A bucket open at the top, and made up of a metal sheet is in the form of a frustum of a cone. The depth of the bucket is $ 24 \mathrm{~cm} $ and the diameters of its upper and lower circular ends are $ 30 \mathrm{~cm} $ and $ 10 \mathrm{~cm} $ respectively. Find the cost of metal sheet used in it at the rate of $ ₹ 10 $ per $ 100 \mathrm{~cm}^{2} $. (Use $ \pi=3.14 $ ).

Given:

A bucket open at the top, and made up of a metal sheet is in the form of a frustum of a cone.

The depth of the bucket is \( 24 \mathrm{~cm} \) and the diameters of its upper and lower circular ends are \( 30 \mathrm{~cm} \) and \( 10 \mathrm{~cm} \) respectively. 

To do:

We have to find the cost of metal sheet used in it at the rate of \( ₹ 10 \) per \( 100 \mathrm{~cm}^{2} \).

Solution:

 Upper radius of the bucket $\mathrm{R}=\frac{30}{2}$

$=15 \mathrm{~cm}$

Lower radius of the bucket $r=\frac{10}{2}$

$=5 \mathrm{~cm}$

Height of the bucket $h=24 \mathrm{~cm}$

Therefore,

Slant height of the bucket $l=\sqrt{h^{2}+(\mathrm{R}-r)^{2}}$

$=\sqrt{24^{2}+(15-5)^{2}}$

$=\sqrt{24^{2}+10^{2}}$

$=\sqrt{576+100}$

$=\sqrt{676}$

$=26$

Curved surface area of the bucket $=\pi(r+\mathrm{R}) l$

$=\pi(15+5) \times 26$

$=520 \pi \mathrm{cm}^{3}$

Area of the base of the bucket $=\pi r^{2}$

$=\pi \times 5^2$

$=25 \pi \mathrm{cm}^{2}$

Total surface area of open frustum $=520 \pi+25 \pi$

$=545 \pi \mathrm{cm}^{2}$

$=545 \times 3.14$

$=1711.3 \mathrm{~cm}^{2}$

Cost of metal sheet per $100 \mathrm{~cm}^{2}=Rs.\ 10$

Total cost of metal sheet used $=Rs.\ \frac{1711.3 \times 10}{100}$

$=Rs.\ 171.13$

The cost of metal sheet used is Rs. 171.