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The adjoining figure shows a model of a solid consisting of a cylinder surmounted by a hemisphere at one end. If the model is drawn to a scale of 1 : 200, find

(i) the total surface area of the solid in π m2.

(ii) the volume of the solid in π litres.

The adjoining figure shows a model of a solid consisting of a cylinder surmounted by a hemisphere at one end. If the model is drawn to a scale of 1 : 200, find the total surface area of the solid and the volume of the solid. Mensuration, ML Aggarwal Understanding Mathematics Solutions ICSE Class 10.

Mensuration

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Answer

(i) In the given figure,

Height of cylindrical portion (H) = 8 cm.

Radius (r) = 3 cm.

Scale = 1 : 200

∴ k = 200.

Total surface area (S) = Surface area of hemisphere + Curved surface area of cylinder = 2πr2 + 2πrH

S=2πr(r+H)=2×π×3×(3+8)=66π cm2.S = 2πr(r + H) \\[1em] = 2 \times π \times 3 \times (3 + 8) \\[1em] = 66π \text{ cm}^2.

∴ Surface area of solid = 66π × k2 = 66π × (200)2

= 66π × 40000 cm2

= 66π×40000100×100\dfrac{66π \times 40000}{100 \times 100} m2

= 264π m2.

Hence, the surface area of solid = 264π m2.

(ii) Volume (V) = Volume of hemisphere + Volume of cylinder = 23πr3+πr2H\dfrac{2}{3}πr^3 + πr^2H

Putting values we get,

V=πr2(23r+H)=πr2(23×3+8)=π×(3)2×(2+8)=9π×10=90π m3.V = πr^2\Big(\dfrac{2}{3}r + H\Big) \\[1em] = πr^2 \Big(\dfrac{2}{3} \times 3 + 8\Big) \\[1em] = π \times (3)^2 \times (2 + 8) \\[1em] = 9π \times 10 \\[1em] = 90π \text{ m}^3.

∴ Volume of solid = 90π × k3 = 90π × (200)3

= 90π × 8000000 = 72000000π cm3

= 720000000π100×100×100 m3\dfrac{720000000π}{100 \times 100 \times 100} \text{ m}^3

= 720π m3.

1 m3 = 1000 litres.

∴ Volume of solid = 720π × 1000 = 720000π litres.

Hence, the volume of solid = 720000π litres.

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