A spherical ball of lead has been melted and made into identical smaller balls with radius equal to half the radius of the original one How many such balls can be made?

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Solution

Let the radius be R
radius of smaller ball be $\frac{R}{2}$
No. of balls $= \frac{v o l u m e o f b i g b a l l}{v o l u m e o f 1 s m a l l e r b a l l}$
$= \frac{\frac{4}{3} π R^{3}}{\frac{4}{3} π {(\frac{R}{2})}^{3}} = \frac{8 R^{3}}{R^{3}} = 8$ balls.

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A spherical ball of lead has been melted and made into identical smaller balls with radius equal to half the radius of the original one How many such balls can be made?

Let the radius be Rradius of smaller ball be R2No. of balls =volumeofbigballvolumeof1smallerball=43πR343π(R2)3=8R3R3=8 balls.

Let the radius be R
radius of smaller ball be $\frac{R}{2}$
No. of balls $= \frac{v o l u m e o f b i g b a l l}{v o l u m e o f 1 s m a l l e r b a l l}$
$= \frac{\frac{4}{3} π R^{3}}{\frac{4}{3} π {(\frac{R}{2})}^{3}} = \frac{8 R^{3}}{R^{3}} = 8$ balls.