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 cm^{2}. (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 cm^{2}

Formula used/Theory.

CSA of frustum = π(r + R) × l

Volume of frustum = h[R^{2} + r^{2} + Rr]

⇒ Volume of Frustum = h[R^{2} + r^{2} + Rr]

= × 30cm × [(30 cm)^{2} + (10 cm)^{2} + 30 cm × 10 cm]

= 31.4 cm × [900cm^{2} + 100cm^{2} + 300cm^{2}]

= 31.4 cm × 1300 cm^{2}

= 40820 cm^{3}

1 cm^{3} = litres

40820 cm^{3} = litres

= 40.820 litres

As milk is Rs. 30 per litre

For 40.82 litres

Rs. 30 × 40.82

= Rs. 1224.6

In frustum

L^{2} = height^{2} +

L^{2} = 30^{2} + [30 – 10]^{2}

L^{2} = 900 cm^{2} + 400 cm^{2} = 1300 cm^{2}

L = √(1300 cm^{2}) = 10√13 cm

⇒ CSA of Frustum = π(30 + 10)10√13 cm

= 3.14 × 40cm × 10√13cm

= 1256√13cm^{2}

⇒ Area of base = πr^{2}

= 3.14 × 10 × 10

= 314 cm^{2}

Area of container = 1256√13 + 314

= 314[4√13 + 1]

= 314[4 × 3.6 + 1]

= 314 × 15.4

= 4835.6 cm^{2}

Cost of 100 cm^{2} metal sheet = Rs. 50

Cost of 1 cm^{2} metal sheet = Rs. 0.5

Cost of 4835.6 cm^{2} metal sheet = Rs. 4835.6 × 0.5

= Rs. 2417.8

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If the radii of a frustum of a cone are 7 cm and 3 cm and the height is 3 cm, then the curved surface area is …………cm^{2}.