Question

An iron pillar has some part in the form of a right circular cylinder and remaining in the form of a right circular cone. The radius of base of each of cone and cylinder is $8cm$. The cylindrical part is $240cm$ high and the conical part is $36cm$ high. FindThe volume of iron used the weight of pillar, if $1c{m}^3$ of iron weighs $10g$.

Answer 1

The volume of cylindrical part $=\pi r^{2}h$

$=\frac{22}{7}\times 8^2\times 240=48274.285$

The volume of conical part $=\frac{1}{3}\times b\times h$

$=\frac{1}{3}\times \frac{22}{7}\times 8^2\times 36=2413.7142$

Total volume $=48274.285+2413.7142$

$50688c{m}^3$.

Given $-10g$

Mass $=Density\times Volume$.

$=10\times 50688=506880g$.

$=506.88kg$