The part of the parabola $y^{2}=4x$ cut off by the latus rectum revolves about the tangent at the vertex. The volume of the reel thus generated is :

Question

Accepted Answer

To solve this problem, we need to find the volume of the solid generated when the part of the parabola $ y^2 = 4x $ cut off by the latus rectum is revolved about the tangent at the vertex.

### Step-by-Step Explanation:

1. **Identify the Parabola and Latus Rectum:**
   - The given parabola is $ y^2 = 4x $. This is a standard form of a parabola that opens to the right.
   - The vertex of this parabola is at the origin $(0,0)$.
   - The latus rectum of a parabola $ y^2 = 4ax $ is a line segment perpendicular to the axis of the parabola and passes through the focus. The length of the latus rectum is $ 4a $. Here, $ a = 1 $, so the length of the latus rectum is 4.
   - The endpoints of the latus rectum are at $(1, 2)$ and $(1, -2)$.

2. **Tangent at the Vertex:**
   - The tangent to the parabola at the vertex $(0,0)$ is the y-axis, $ x = 0 $.

3. **Volume of Revolution:**
   - We need to revolve the segment of the parabola cut off by the latus rectum about the tangent at the vertex, which is the y-axis.
   - The segment of the parabola from $(1, -2)$ to $(1, 2)$ is revolved around the y-axis.

4. **Using the Disk Method:**
   - The volume $ V $ of the solid of revolution generated by revolving a curve $ x = f(y) $ from $ y = c $ to $ y = d $ about the y-axis is given by:
     $$
     V = \pi \int_{c}^{d} [f(y)]^2 \, dy
     $$
   - For the parabola $ y^2 = 4x $, we have $ x = \frac{y^2}{4} $.
   - The limits of integration are from $ y = -2 $ to $ y = 2 $.

5. **Calculate the Volume:**
   $$
   V = \pi \int_{-2}^{2} \left(\frac{y^2}{4}\right)^2 \, dy = \pi \int_{-2}^{2} \frac{y^4}{16} \, dy
   $$
   $$
   V = \frac{\pi}{16} \int_{-2}^{2} y^4 \, dy
   $$
   - The integral of $ y^4 $ is $\frac{y^5}{5}$.
   $$
   V = \frac{\pi}{16} \left[ \frac{y^5}{5} \right]_{-2}^{2}
   $$
   $$
   V = \frac{\pi}{16} \left( \frac{32}{5} - \left(-\frac{32}{5}\right) \right)
   $$
   $$
   V = \frac{\pi}{16} \times \frac{64}{5} = \frac{4\pi}{5}
   $$

### Conclusion:

- **Option 2: $\frac{4\pi}{5}$** is the correct answer because the calculated volume of the solid generated by revolving the segment of the parabola about the y-axis is indeed $\frac{4\pi}{5}$.

### Explanation for Incorrect Options:

- **Option 1: $4\pi$** is incorrect because it does not account for the division by 5 in the integration process.
- **Option 3: $\frac{\pi}{4}$** is incorrect because it results from an incorrect calculation of the integral.
- **Option 4: $\frac{\pi}{5}$** is incorrect because it underestimates the volume by a factor of 4.

Thus, the correct answer is Option 2: $\frac{4\pi}{5}$.