A metallic sphere of radius 21 ,cm is melted | vedclass

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(D) Radius of the sphere $R = 21 \,cm$.Volume of the sphere $= \frac{4}{3} \pi R^3 = \frac{4}{3} \pi (21)^3 \,cm^3$.For the cylindrical wire,diameter

Vedclass · Accepted Answer

$1975.68$

Vedclass · Answer

(D) Radius of the sphere $R = 21 \,cm$.Volume of the sphere $= \frac{4}{3} \pi R^3 = \frac{4}{3} \pi (21)^3 \,cm^3$.For the cylindrical wire,diameter $= 0.5 \,cm$,so radius $r = \frac{0.5}{2} = 0.25 = \frac{1}{4} \,cm$.Let the length of the wire be $h \,cm$.Volume of the wire $= \pi r^2 h = \pi (\frac{1}{4})^2 h \,cm^3$.Since the sphere is recast into a wire,their volumes are equal:$\pi (\frac{1}{4})^2 h = \frac{4}{3} \pi (21)^3$$\frac{1}{16} h = \frac{4}{3} 	imes 21 	imes 21 	imes 21$$h = 16 	imes 4 	imes 7 	imes 21 	imes 21$$h = 64 	imes 7 	imes 441 = 197568 \,cm$.To convert into meters,divide by $100$:$h = \frac{197568}{100} = 1975.68 \,m$.Thus,the length of the wire is $1975.68 \,m$.

$A$ metallic sphere of radius $21 \,cm$ is melted and recast to form a wire with diameter $0.5 \,cm$. Find the length of the wire in $m$.

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