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(D) The volume of a sphere is given by $V = \frac{4}{3} \pi r^3$.Total volume of the two solid spheres $= \frac{4}{3} \pi (1^3 + 6^3) = \frac{4}{3} \p

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$1$

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(D) The volume of a sphere is given by $V = \frac{4}{3} \pi r^3$.Total volume of the two solid spheres $= \frac{4}{3} \pi (1^3 + 6^3) = \frac{4}{3} \pi (1 + 216) = \frac{4}{3} \pi (217) \, cm^3$.Let the internal radius of the hollow sphere be $r \, cm$. The volume of the material of the hollow sphere is given by $V = \frac{4}{3} \pi (R^3 - r^3)$,where $R = 9 \, cm$ is the external radius.Equating the volumes: $\frac{4}{3} \pi (217) = \frac{4}{3} \pi (9^3 - r^3)$.$217 = 729 - r^3$.$r^3 = 729 - 217 = 512$.$r = \sqrt[3]{512} = 8 \, cm$.Thickness of the hollow sphere $=$ External radius $-$ Internal radius $= 9 \, cm - 8 \, cm = 1 \, cm$.

The radii of two solid iron spheres are $1 \, cm$ and $6 \, cm$ respectively. $A$ hollow sphere is made by melting the two spheres. If the external radius of the hollow sphere is $9 \, cm$,then its thickness (in $cm$) is

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