Q. 25.0( 4 Votes )
A container, open from the top and made up of a metal sheet is the form of frustum of a cone of height 30 cm with radii 30 cm and 10 cm. Find the cost of the milk which can completely fill container at the rate of Rs. 30 per litre. Also find the cost of metal sheet used to make the container, if it costs Rs. 50 per 100 cm2. (π = 3.14)
Answer :
Given.
Radii of frustum is 10 cm and 30 cm
Height of frustum is 30 cm
Rate of milk is Rs. 30 per litre
Rate metallic sheet is Rs . 50 per 100 cm2
Formula used/Theory.
CSA of frustum = π(r + R) × l
Volume of frustum = 
h[R2 + r2 + Rr]
⇒ Volume of Frustum = 
h[R2 + r2 + Rr]
= 
× 30cm × [(30 cm)2 + (10 cm)2 + 30 cm × 10 cm]
= 31.4 cm × [900cm2 + 100cm2 + 300cm2]
= 31.4 cm × 1300 cm2
= 40820 cm3
1 cm3 = 
litres
40820 cm3 = 
litres
= 40.820 litres
As milk is Rs. 30 per litre
For 40.82 litres
Rs. 30 × 40.82
= Rs. 1224.6
In frustum
L2 = height2 + 
L2 = 302 + [30 – 10]2
L2 = 900 cm2 + 400 cm2 = 1300 cm2
L = √(1300 cm2) = 10√13 cm
⇒ CSA of Frustum = π(30 + 10)10√13 cm
= 3.14 × 40cm × 10√13cm
= 1256√13cm2
⇒ Area of base = πr2
= 3.14 × 10 × 10
= 314 cm2
Area of container = 1256√13 + 314
= 314[4√13 + 1]
= 314[4 × 3.6 + 1]
= 314 × 15.4
= 4835.6 cm2
Cost of 100 cm2 metal sheet = Rs. 50
Cost of 1 cm2 metal sheet = Rs. 0.5
Cost of 4835.6 cm2 metal sheet = Rs. 4835.6 × 0.5
= Rs. 2417.8
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