Question

An iron pillar consists of a cylindrical portion of 2.8m height and 20cm. in diameter and a cone of 45 cm height surmounting it find the weight of the piller $1c{m}^3$ of iron weight $7.5g$

Answer 1

The volume of the iron pillar=Volume of the cylindrical part+Volume of cone

$V=V_{cyl}+V_{con}$

$\therefore V=\pi r^2{h}_{cyl}+\frac{1}{3}\pi r^2{h}_{con}$

Now, $r=\frac{d}{2}=\frac{20}{2}=10cm$

$h_{cyl}=2.8m=280cm$

$h_{con}=45cm$

$\therefore V=\pi (10{)}^2\times 280+\frac{1}{3}\pi (10{)}^2\times 45$

$V=\pi (100\times 280+100\times 15)$

$V=\pi (100\times (280+15\left)\right)$

$V=\pi (100\times 295)=\pi \times =29500c{m}^3$

Thus Weight$=V\times \text{weight per cm^3}$

$w=29500\times 7.5=221250g=221.25kg$