Question

A bucket made up of a metal sheet is in the form of a frustum of a cone of height $16cm$ with diameters of its lower and upper ends as $16cm$ and $40cm$ respectively. Find the volume of the bucket. Also, find the cost of the bucket if the cost of metal sheet used is Rs. $20$ per $100{cm}^2$. (Use $\pi =3.14$)

Answer 1

Given,

Hight of frustum of cone $h=16cm$

Diameter of lower circular end $d=16cm$

Diameter of upper circular end $D=40cm$

Hence,

Radius of lower circular end $r=\frac{d}{2}=8cm$

Radius of upper circular end $R=\frac{D}{2}=20cm$

Now,

Volume of frustum of cone $V=\frac{\pi }{3}h(R^2+r^2+R\times r)$

$=\frac{\pi }{3}\times 16\times ({20}^2+8^2+20\times 8)c{m}^3$

$=\frac{\pi }{3}\times 16\times (400+64+160)c{m}^3$

$=\frac{3.14}{3}\times 16\times 624c{m}^3$

$=\frac{31349.76}{3}c{m}^3$

Volume of frustum of cone $V=10449.92c{m}^3$

$\text{Surface area of bucket (S) = curved surface area of a frustum of cone + area of lower circular end}$

$S=\pi (R+r)\sqrt{(R-r{)}^2+h^2}+\pi r^2$

$=((R+r)\sqrt{(R-r{)}^2+h^2}+r^2)\pi$

$=((20+8)\sqrt{(20-8{)}^2+{16}^2}+8^2)\pi c{m}^2$

$=(\left(28\right)\sqrt{(12{)}^2+256}+64)\pi c{m}^2$

$=(\left(28\right)\sqrt{(144+256)}+64)\pi c{m}^2$

$=(\left(28\right)\sqrt{(400}+64)\pi c{m}^2$

$=(28\times 20+64)\pi c{m}^2$

$=(560+64)\times 3.14c{m}^2$

$=624\times 3.14c{m}^2$

$=1959.36c{m}^2$

$\text{Total surface area of meatal sheet used}=1959.36c{m}^2$

Cost of $100c{m}^2$ metal sheet$=20Rs.$

Cost of $1c{m}^2$ metal sheet$=\frac{20}{100}Rs=0.2Rs..$

Cost of $1959.36c{m}^2$ metal sheet$=0.2\times 1959.36Rs.=391.872Rs.$

Hence, cost of metal sheet used to make bucket $391.872Rs.$