A milk container is made of metal sheet in the shape of frustum of a cone whose volume is 10459 3/7cm3. The radii of its lower and upper ends are 8cm and 20 cm respectively. Find the cost of metal sheet used in making the container at the rate of Rs 1.40 per square centimeter.
Answers
Answer:
Rs. 2462.60
Step-by-step explanation:
Upper radius (r) = 8 cm
Lower diameter (R)= 20 cm
The volume of frustum = 10459 3/7 cm³ = 10459.4285 cm³
Therefore height of frustum(h) = ?
Volume = π/3. h (R² + r² + R.r)
10459.4285 = π/3. h (20² + 8² + 160)
∴ h = 16.01 cm ≅ 16 cm
The surface area of container = π (R+r) √(R-r)² + h²
= π (20 + 8) √(12² + 16²) ≅ 1759 cm²
Cost of metal for making container = ₹ 1.4 /cm²
→ Cost of metal = 1.4 . 1749 = Rs. 2462.60
Answer:
Step-by-step explanation:
R = 20 cm
r = 8 cm
h = Height in cm
Volume = (1/3) π h (R² + r² + R.r)
=> 10459 3/7 = (1/3) (22/7) h ( 20² + 8² + 20*8)
=> 73216/7 = (22h/21) (400 + 64 + 160)
=> 9984 = h (624)
=> h = 16
Height = 16 cm
Surface Area = π(R + r)( √(R-r)² + h²) + πR² + πr²
= (22/7) (20 + 8) (√(20-8)² + 16²) + (22/7)20² + (22/7)8²
=(22/7) (28) (20) + (22/7)( 400 + 64)
= 1760 + 10208/7
= 22528/7 cm²
Cost = 1.4 * 22528/7 = Rs 4505.6
If it is open from top then surface Area
Surface Area = π(R + r)( √(R-r)² + h²) + πR²
= (22/7) (28) (20) + (22/7)( 400)
= 1760 + 8800/7
= 21120/7 cm²
Cost = 1.4 * 21120/7 = Rs 4224