A bucket made up of a metal sheet is in the form of frustum of a cone.
Question:

A bucket made up of a metal sheet is in the form of frustum of a cone. Its depth is 24 cm and the diameters of the top and bottom are 30 cm and 10 cm, respectively. Find the cost of completely filling the bucket with milk at the rate of Rs 20 per litre and the cost of metal sheet used if it costs Rs 10 per 100 cm2.

Solution:

Greater diameter of the bucket = 30 cm
Radius of the bigger end of the bucket = R = 15 cm
Diameter of the smaller end of the bucket = 10 cm
Radius of the smaller end of the bucket = r = 5 cm
Height of the bucket = 24 cm

Slant height, l

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

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

$=\sqrt{576+(10)^{2}}$

$=\sqrt{576+100}$

 

$=\sqrt{676}=26 \mathrm{~cm}$

Capacity of the frustum

$=\frac{1}{3} \pi h\left(R^{2}+r^{2}+R r\right)$

$=\frac{1}{3} \times 3.14 \times 24\left(15^{2}+5^{2}+15 \times 5\right)$

$=3.14 \times 8 \times 325$

 

$=8164 \mathrm{~cm}^{3}=8.164$ litres

A litre of milk cost Rs 20.

So, total cost of filling the bucket with milk $=8.164 \times 20=$ Rs $163.28$

Surface area of the bucket

$=\pi r^{2}+\pi l(R+r)$

$=\pi\left[5^{2}+26(15+5)\right]$

$=3.14[25+26(20)]$

$=3.14[25+520]$

 

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

Cost of 100 cm2 of metal sheet is Rs 10.

So, cost of metal used for making the bucket $=\frac{1711.3}{100} \times 10=$ Rs $171.13$

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