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 represented by the equation of continuity for fluid flow.

### Step-by-Step Solution:

1. **Understand the Setup**:
- We have two tubes, A and B, in series.
- The radius of tube A is R.
- The radius of tube B is 2R.
- The velocity of water flowing through tube A is v.

2. **Apply the Equation of Continuity**:
- The equation of continuity states that the product of the cross-sectional area (A) and the velocity (v) at any two points in a fluid flow must be equal:
\[ A_1 v_1 = A_2 v_2 \]
- Here, \(A_1\) is the cross-sectional area of tube A, \(v_1\) is the velocity in tube A, \(A_2\) is the cross-sectional area of tube B, and \(v_2\) is the velocity in tube B.

3. **Calculate the Areas**:
- The cross-sectional area of tube A (radius R):
\[ A_1 = \pi R^2 \]
- The cross-sectional area of tube B (radius 2R):
\[ A_2 = \pi (2R)^2 = \pi (4R^2) = 4\pi R^2 \]

4. **Set Up the Equation**:
- Substitute the areas and velocities into the equation of continuity:
\[ A_1 v_1 = A_2 v_2 \]
\[ \pi R^2 \cdot v = 4\pi R^2 \cdot v_2 \]

5. **Simplify the Equation**:
- Cancel \(\pi R^2\) from both sides:
\[ v = 4 v_2 \]

6. **Solve for \(v_2\)**:
- Rearranging the equation gives:
\[ v_2 = \frac{v}{4} \]

### Final Answer:
The velocity of water flowing through tube B is:
\[ v_2 = \frac{v}{4} \]