Answer
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Hint: When something is melted and converted to another shape then its volume will not change.
The problem statement says, the cuboid is melted so the hack here is, volume will not be changed. So, the volume of cuboid will be equal to the volume of solid cone. We know that, Volume of cuboid$ = l \times b \times h$where,$l = length,b = breath\;{\text{and }}h = height$. Also, volume of cone$ = \dfrac{1}{3}\pi {r^2}h$.
Here we have given length, breadth and height of a solid metallic cuboid and height of the cone. The only remaining unknown in both the volume formulas is radius. Let’s calculate it.
$
l \times b \times h = \dfrac{1}{3}\pi {r^2}h \\
\Rightarrow 44 \times 21 \times 12 = \dfrac{1}{3} \times \dfrac{{22}}{7}{r^2} \times 24{\text{ }}[l = 44cm,b = 21cm,{h_{cuboid}} = 12cm,{h_{cone}} = 24cm] \\
\Rightarrow 44 \times 21 = \dfrac{1}{3} \times \dfrac{{22}}{7}{r^2} \times 2 \\
\Rightarrow 21 = \dfrac{1}{3} \times \dfrac{{{r^2}}}{7} \\
\Rightarrow {r^2} = 21 \times 3 \times 7 \\
\Rightarrow {r^2} = 21 \times 21 \\
\Rightarrow r = 21cm \\
$
But we need to find diameter. So, we’ll use$d = 2r$. That is$d = 2 \times 21 = 42cm$.
Hence the required diameter of the cone is$42cm$.
Note: The hack here in this question was to understand, when something is melted and converted to another shape then its volume will not change then using the correct formula will lead us to the answer.
The problem statement says, the cuboid is melted so the hack here is, volume will not be changed. So, the volume of cuboid will be equal to the volume of solid cone. We know that, Volume of cuboid$ = l \times b \times h$where,$l = length,b = breath\;{\text{and }}h = height$. Also, volume of cone$ = \dfrac{1}{3}\pi {r^2}h$.
Here we have given length, breadth and height of a solid metallic cuboid and height of the cone. The only remaining unknown in both the volume formulas is radius. Let’s calculate it.
$
l \times b \times h = \dfrac{1}{3}\pi {r^2}h \\
\Rightarrow 44 \times 21 \times 12 = \dfrac{1}{3} \times \dfrac{{22}}{7}{r^2} \times 24{\text{ }}[l = 44cm,b = 21cm,{h_{cuboid}} = 12cm,{h_{cone}} = 24cm] \\
\Rightarrow 44 \times 21 = \dfrac{1}{3} \times \dfrac{{22}}{7}{r^2} \times 2 \\
\Rightarrow 21 = \dfrac{1}{3} \times \dfrac{{{r^2}}}{7} \\
\Rightarrow {r^2} = 21 \times 3 \times 7 \\
\Rightarrow {r^2} = 21 \times 21 \\
\Rightarrow r = 21cm \\
$
But we need to find diameter. So, we’ll use$d = 2r$. That is$d = 2 \times 21 = 42cm$.
Hence the required diameter of the cone is$42cm$.
Note: The hack here in this question was to understand, when something is melted and converted to another shape then its volume will not change then using the correct formula will lead us to the answer.
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