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 dm² . (Use π = 3.14)
Answers
Let the radius of the top part be 'R' and radius of the bottom part be 'r' respectively.
Diameter (Top) = 40 cm
Radius (R) = 40/2 = 20 cm
10 cm = 1 dm
20 cm = 2dm
Diameter (Bottom) = 20 cm
Radius (r) = 20/2 = 10 cm
10 cm = 1 dm
Height = 12 cm = 1.2 dm
Slant height ('l') = √H² + (R - r)²
= √(1.2)² + (2 - 1)²
= √1.44 + 1
= √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
= π(R + r)*l + πr²
⇒ 3.14*(2 + 1)*1.56 + 3.14*1²
⇒(3.14*3*1.56) + 3.14
⇒ 14.6952 + 3.14
⇒ 17.8352 sq dm
CSA of the tin sheet used = 17.8352 sq dm
Rate of tin sheet = Rs. 1.2 per sq dm
Total cost of the tin sheet used for making the bucket = 17.8352*1.2
= Rs. 21.40
Volume = 1/3πh(R² + r² + R*r)
= 1/3*3.14*1.2*(2² + 1² + 2*1)
= 1/3*3.14*1.2*7
= 26.376/3
Volume = 8.792 cu dm
here is ur ans....
Diameter (Top) = 40 cm
Radius (R) = 40/2 = 20 cm
10 cm = 1 dm
20 cm = 2dm
Diameter (Bottom) = 20 cm
Radius (r) = 20/2 = 10 cm
10 cm = 1 dm
Height = 12 cm = 1.2 dm
Slant height ('l') = √H² + (R - r)²
= √(1.2)² + (2 - 1)²
= √1.44 + 1
= √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
= π(R + r)*l + πr²
⇒ 3.14*(2 + 1)*1.56 + 3.14*1²
⇒(3.14*3*1.56) + 3.14
⇒ 14.6952 + 3.14
⇒ 17.8352 sq dm
CSA of the tin sheet used = 17.8352 sq dm
Rate of tin sheet = Rs. 1.2 per sq dm
Total cost of the tin sheet used for making the bucket = 17.8352*1.2
= Rs. 21.40
Volume = 1/3πh(R² + r² + R*r)
= 1/3*3.14*1.2*(2² + 1² + 2*1)
= 1/3*3.14*1.2*7
= 26.376/3
Volume = 8.792 cu