Question

8. A hemispherical bowl has radius of its rim equal to 2.1 cm . The total capacity
such bowls is nearly equal to (cm)

Answer 1

$\huge\bf{\underline{\pink{Answer:-}}}$

$\sf Volume=19.404cm{}^{3}$

$\mathbb{\underline{\purple{EXPLANATION:-}}}$

$\bold{Given}\begin{cases}\sf{Radius\:of\: hemisphere\: is 2.1cm}\end{cases}$

$\bf{To\:find}$

We have to find the capacity of bowls means find the volumeof bowl

$\bf\underline{\underline{\red{According\:To\: Question:-}}}$

$\sf Volume \:of\: hemisphere=\frac{1}{2}\times Volume \:of\: sphere$

$\bold{we\:know}\begin{cases}\sf{Volume\:of\:sphere =\frac{4}{3}\pi r{}^{3}}\end{cases}$

$\sf Volume\:of\: hemisphere=\frac{1}{\cancel 2}\times \frac{\cancel{4}}{3}\pi r{}^{3}$

$\sf Volume\:of\: hemisphere=\frac{2}{3}\pi r{}^{3}$

$\bold{Radius}\begin{cases}\sf{ 2.1cm}\end{cases}$

$\sf Volume=\frac{2}{3}\pi r{}^{3}\\ \\ \bf \implies Take\pi=\frac{22}{7}\\ \\ \sf\leadsto Volume=\frac{2}{\cancel3}\times \frac{22}{\cancel7}\times \cancel{2.1}\times \cancel{2.1}\times 2.1\\ \\ \sf Volume=2\times 22\times 0.3\times 0.7\times 2.1\\ \\ \sf\leadsto Volume=44\times 0.21\times 2.1\\ \\ \sf\leadsto Volume=44\times 0.441\\ \\ \sf\implies Volume = 19.404cm{}^{3}$

The volume of bowl :-

$\sf\implies {\red{19.404cm{}^{3}}}$

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