Question

A solid iron pole having cylindrical portion 110 cm high and of base diameter 12 cm is surmounted by a cone 9 cm high find the mass of the pole,given that the mass of 1$c{m}^3$ of iron is 8 gm.

Answer 1

Solution:

As per the data given in the question, we have

Base diameter of the cylinder = 12 cm

Radius of the cylinder = 6 cm = r (as we know that the radius os half of the diameter)

Height of the cylinder = 110 cm = h

Length of the cone = 9 cm = L

Now,

The volume of the cylinder = $\pi r^{2}h=\pi \times 6^2\times 110----\left(1\right)$

The volume of the cone = V₂ = $\frac{1}{3}\pi r^{2}L=\frac{1}{3}\times \pi \times 6^2\times 12=108\pi c{m}^3$

Volume of the pole = volume of the cylinder + volume of the cone = V₁ + V₂ = V

V = 108π + π (6)² 110

V = 12785.14 cm³

Given mass of 1 cm³ of the iron pole = 8 gm

Then, mass of 12785.14 cm³ of the iron pole = 8 $\times$ 12785.14 = 102281.12 gm = 102.2 kg

Therefore, the mass of the iron pole = 102.2 kg