Question

What is the ratio of the sum of volumes of two-cylinder of radius 1 cm and height 2 cm each

to the volume of a sphere of radius 3 cm?

Need help with this question, the answer I keep getting is 1:9​

Answer 1

$\frak{Given} \begin{cases} \sf Radius\:of\:two\:cylinders\: = \frak{1\:cm} & \\ \\\sf Height\:of\:two\:cylinders\: = \frak{2\:cm}& \\ \\ \sf Radius\:of\:sphere\: = \frak{31\:cm}&\end{cases}$

Need to find: The ratio of the sum of volumes of two cylinder to the volume of a sphere.

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$\star$ As we know that,

Volume of Cylinder = $\frak{\pi r^2 h}$

Volume of Sphere = $\frak{\dfrac{4}{3} \pi r^3}$

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$\bigstar\:{\underline{\sf{\pmb{According\:to\:the\:Question\::}}}}$

$\star$There are two cylinder of equal radius and equal Height. So, We can say that there volume is also equal.

$\star$Now, We can calculate the ratio of volume of those two cylinders to the volume of sphere as,

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$:\implies \frak{\dfrac{2 \times (Volume\:of\:cylinder)}{ (Volume\:of\:sphere)}}\\\\\\ :\implies\sf \dfrac{2 \times \bigg( \dfrac{22}{7} \times (1)^2 \times 2 \bigg)}{ \dfrac{4}{3} \times \dfrac{22}{7} \times (3)^3}\\\\\\ :\implies\sf \dfrac{2 \times \bigg( \dfrac{22}{7} \times 1 \times 2 \bigg)}{ \dfrac{4}{\cancel{3}} \times \dfrac{22}{7} \times \cancel{27}}\\\\\\ :\implies\sf \dfrac{2 \times \dfrac{22}{7} \times 2}{4 \times \dfrac{22}{7} \times 9}\\\\\\:\implies\sf \dfrac{4 \times \cancel{\dfrac{22}{7}}}{4 \times \cancel{\dfrac{22}{7}} \times 9}\\\\\\ :\implies\sf \dfrac{\cancel{4}}{\cancel{4} \times 9}\\ \\\\ :\implies{\boxed{\underline{ \frak{ \purple{ \dfrac{1}{9}}}}}}\:\bigstar$

$\therefore\:{\underline{\sf{Hence\:the\:required\:ratio\:is\: {\textsf{\textbf{1:9}}}.}}}$

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$\star$ The answer you're getting is correct mate! :)