A bucket has top and bottom diameter of 40 cm and 20 cm respectively.

Question:

A bucket has top and bottom diameter of 40 cm and 20 cm respectively. Find the volume of the bucket if its depth is 12 cm. Also, find the cost of tin sheet used for making the bucket at the rate of Rs. 1.20 per dm. (Use π = 3.14)

Solution:

The radii of the top and bottom circles are r1 = 20 cm and r2 = 10 cm respectively. The height of the bucket is = 12 cm. Therefore, the volume of the bucket is

$V=\frac{1}{3} \pi\left(r_{1}^{2}+r_{1} r_{2}+r_{2}^{2}\right) \times h$

$=\frac{1}{3} \pi\left(20^{2}+20 \times 10+10^{2}\right) \times 12$

$=\frac{1}{3} \times \frac{22}{7} \times 700 \times 12$

$=8800 \mathrm{~cm}^{3}$

The slant height of the bucket is

$l=\sqrt{\left(r_{1}-r_{2}\right)^{2}+h^{2}}$

$=\sqrt{(20-10)^{2}+12^{2}}$

$=2 \sqrt{61} \mathrm{~cm}$

The total surface area of the bucket is

$=\pi\left(r_{1}+r_{2}\right) \times l+\pi r_{2}^{2}$

$=\frac{22}{7} \times(20+10) \times 2 \sqrt{61}+\frac{22}{7} \times 10^{2}$

$=\frac{1320 \sqrt{61}+2200}{7} \mathrm{~cm}^{2}$

$=\frac{1320 \sqrt{61}+2200}{7 \times 100} \mathrm{dm}^{2}$

The total cost of tin sheet used for making the bucket is

$=1.20 \times\left(\frac{1320 \sqrt{61}+2200}{7 \times 100}\right)$

$=21.40$

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