Question

A bucket has top and bottom diameters of 40 cm respectively. Find the volume of the bucket if its depth is 12 cm. Also, find the cost of tin used for making the bucket at the rater of Rs.1.20 per $d{m}^2$ (Use $\pi$= 3.14)

Answer 1

Solution:-

Let the radius of the top part be 'R' and radius of the bottom part be 'r' respectively.

Diameter (Top) = 40 cm
Radius (R) = $\frac{40}{2}$ = 20 cm
10 cm = 1 dm
20 cm = 2dm

Diameter (Bottom) = 20 cm
Radius (r) =$\frac{20}{2}$ = 10 cm
10 cm = 1 dm

Height = 12 cm = 1.2 dm

Slant height

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

$=\sqrt{{1.2}^2+(2-1{)}^2}$

$=\sqrt{1.44+1}$

$=\sqrt{2.44}=1.56$

So, slant height (l) is 1.56 dm

Area of the tin sheet used for bucket = Curved surface area of the bucket + Area of the base

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

$=>3.14\times (2+1)\times 1.56+3.14\times 1^2$

$=>3.14\times 3\times 1.56+3.14$

$=>14.6952+3.14$

$=>17.8352d{m}^2$

CSA of the tin sheet used = 17.8352 $d{m}^2$
Rate of tin sheet = Rs. 1.2 per $d{m}^2$
Total cost of the tin sheet used for making the bucket = 17.8352$\times$1.2
= Rs. 21.40
Answer.
Volume

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

$=\frac{1}{3}\times 3.14\times 1.2\times (2^2+1^2+2\times 1)$

$=\frac{1}{3}\times 3.14\times 1.2\times 7=\frac{26.376}{3}=8.792d{m}^3$