A solid piece of iron of dimensions 66 cm - 49 cm - 12 cm is moulded into a sphere- The radius of the sphere is

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Volume of a Sphere

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Question

A solid piece of iron of dimensions $66 c m \times 49 c m \times 12 c m$ is moulded into a sphere. The radius of the sphere is

7 c m

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3 c m

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21 c m

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14 c m

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Solution

The correct option is C $21 c m$
Volume of sphere $=$ volume of solid iron of cuboid shape
$\frac{4}{3} \times \frac{22}{7} \times r^{3} = 66 \times 49 \times 12$
$r^{3} = \frac{66 \times 49 \times 12 \times 7 \times 3}{4 \times 22}$
$=> r^{3} = 21^{3}$
$r = 21 c m$

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A solid piece of iron of dimensions 66cm×49cm×12cm is moulded into a sphere. The radius of the sphere is

The correct option is C 21cmVolume of sphere = volume of solid iron of cuboid shape43×227×r3=66×49×12r3=66×49×12×7×34×22=>r3=213r=21cm

A solid piece of iron of dimensions $66 c m \times 49 c m \times 12 c m$ is moulded into a sphere. The radius of the sphere is

The correct option is C $21 c m$
Volume of sphere $=$ volume of solid iron of cuboid shape
$\frac{4}{3} \times \frac{22}{7} \times r^{3} = 66 \times 49 \times 12$
$r^{3} = \frac{66 \times 49 \times 12 \times 7 \times 3}{4 \times 22}$
$=> r^{3} = 21^{3}$
$r = 21 c m$