Q6:

A solid iron pole consists of a cylinder of height $220$ cm and base diameter $24$ cm, which is surmounted by another cylinder of height $60$ cm and radius $8$ cm. Find the mass of the pole, given that $1$ cm$^3$ of iron has approximately $8$g mass. (Use $\pi = 3.14$)

Given:

A solid iron pole composed of two cylinders:

• Cylinder 1 (Base): Height ($h_1$) = $220$ cm, Diameter ($d_1$) = $24$ cm.
• Cylinder 2 (Top): Height ($h_2$) = $60$ cm, Radius ($r_2$) = $8$ cm.
• Density of iron: $1$ cm$^3$ = $8$ g.
• Constant: $\pi = 3.14$.

To Find:

The total mass of the iron pole in grams (or kilograms).

Step 1: Determine the dimensions of the cylinders.

For the base cylinder (Cylinder 1):

Radius ($r_1$) = $\frac{\text{Diameter}}{2} = \frac{24 \text{ cm}}{2} = 12$ cm.

Height ($h_1$) = $220$ cm.

For the top cylinder (Cylinder 2):

Radius ($r_2$) = $8$ cm.

Height ($h_2$) = $60$ cm.

Step 2: Calculate the volume of the pole.

The volume of a cylinder is given by the formula: $V = \pi r^2 h$.

Total Volume ($V_{total}$) = Volume of Cylinder 1 + Volume of Cylinder 2

$V_{total} = (\pi \cdot r_1^2 \cdot h_1) + (\pi \cdot r_2^2 \cdot h_2)$

$V_{total} = \pi [ (12)^2 \cdot 220 + (8)^2 \cdot 60 ]$

$V_{total} = 3.14 [ (144 \cdot 220) + (64 \cdot 60) ]$

$V_{total} = 3.14 [ 31680 + 3840 ]$

$V_{total} = 3.14 [ 35520 ]$

$V_{total} = 111532.8$ cm$^3$.

Step 3: Calculate the mass of the pole.

Given that $1$ cm$^3$ of iron has a mass of $8$ g.

Total Mass = $V_{total} \times 8$ g/cm$^3$

Total Mass = $111532.8 \times 8$

Total Mass = $892262.4$ g.

Step 4: Convert to kilograms (optional but standard).

Since $1000$ g = $1$ kg:

Total Mass = $\frac{892262.4}{1000} = 892.2624$ kg.

Final Answer: The mass of the pole is 892262.4 g or approximately 892.26 kg.