A right circular metal cone (solid) is 42 cm high and its radius is $\frac{21}{2}$ cm. It is melted and recast into a sphere. Then the radius of the sphere will be:

When a solid object is melted and recast into another shape, the volume remains the same.

Volume of Cone = Volume of Sphere

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For the Cone:

• Height $(h) = 42$ cm
• Radius $(r) = \frac{21}{2}$ cm

For the Sphere:

• Radius $(R) = ?$

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Volume of Cone $= \frac{1}{3}\pi r^2 h$

Volume of Sphere $= \frac{4}{3}\pi R^3$

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Setting the volumes equal:

$\frac{1}{3}\pi r^2 h = \frac{4}{3}\pi R^3$

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Substituting the values:

$\frac{1}{3} \times \pi \times \left(\frac{21}{2}\right)^2 \times 42 = \frac{4}{3} \times \pi \times R^3$

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Canceling $\pi$ from both sides:

$\frac{1}{3} \times \left(\frac{21}{2}\right)^2 \times 42 = \frac{4}{3} \times R^3$

$\left(\frac{21}{2}\right)^2 = \frac{441}{4}$

$\frac{1}{3} \times \frac{441}{4} \times 42 = \frac{4}{3} \times R^3$

Multiplying both sides by $3$:

$\frac{441}{4} \times 42 = 4R^3$

$\frac{441 \times 42}{4} = 4R^3$

$\frac{18522}{4} = 4R^3$

$4630.5 = 4R^3$

$R^3 = 1157.625$

$R = 10.5$ cm $= \frac{21}{2}$ cm

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Therefore, the radius of the sphere is $\frac{21}{2}$ cm.