Two tubes A and B are in series. The radius of A is R and that of B is 2R. If water flows through A with velocity v then the velocity of water through B is

Options

A. `v/2`
B. v
C. `v/4`
D. `v/8`

Answer

To solve the problem, we will use the principle of conservation of mass, which is expressed through the continuity equation. The continuity equation states that for an incompressible fluid flowing through a pipe, the product of the cross-sectional area (A) and the velocity (v) of the fluid remains constant along the flow.

### Step-by-Step Solution:

1. **Identify the Areas of the Tubes**:
- For tube A, the radius is \( R \).
- The cross-sectional area \( A_1 \) of tube A is given by:
\[ A_1 = \pi R^2 \]
- For tube B, the radius is \( 2R \).
- The cross-sectional area \( A_2 \) of tube B is given by:
\[ A_2 = \pi (2R)^2 = \pi \cdot 4R^2 = 4\pi R^2 \]

2. **Apply the Continuity Equation**:
- According to the continuity equation:
\[ A_1 V_1 = A_2 V_2 \]
- Where \( V_1 \) is the velocity of water in tube A (given as \( v \)), and \( V_2 \) is the velocity of water in tube B (which we need to find).

3. **Substitute the Known Values**:
- Substitute \( A_1 \), \( V_1 \), and \( A_2 \) into the continuity equation:
\[ (\pi R^2)(v) = (4\pi R^2)(V_2) \]

4. **Simplify the Equation**:
- Cancel \( \pi R^2 \) from both sides:
\[ v = 4 V_2 \]

5. **Solve for \( V_2 \)**:
- Rearranging the equation to find \( V_2 \):
\[ V_2 = \frac{v}{4} \]

### Conclusion:
The velocity of water through tube B is:
\[ V_2 = \frac{v}{4} \]