The difference between the outer and the inner curved surface areas of a hollow cylinder, 14cm long, is 88sq. cm. Find the outer and the inner radii of the cylinder, given that the volume of metal used is 176cu.cm.

Answer

To solve the problem, we need to find the outer and inner radii of a hollow cylinder given the difference in curved surface areas and the volume of the metal used.

### Step-by-Step Solution:

1. **Understand the Problem:**
We have a hollow cylinder with:
- Length (height) \( h = 14 \) cm
- Difference in outer and inner curved surface areas = \( 88 \) sq. cm
- Volume of metal used = \( 176 \) cu. cm

Let:
- \( R \) = outer radius
- \( r \) = inner radius

2. **Curved Surface Area of a Cylinder:**
The formula for the curved surface area (CSA) of a cylinder is given by:
\[ \text{CSA} = 2 \pi r h \]
Therefore, the outer curved surface area (OCSA) and inner curved surface area (ICSA) can be expressed as:
- OCSA = \( 2 \pi R h \)
- ICSA = \( 2 \pi r h \)

3. **Set Up the Equation for the Difference in Surface Areas:**
According to the problem:
\[ OCSA - ICSA = 88 \]
Substituting the formulas:
\[ 2 \pi R h - 2 \pi r h = 88 \]
Factor out \( 2 \pi h \):
\[ 2 \pi h (R - r) = 88 \]

4. **Substituting Values:**
Substitute \( h = 14 \) cm and \( \pi \approx \frac{22}{7} \):
\[ 2 \times \frac{22}{7} \times 14 (R - r) = 88 \]
Simplifying:
\[ \frac{44}{7} \times 14 (R - r) = 88 \]
\[ 88 (R - r) = 88 \]
Dividing both sides by 88:
\[ R - r = 1 \quad \text{(Equation 1)} \]

5. **Volume of the Metal Used:**
The volume of the metal used is the volume of the outer cylinder minus the volume of the inner cylinder:
\[ V = \pi R^2 h - \pi r^2 h = 176 \]
Factoring out \( \pi h \):
\[ \pi h (R^2 - r^2) = 176 \]
Substituting \( h = 14 \) cm and \( \pi \approx \frac{22}{7} \):
\[ \frac{22}{7} \times 14 (R^2 - r^2) = 176 \]
Simplifying:
\[ 44 (R^2 - r^2) = 176 \]
Dividing both sides by 44:
\[ R^2 - r^2 = 4 \quad \text{(Equation 2)} \]

6. **Using the Difference of Squares:**
We can express \( R^2 - r^2 \) as:
\[ R^2 - r^2 = (R - r)(R + r) \]
Substituting \( R - r = 1 \) from Equation 1:
\[ 1(R + r) = 4 \]
Thus:
\[ R + r = 4 \quad \text{(Equation 3)} \]

7. **Solving the System of Equations:**
Now we have two equations:
- \( R - r = 1 \) (Equation 1)
- \( R + r = 4 \) (Equation 3)

Adding these two equations:
\[ (R - r) + (R + r) = 1 + 4 \]
\[ 2R = 5 \implies R = 2.5 \text{ cm} \]

Substituting \( R \) back into Equation 1:
\[ 2.5 - r = 1 \implies r = 2.5 - 1 = 1.5 \text{ cm} \]

8. **Final Answer:**
The outer radius \( R \) is \( 2.5 \) cm and the inner radius \( r \) is \( 1.5 \) cm.