A cubical block of side $10\ cm$ is surmounted by a hemisphere. What is the largest diameter that the hemisphere can have? Find the cost of painting the total surface area of the solid so formed, at the rate of Rs. $5$ per sq. cm. [Use $ \pi \ =\ 3.14$]

Given: A cubical block of side $10\ cm$ is surmounted by a hemisphere.

To do: To find the cost of painting the total surface area of the solid so formed, at the rate of Rs. 5 per sq. cm.

Solution:

Side of the cubical block, $a = 10\ cm$

Longest diagonal of the cubical block $= a\sqrt{2}\ cm=10\sqrt{2}\ cm$

Since the cube is surmounted by a hemisphere,

Therefore the side of the cube should be equal to the diameter of the hemisphere.

Diameter of the sphere$= 10\ cm$

Radius of the sphere, $r = 5\ cm$

Total surface area of the solid $=$Total surface area of the cube$ –$ Inner cross‐section area of the hemisphere $+$ Curved surface area of the hemisphere

$=6a^{2} -\pi r^{2} +2\pi r^{2}$

$=6a^{2} +\pi r^{2}$

$=6\times 10\times 10+3.14\times 5\times 5$

$=600+78.5$

$=678.5\ cm^{2}$

Rate to form such a solid$=Rs.\ 5\ sq.\ cm.$

Total cost to form such the solid$=678.5\times 5$

 $=Rs.\ 3392.5$

Therefore the largest diameter the hemisphere can have $=$side of the cube$=10\ cm$.

And the cost to form such a solid$=Rs.3392.5$
 

Updated on: 10-Oct-2022

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