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# A solid metallic sphere of radius 5.6 cm is melted and solid cones each of radius 2.8 cm and height 3.2 cm are made. Find the number of such cones formed.

Last updated date: 23rd May 2024
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Hint: In this question the sphere is melted and cones are made, so the volume of the sphere should be equal to the sum of all the volumes of the cones, so use this concept to reach the solution of the question.

It is given that the radius (r) of a solid metallic sphere is 5.6 cm.
Now as we know that the volume (${V_s}$) of a solid metallic sphere is $\dfrac{4}{3}\pi {r^3}$.
$\Rightarrow {V_s} = \dfrac{4}{3}\pi {\left( {5.6} \right)^3}{\text{ c}}{{\text{m}}^3}$.
Now it is given that radius $\left( {{r_1}} \right)$ and the height (h) of the cone is 2.8 and 3.2 cm respectively.
As we know that the volume (${V_c}$) of cone is $\dfrac{1}{3}\pi {r_1}^2h$
$\Rightarrow {V_c} = \dfrac{1}{3}\pi {\left( {2.8} \right)^2}\left( {3.2} \right){\text{ c}}{{\text{m}}^3}$
Now it is given that the sphere is melted and cones are made.
Let the number of cones be $x$.
So, the volume of the sphere should be equal to the $x$ multiplied by the volume of a single cone.
$\Rightarrow {V_s} = x\left( {{V_c}} \right)$
$\Rightarrow \dfrac{4}{3}\pi {\left( {5.6} \right)^3} = x\left( {\dfrac{1}{3}\pi {{\left( {2.8} \right)}^2}\left( {3.2} \right)} \right)$
$\Rightarrow 4{\left( {5.6} \right)^3} = x{\left( {2.8} \right)^2}\left( {3.2} \right) \\ \Rightarrow 4{\left( {2 \times 2.8} \right)^3} = x{\left( {2.8} \right)^2}\left( {3.2} \right) \\ \Rightarrow 4\left( {{2^3} \times 2.8} \right) = x\left( {3.2} \right) \\ \Rightarrow x = \dfrac{{4 \times 8 \times 2.8}}{{3.2}} = \dfrac{{32 \times 28}}{{32}} = 28 \\$
So, the required number of cones is 28.

Note: In such types of questions the key concept we have to remember is that always recall the formulas of sphere and cone which is stated above, then according to given condition equate them as above and simplify, we will get the required number of cones which are made after melting the solid sphere.