Q. 9

Answer :

Given: depth of bucket = height of bucket/frustum = h = 24 cm

Diameter of lower circular end = 10 cm

Diameter of upper circular end = 30 cm

∴ Radius of lower circular end = r = 10/2 = 5 cm

∴ Radius of lower circular end = R = 30/2 = 15 cm

Cost of 100 cm^{2} metal sheet = 10 Rs

∴ Cost of 1 cm^{2} metal sheet = 10/100 = 0.1 Rs

Cost of 1 litre milk = 20 Rs

Total surface area of frustum = πr^{2} + πR^{2} + π(R + r)l cm^{2}

Where l = slant height

∴ l = 26 cm

Since the top is open we need to subtract the area of top/upper circle from total surface area of frustum because we don’t require a metal plate for top.

Radius of top/upper circle = R

Area of upper circle = πR^{2}

∴ area of metal sheet used = (total surface area of frustum)-πR^{2}

= πr^{2} + πR^{2} + π(R + r)l- πR^{2} cm^{2}

= πr^{2} + π(R + r)l cm^{2}

= π × (5^{2} + (15 + 5)26) cm^{2}

= 3.14 × 545 cm^{2}

= 1711.3 cm^{2}

∴ 1711.3 cm^{2} metal sheet is required to make the container.

∴ Cost of 1711.3 cm^{2} metal sheet = 1711.3 × cost of 1 cm^{2} metal sheet

= 1711.3 × 0.1 Rs

= 171.13 Rs

∴ Cost of metal sheet required to make container = 171.13 Rs

Now,

Volume of milk which can completely fill the bucket = volume of frustum

= (1/3) × 3.14 × 26 × 325 cm^{3}

= 26533/3 cm^{3}

= 8844.33 cm^{3}

Now 1 litre = 1000 cm^{3}

∴ 8844.33 cm^{3} = 8844.33/1000 litres

= 8.84433 litres

∴ Volume of milk which can completely fill the bucket = 8.84433 litres

∴ Cost of milk which can completely fill the bucket = volume of milk which can completely

fill the bucket × cost of 1 litre milk Rs

= 8.84433 × 20 Rs

= 176.8866 Rs

Cost of milk which can completely fill the bucket = 176.8866 Rs

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