A lead pipe is 35 cm long its external diameter is 2.4 cm and its thickness is 2 mm . If 1 cubic cm of lead weight 5 gms . The weight of pipe is -----

Options

A. 220 gms
B. 242 gms
C. 420 gms
D. 484 gms

Answer

To find the weight of the lead pipe, we will follow these steps:

### Step 1: Identify the given values
- Length of the pipe (H) = 35 cm
- External diameter (D) = 2.4 cm
- Thickness of the pipe = 2 mm = 0.2 cm (since 1 mm = 0.1 cm)
- Weight of lead per cubic cm = 5 g

### Step 2: Calculate the external radius (R)
The external radius (R) can be calculated using the external diameter:
\[ R = \frac{D}{2} = \frac{2.4 \text{ cm}}{2} = 1.2 \text{ cm} \]

### Step 3: Calculate the internal radius (r)
The internal radius (r) can be calculated by subtracting the thickness from the external radius:
\[ r = R - \text{thickness} = 1.2 \text{ cm} - 0.2 \text{ cm} = 1.0 \text{ cm} \]

### Step 4: Calculate the volume of the hollow cylinder (pipe)
The volume (V) of a hollow cylinder is given by the formula:
\[ V = \pi H (R^2 - r^2) \]
Substituting the values:
- \( H = 35 \text{ cm} \)
- \( R = 1.2 \text{ cm} \)
- \( r = 1.0 \text{ cm} \)

Calculating \( R^2 \) and \( r^2 \):
\[ R^2 = (1.2)^2 = 1.44 \text{ cm}^2 \]
\[ r^2 = (1.0)^2 = 1.0 \text{ cm}^2 \]

Now substituting back into the volume formula:
\[ V = \pi \times 35 \times (1.44 - 1.0) \]
\[ V = \pi \times 35 \times 0.44 \]
Using \( \pi \approx 3.14 \):
\[ V \approx 3.14 \times 35 \times 0.44 \]
\[ V \approx 3.14 \times 15.4 \]
\[ V \approx 48.436 \text{ cm}^3 \]

### Step 5: Calculate the weight of the pipe
The weight (W) of the pipe can be calculated using the volume and the weight of lead per cubic cm:
\[ W = V \times \text{weight per cubic cm} \]
\[ W = 48.436 \text{ cm}^3 \times 5 \text{ g/cm}^3 \]
\[ W \approx 242.18 \text{ g} \]

### Final Answer
The weight of the lead pipe is approximately **242.18 grams**.

---