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A tent of height $3.3 m$ is in the form of a right circular cylinder of diameter $12 m$ and height $2.2 m$ , surmounted by a right circular cone of the same diameter. Find the cost of the canvas of the tent at the rate of $Rs . 500 per m^{2}$ .

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Rs . 4900

Rs . 2000

Rs . 99000

Rs . 5600

answer is C.

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Detailed Solution

Height of tent

= 3.3 m

The height of cylindrical proportion

∴ T h e h eig h t of conical portion = Heig h t of tent - Heig h t of cylindrical portion

= 3.3 - 2.2

= 1.1 m

Given, diameter of the cylinder,

d = 12 m

∴ Radius

= \frac{12}{2} = 6 m

CSA of cylindrical portion

= 2 πr h

= 2 \times \frac{22}{7} \times 6 \times 2.2

= 82.971 m^{2}

Firstly, we have to find the slant height

(l)

of the conical portion

\Rightarrow l^{2} = h^{2} + r^{2}

\Rightarrow l^{2} = {(1.1)}^{2} + {(6)}^{2}

\Rightarrow l^{2} = 1.21 + 36

\Rightarrow l^{2} = 37.21

\Rightarrow l = \sqrt{37.21}

\Rightarrow l = 6.1 m

∴ CSA of the conical portion

= πrl

= \frac{22}{7} \times 6 \times 6.1

= 115.029 m^{2}

So,

Total Surface area of tent = Surface area of conical portion + Surface area of cylindrical portion

= 115.029 + 82.971

= 198 m^{2}

∴ Canvas required to make the tent

= 198 m^{2}

Cost of

1 m^{2}

canvas

= Rs . 500

Cost of

198 m^{2}

canvas

= Rs . 500 \times 198

= Rs . 99000

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