A cone is made from a circular sheet of radius `sqrt3` by cutting out a sector and giving the cut edges of the remaining piece together. The maximum volume attainable for the cone is (A) `pi/3` (B) `pi/6` (C) `(2pi)/3` (D) `3sqrt(3)pi`

Answer

To find the maximum volume of a cone formed from a circular sheet of radius \(\sqrt{3}\), we will follow these steps:

### Step 1: Understand the problem
We have a circular sheet with a radius \( R = \sqrt{3} \). We cut out a sector from this sheet and form a cone by bringing the cut edges together. We need to find the maximum volume of this cone.

### Step 2: Volume of the cone formula
The volume \( V \) of a cone is given by the formula:
\[ V = \frac{1}{3} \pi r^2 h \]
where \( r \) is the radius of the base of the cone and \( h \) is the height of the cone.

### Step 3: Relate the cone's dimensions to the circular sheet
When the sector is cut out, the radius of the cone's base \( r \) and its height \( h \) can be related to the radius of the original circular sheet \( R \) using the Pythagorean theorem:
\[ R^2 = r^2 + h^2 \]
From this, we can express \( r^2 \) as:
\[ r^2 = R^2 - h^2 \]

### Step 4: Substitute \( r^2 \) into the volume formula
Substituting \( r^2 \) into the volume formula gives:
\[ V = \frac{1}{3} \pi (R^2 - h^2) h \]
Expanding this, we have:
\[ V = \frac{1}{3} \pi (R^2 h - h^3) \]

### Step 5: Differentiate the volume with respect to height \( h \)
To find the maximum volume, we differentiate \( V \) with respect to \( h \):
\[ \frac{dV}{dh} = \frac{1}{3} \pi (R^2 - 3h^2) \]

### Step 6: Set the derivative to zero to find critical points
Setting \(\frac{dV}{dh} = 0\) gives:
\[ R^2 - 3h^2 = 0 \implies R^2 = 3h^2 \implies h^2 = \frac{R^2}{3} \implies h = \frac{R}{\sqrt{3}} \]

### Step 7: Find the radius \( r \) using \( h \)
Now, substitute \( h \) back to find \( r \):
\[ r^2 = R^2 - h^2 = R^2 - \frac{R^2}{3} = \frac{2R^2}{3} \implies r = R \sqrt{\frac{2}{3}} \]

### Step 8: Substitute \( R \) into the volume formula
Substituting \( R = \sqrt{3} \):
\[ V = \frac{1}{3} \pi \left(\frac{2R^2}{3}\right) \left(\frac{R}{\sqrt{3}}\right) \]
Calculating \( R^2 \):
\[ R^2 = (\sqrt{3})^2 = 3 \]
Thus,
\[ V = \frac{1}{3} \pi \left(\frac{2 \cdot 3}{3}\right) \left(\frac{\sqrt{3}}{\sqrt{3}}\right) = \frac{1}{3} \pi (2) (1) = \frac{2\pi}{3} \]

### Final Answer
The maximum volume attainable for the cone is:
\[ \boxed{\frac{2\pi}{3}} \]