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dk_ch Share
Feb 10, 2018

The surface area of the cone=area of the given metal lamina in the shape of a sector of radius r is given by

$= \frac{360 - 144}{360} \times \pi {r}^{2} = \frac{216}{360} \times \pi r = \frac{3}{5} \pi {r}^{2}$

The slant height of the cone will be radius of the sector $r$

If the radius of the cone be ${r}_{c}$ then
its surface area will be $= \pi {r}_{c} \times r = \frac{3}{5} \pi {r}^{2}$
$\implies {r}_{c} = \frac{3}{5} r$

Height of the cone $h = \sqrt{{\text{slant height"^2-"radius}}^{2}}$

$= \sqrt{{r}^{2} - {r}_{c}^{2}} = \sqrt{{r}^{2} - \frac{9}{25} {r}^{2}} = \frac{4}{5} r$

So volume of the cone $V = \frac{1}{3} \pi {r}_{c}^{2} h = \frac{1}{3} \pi {\left(\frac{3}{5} r\right)}^{2} \times \frac{4}{5} r$

Volume of n pieces of spheres each of radius $a$ will be

$= \frac{4}{3} \pi {a}^{3} n$

By the condition of the problem we have

$\frac{4}{3} \pi {a}^{3} n = \frac{1}{3} \pi {\left(\frac{3}{5} r\right)}^{2} \times \frac{4}{5} r = \frac{3 \times 4}{125} \pi {r}^{3}$

$\implies 125 n {a}^{3} = 9 {r}^{3}$

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