Question

a spherical iron shell with 8cm external diameter wieghs 1860 4/7g.find the thickness of the shell of the density of metal 12g/cubic cm.

Answer 1

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Answer 2

$Given \: external \: diameter \: of \: a \\spherical \: iron \: shell (d) = 8 \:cm$

$Let \: external \: radius (R) = \frac{d}{2} \\= \frac{8}{2} \\= 4 \:cm$

$Let \: internal \: radius \: of \: the \:shell = r\:cm$

$i) Volume \: of \:the \:shell (V) \\= Volume_{(external ) }- Volume_{(internal)} \\= \frac{4}{3} \pi R^{3} - \frac{4}{3} \pi r^{3}\\= \frac{4}{3}\pi (R^{3} - r^{3} ) \\= \frac{4}{3} \pi (4^{3} - r^{3} )\: --(1)$

$Weight \: of \: the \:shell (W)\\ = 1860\frac{4}{7} \:g\\= \frac{13024}{7} \:g$

$Density \: of \:the \:shell (D) = 12 \:g/cm^{3}$

$\boxed { \pink { Volume = \frac{Weight}{Density }}}$

$\implies \frac{4}{3} \pi(4^{3} - r^{3} ) = \frac{\frac{13020}{7}}{12}$

$\implies 64 - r^{3} = \frac{13024}{7\times 12}\times \frac{3}{4}\times \frac{7}{22}$

$\implies 64 - r^{3} = 37$

$\implies 64 - 37 = r^{3}$

$\implies 27 = r^{3}$

$\implies r^{3} = 3^{3}$

$\implies r = 3 \: cm$

$\implies Thickness \: of \: the \:shell = R - r \\= 4 \:cm - 3 \:cm \\= 1 \: cm$

Therefore.,

$\red {Thickness \: of \: the \:shell} \green {= 1 \: cm}$

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