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A hemispherical brass bowl has inner-diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of ₹ 16 per 100 cm2.

Mensuration

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Answer

Given, internal diameter = 10.5 cm.

Internal radius = Internal diameter2\dfrac{\text{Internal diameter}}{2}

= 10.52=5.25\dfrac{10.5}{2} = 5.25 cm.

Internal curved surface area of hemispherical shell = 2πr2.

Putting values in equation we get,

Internal curved surface area of hemisphere = 2 x 227\dfrac{22}{7} x (5.25)2

=2×22×27.56257=1212.757=173.25 cm2.= \dfrac{2 \times 22 \times 27.5625}{7} \\[1em] = \dfrac{1212.75}{7} = 173.25 \text{ cm}^2.

The cost of tin-plating it on the inside at the rate of ₹ 16/100 cm2 or ₹ 0.16/cm2.

∴ Cost of tin-plating 173.25 cm2 = 173.25 × 0.16 = ₹ 27.72.

Hence, the cost of tin-plating = ₹ 27.72.

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