a hemispherical bowl made of brass …
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Preeti Dabral 1 year, 9 months ago
Inner diameter of bowl = 10.5 cm
{tex}\therefore {/tex} Inner radius of bowl {tex}\left( r \right)=\frac{10.5}{2}{/tex} = 5.25 cm
Now, Inner surface area of bowl = {tex}2\pi {{r}^{2}}{/tex}
{tex}=2\times \frac{22}{7}\times 5.25\times 5.25{/tex}
{tex}=2\times \frac{22}{7}\times \frac{21}{4}\times \frac{21}{4}{/tex}
={tex}\frac{693}{4}c{{m}^{2}}{/tex}
{tex}\because {/tex} Cost of tin-plating per{tex} 100\text{ }c{{m}^{2}}{/tex} = ₹ 16
{tex}\therefore {/tex} Cost of tin-plating per{tex} 1\text{ }c{{m}^{2}}=\frac{16}{100}{/tex}
{tex}\therefore {/tex} Cost of tin-plating per {tex}\frac{693}{4}c{{m}^{2}}=\frac{16}{100}\times \frac{693}{4}{/tex} = ₹27.72
0Thank You