

where:
- \( \Delta P \) is the pressure difference,
- \( r \) is the radius of the tube,
- \( \eta \) is the dynamic viscosity of the fluid,
- \( l \) is the length of the tube.

### Step 1: Establish the parameters for the first tube
For the first tube:
- Radius \( r_1 = a \) cm
- Length \( l_1 = l \) cm
- Pressure difference \( \Delta P_1 = h \rho g \)

The flow rate through the first tube is given as:
\[ q_1 = 16 \, \text{cm}^3/\text{s} \]

### Step 2: Write the flow rate equation for the first tube
Using the formula for flow rate:
\[ q_1 = \frac{\pi \Delta P_1 r_1^4}{8 \eta l_1} \]
Substituting the known values:
\[ 16 = \frac{\pi (h \rho g) a^4}{8 \eta l} \]

### Step 3: Establish the parameters for the second tube
For the second tube:
- Radius \( r_2 = \frac{a}{2} \) cm
- Length \( l_2 = l \) cm
- Pressure difference \( \Delta P_2 = h \rho g \) (same as the first tube)

### Step 4: Write the flow rate equation for the second tube
Using the same flow rate formula for the second tube:
\[ q_2 = \frac{\pi \Delta P_2 r_2^4}{8 \eta l_2} \]
Substituting the known values:
\[ q_2 = \frac{\pi (h \rho g) \left(\frac{a}{2}\right)^4}{8 \eta l} \]

### Step 5: Simplify the expression for \( q_2 \)
Calculating \( r_2^4 \):
\[ r_2^4 = \left(\frac{a}{2}\right)^4 = \frac{a^4}{16} \]
Thus, substituting this back into the equation for \( q_2 \):
\[ q_2 = \frac{\pi (h \rho g) \frac{a^4}{16}}{8 \eta l} \]
This simplifies to:
\[ q_2 = \frac{1}{16} \cdot \frac{\pi (h \rho g) a^4}{8 \eta l} \]

### Step 6: Relate \( q_2 \) to \( q_1 \)
From the first tube's equation, we know:
\[ \frac{\pi (h \rho g) a^4}{8 \eta l} = 16 \]
Thus:
\[ q_2 = \frac{1}{16} \cdot 16 = 1 \, \text{cm}^3/\text{s} \]

### Final Answer
The quantity of water flowing through the tube per second when the radius is \( \frac{a}{2} \) cm is:
\[ \boxed{1 \, \text{cm}^3/\text{s}} \]