A cylindrical water vessel has a base radius of 7 cm and a curved surface area of 1408 cm². A cone-shaped container has the same base radius, and its height is 25% more than the height of the cylindrical vessel. Using π = 22/7, find the volume of the cone-shaped container.

The correct answer is $\frac{6160}{3}\text{cm}{3}^{}$.

Step 1: Calculate the height of the cylindrical water vessel.

We are given the following properties of the cylinder:

Radius of the base of the cylinder ($r$) = 7 cm

Curved surface area of the cylinder = 1408 cm²

The formula for the curved surface area of a cylinder is:

$\text{Curved Surface Area}=2\pi rh$

Substitute the given values into the formula using $\pi =\frac{22}{7}$:

$2\times \frac{22}{7}\times 7\times h=1408$

Simplifying the left side by canceling out 7 in the numerator and denominator:

$44\times h=1408$

Solving for the height of the cylinder ($h$):

$h=\frac{1408}{44}=32\text{cm}$

Step 2: Determine the height and radius of the cone-shaped container.

The problem states that the cone-shaped container has the same base radius as the cylindrical vessel:

Radius of the cone ($r$) = 7 cm

The height of the cone ($H$) is 25% more than the height of the cylinder ($h$):

$H=h+0.25h=1.25h$

$H=1.25\times 32=40\text{cm}$

Step 3: Calculate the volume of the cone-shaped container.

The formula for the volume of a cone is:

$\text{Volume}=\frac{1}{3}\pi r^{2}H$

Substitute $r=7\text{cm}$, $H=40\text{cm}$, and $\pi =\frac{22}{7}$ into the volume formula:

$\text{Volume}=\frac{1}{3}\times \frac{22}{7}\times 7^2\times 40$

$\text{Volume}=\frac{1}{3}\times \frac{22}{7}\times 49\times 40$

Simplifying $\frac{49}{7}=7$:

$\text{Volume}=\frac{1}{3}\times 22\times 7\times 40$

$\text{Volume}=\frac{1}{3}\times 6160$

$\text{Volume}=\frac{6160}{3}\text{cm}{3}^{}$

Therefore, the volume of the cone-shaped container is $\frac{6160}{3}\text{cm}{3}^{}$.