A hollow cylindrical tube 20 cm long is made of iron and its external diameter is 8 cm. The volume of iron used in making the tube is `440" cm"^(3)`. What is the thickness of the tube ?

Options

A. 1 cm
B. `0.5` cm
C. 2 cm
D. `1.5` cm

Answer

To find the thickness of the hollow cylindrical tube, we will follow these steps:

### Step 1: Identify the given values
- Length of the tube (h) = 20 cm
- External diameter (d2) = 8 cm
- Volume of iron used (V_iron) = 440 cm³

### Step 2: Calculate the external radius (R2)
The external radius (R2) can be calculated from the external diameter:
\[ R2 = \frac{d2}{2} = \frac{8 \, \text{cm}}{2} = 4 \, \text{cm} \]

### Step 3: Set up the volume equations
The volume of the hollow cylinder can be calculated using the formula:
\[ \text{Volume of the hollow cylinder} = \text{Volume of outer cylinder} - \text{Volume of inner cylinder} \]
The volume of the outer cylinder (with radius R2) is:
\[ V_{outer} = \pi R2^2 h = \pi (4 \, \text{cm})^2 (20 \, \text{cm}) = \pi \cdot 16 \cdot 20 = 320\pi \, \text{cm}^3 \]
The volume of the inner cylinder (with radius R1) is:
\[ V_{inner} = \pi R1^2 h \]

### Step 4: Set up the equation for the volume of iron
The volume of iron used in making the tube is given by:
\[ V_{iron} = V_{outer} - V_{inner} \]
Substituting the volumes we calculated:
\[ 440 \, \text{cm}^3 = 320\pi - \pi R1^2 \]
Factoring out \(\pi\):
\[ 440 = \pi (320 - R1^2) \]

### Step 5: Solve for R1^2
To isolate R1^2, we first need to calculate \(\pi\):
Using \(\pi \approx \frac{22}{7}\):
\[ 440 = \frac{22}{7} (320 - R1^2) \]
Multiplying both sides by 7:
\[ 3080 = 22(320 - R1^2) \]
Dividing by 22:
\[ 140 = 320 - R1^2 \]
Rearranging gives:
\[ R1^2 = 320 - 140 = 180 \]

### Step 6: Calculate R1
Taking the square root:
\[ R1 = \sqrt{180} = \sqrt{36 \times 5} = 6\sqrt{5} \approx 13.42 \, \text{cm} \]

### Step 7: Calculate the thickness of the tube
The thickness (t) of the tube is given by:
\[ t = R2 - R1 = 4 \, \text{cm} - 6\sqrt{5} \approx 4 \, \text{cm} - 13.42 \, \text{cm} \approx -9.42 \, \text{cm} \]
This indicates a miscalculation in the earlier steps.

### Correcting the calculation
After reviewing the calculations, we find:
\[ R1^2 = 16 - 7 = 9 \implies R1 = 3 \, \text{cm} \]
Thus, the thickness is:
\[ t = R2 - R1 = 4 \, \text{cm} - 3 \, \text{cm} = 1 \, \text{cm} \]

### Final Answer
The thickness of the tube is **1 cm**.

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