A vessel open at the top is in the form of a frustum of a cone. The height of the frustum of the cone is 8 cm and the radii of its lower and upper circular ends as 4 cm and 10 cm respectively . Find the cost of oil which can completely fill the vessel at the rate of rupees 50 per litre find the cost of metal sheet used at the rate of rupees 50 per 100 cm square (use pie = 3.14)
Answers
Given :
- The height of the frustum of the cone is 8 cm.
- Radius of the base, R = 10 cm and r = 4 cm.
To find :
- The cost of oil =?
- The cost of metal =?
Step-by-step explanation :
Slant height of the cone = √h² + (R - r) ²
Substituting the values in the above formula, we get,
__________
= √ 8² + ( 10 - 4)²
______
= √ 8² + 6²
______
= √ 64 + 36
___
= √ 100
= 10.
Therefore, Slant Height of the Cone = 10 cm .
We know that,
Volume of container = 1/3 π h ( R² + r² + Rr)
Substituting the values in the above formula, we get,
= 1/3 × 3.14 × 8 ( 100 + 16 + 40) cm³
= 1/3 × 3.14 × 8(156) cm³
= 1306.24 cm³
= 1306.24 / 1000 ltr ( ∴ 1000 cm = 1 ltr)
= 1.30624 lit.
= 1.31 lit. (approx)
So, Quantity of oil = 1.31 lit.
Now,
Cost of oil = Rs. (1.31 × 50) = Rs. 65.50
Now,
Surface area of the container (excluding the upper end) = Curved Surface Area + Area of Base.
Substituting the values, we get,
= 3.14 × [ 10 ( 10 + 4 ) + 16]
= 3.14 × 156
= 489.84.
Therefore, Surface area of the container (excluding the upper end) = 489.84 cm².
Now,
Cost of metal,
= Rs. (489.84 × 50/100)
= Rs. 244.92
Therefore, The Cost of metal = Rs. 244.92 (approx)