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A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of Rs 16 per 100 cm^2.

Q : 5    A hemispherical bowl made of brass has inner diameter \small 10.5\hspace{1mm}cm. Find the cost of tin-plating it on the inside at the rate of  Rs 16 per \small 100\hspace{1mm}cm^2

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Given,

The inner radius of the hemispherical bowl = r = \frac{10.5}{2}\ cm

We know,

The curved surface area of a hemisphere = 2\pi r^2

\therefore The surface area of the hemispherical bowl = 2\times\frac{22}{7}\times\left (\frac{10.5}{2} \right )^2

=11\times1.5\times10.5

= 173.25 \ cm^2

Now,

Cost of tin-plating \small 100\hspace{1mm}cm^2 =  Rs 16

\therefore Cost of tin-plating \small 33\hspace{1mm}cm^2 =  \small \\ Rs. \left (\frac{16}{100}\times173.25 \right )

\small = Rs. 27.72

Therefore, the cost of tin-plating it on the inside is \small Rs. 27.72

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