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3080 cm3 of water is required to fill a cylindrical vessel completely and 2310 cm3 of water is required to fill it upto 5 cm below the top. Find :

(i) radius of the vessel.

(ii) height of the vessel.

(iii) wetted surface area of the vessel when it is half-filled with water.

Mensuration

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Answer

Let radius of vessel be r cm and height be h cm.

Given,

Volume of cylindrical vessel = 3080 cm3

∴ πr2h = 3080 ………..(1)

Given,

It takes 2310 cm3 of water to fill cylinder upto 5 cm below the top.

∴ πr2(h - 5) = 2310 ………..(2)

Dividing (1) by (2) we get,

πr2hπr2(h5)=30802310hh5=308231231h=308(h5)231h=308h1540308h231h=154077h=1540h=154077h=20 cm\Rightarrow \dfrac{πr^2h}{πr^2(h - 5)} = \dfrac{3080}{2310} \\[1em] \Rightarrow \dfrac{h}{h - 5} = \dfrac{308}{231} \\[1em] \Rightarrow 231h = 308(h - 5) \\[1em] \Rightarrow 231h = 308h - 1540 \\[1em] \Rightarrow 308h - 231h = 1540 \\[1em] \Rightarrow 77h = 1540 \\[1em] \Rightarrow h = \dfrac{1540}{77} \\[1em] \Rightarrow h = 20 \text{ cm}

(i) Substituting value of h in equation (1), we get :

πr2×20=3080227×r2×20=3080r2=3080×720×22r2=49r=7 cm.\Rightarrow πr^2 \times 20 = 3080 \\[1em] \Rightarrow \dfrac{22}{7} \times r^2 \times 20 = 3080 \\[1em] \Rightarrow r^2 = \dfrac{3080 \times 7}{20 \times 22} \\[1em] \Rightarrow r^2 = 49 \\[1em] \Rightarrow r = 7 \text{ cm}.

Hence, radius of vessel = 7 cm.

(ii) From above,

h = 20 cm.

Hence, height of vessel = 20 cm.

(iii) When vessel is half-filled, water will be filled upto 202\dfrac{20}{2} = 10 cm.

Wetted surface area = 2πrh + πr2

= πr(2h + r)

= 227×7×(2×10+7)\dfrac{22}{7} \times 7 \times (2 \times 10 + 7)

= 22×2722 \times 27

= 594 cm2.

Hence, wetted surface area = 594 cm2.

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