# Question #4ef82

Feb 5, 2018

It is revealed from the above figure that surface area of the cyndrical solid with a conical cavity will be

$A = 2 \pi r h + \pi r l + \pi {r}^{2}$

$\implies A = \pi r \left(2 h + l + r\right)$

Given
Total surface area of the solid $A = 904.32 {\mathrm{dm}}^{2}$
Height of the solid $h = 16 \mathrm{dm}$

Radius $r = 6 \mathrm{dm}$

Density $\rho = 7.5 g \text{/"cm^3=7.5kg"/} {\mathrm{dm}}^{3}$

The slant height $\left(l\right)$ of the cone is not known.

So
$904.32 = 3.14 \times 6 \left(2 \times 16 + l + 6\right)$

$\implies l + 38 = \frac{904.32}{3.14 \times 6} = 48$

$\implies l = 48 - 38 = 10 \mathrm{dm}$

So height of the conical cavity will be

${h}_{\text{cone}} = \sqrt{{l}^{2} - {r}^{2}}$

$\implies {h}_{\text{cone}} = \sqrt{{10}^{2} - {6}^{2}} = 8 \mathrm{dm}$

Now volume of the solid

$V = \text{volume of cylinder"-"volume of cavity}$

$= \pi {r}^{2} h - \frac{1}{3} \pi {r}^{2} {h}_{\text{cone}}$

$= \pi {r}^{2} \left(h - {h}_{\text{cone}} / 3\right)$

So weight (mass) of the solid will be

$W = V \times \rho$

$\implies W = V \times \rho = \pi {r}^{2} \left(h - {h}_{\text{cone}} / 3\right) \rho$

$\implies W = 3.14 \times {6}^{2} \left(16 - \frac{8}{3}\right) \times 7.5 k g$

$\implies W = 3.14 \times 36 \times \frac{40}{3} \times 7.5 k g = 11304 k g$