If the curved surface area of a sphere increased by 44%, then by what percentage would its volume increase?

Options

A. 0.2 %
B. 0.44 %
C. CND
D. `72.8% `

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Understand the relationship between the curved surface area (CSA) and the radius of the sphere.
The curved surface area (CSA) of a sphere is given by the formula:
\[ \text{CSA} = 4\pi r^2 \]
where \( r \) is the radius of the sphere.

### Step 2: Set up the initial and new CSA.
Let the initial CSA be represented as:
\[ \text{Old CSA} = 4\pi r_1^2 \]
According to the problem, the CSA increases by 44%, so the new CSA becomes:
\[ \text{New CSA} = \text{Old CSA} + 0.44 \times \text{Old CSA} = 1.44 \times \text{Old CSA} \]
Thus, we have:
\[ \text{New CSA} = 4\pi r_2^2 = 1.44 \times 4\pi r_1^2 \]

### Step 3: Simplify the equation to find the ratio of the radii.
Dividing both sides by \( 4\pi \):
\[ r_2^2 = 1.44 r_1^2 \]
Taking the square root of both sides:
\[ \frac{r_2}{r_1} = \sqrt{1.44} = 1.2 \]
This implies:
\[ r_2 = 1.2 r_1 \]

### Step 4: Calculate the volumes of the spheres.
The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3}\pi r^3 \]
Thus, the old volume is:
\[ \text{Old Volume} = \frac{4}{3}\pi r_1^3 \]
And the new volume is:
\[ \text{New Volume} = \frac{4}{3}\pi r_2^3 = \frac{4}{3}\pi (1.2 r_1)^3 \]

### Step 5: Expand the new volume expression.
Calculating \( (1.2 r_1)^3 \):
\[ (1.2 r_1)^3 = 1.728 r_1^3 \]
Thus, the new volume becomes:
\[ \text{New Volume} = \frac{4}{3}\pi (1.728 r_1^3) = \frac{4}{3}\pi \cdot 1.728 r_1^3 \]

### Step 6: Find the increase in volume.
The increase in volume is:
\[ \text{Increase in Volume} = \text{New Volume} - \text{Old Volume} \]
\[ = \frac{4}{3}\pi (1.728 r_1^3) - \frac{4}{3}\pi r_1^3 \]
\[ = \frac{4}{3}\pi (1.728 - 1) r_1^3 = \frac{4}{3}\pi (0.728) r_1^3 \]

### Step 7: Calculate the percentage increase in volume.
The percentage increase in volume is given by:
\[ \text{Percentage Increase} = \frac{\text{Increase in Volume}}{\text{Old Volume}} \times 100 \]
Substituting the values:
\[ = \frac{\frac{4}{3}\pi (0.728) r_1^3}{\frac{4}{3}\pi r_1^3} \times 100 \]
The \( \frac{4}{3}\pi r_1^3 \) cancels out:
\[ = 0.728 \times 100 = 72.8\% \]

### Conclusion
Thus, the volume of the sphere increases by **72.8%**.

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