Question

plzzz solve kal exam hai..... ​

Answer 1

${ \bold { \underline{\large{Given : - }}}} \:$

$\bold{The\:diameter\:of\: moon \:is \:\dfrac{1}{4}\: of\: diameter \:of \:earth.}$

${ \bold { \underline{\large{To\:Find : - }}}} \:$

◍ $\bold{Ratio\: of\: their\: surface\: areas.}$

◍ $\bold{Ratio\: of\: their\: volume.}$

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$\large\boxed{\boxed{\tt{So\:let's \:do\:it\:!}}}$

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${ \bold { \underline{\large{Solution : - }}}} \:$

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$\bold{Let \:diameter\:of\:earth\:\leadsto\:8x}$

$\therefore{\bold{Radius(r_{1}) \:of\: earth \:\leadsto\: \dfrac{8x}{2}\:\leadsto\:\cancel{\dfrac{8x}{2}}\:\leadsto\: \boxed{4x}}}$

Now,

$\bold{Diameter\:of\:moon\:is \:\dfrac{1}{4}\: of\: diameter \:of \:earth}$

$\therefore{\bold{Diameter\:of\:moon\:\leadsto\:\dfrac{8x}{4}\:\leadsto\:\cancel{\dfrac{8x}{4}}\:\leadsto\:\boxed{2x}}}$

$\bold{It's \:radius(r_{2})\:\leadsto\:\dfrac{2x}{2} \:\leadsto\:\cancel{\dfrac{2x}{2}}\:\leadsto\:\boxed{x}}$

◍ ${ \bold { \underline{Ratio\:of\:their\:surface\:areas : - }}} \:$

ㅤㅤㅤ$\longmapsto$ $\bold{\dfrac{Surface\:area_{(moon)}}{Surface\:area_{(Earth)}}}$

ㅤㅤㅤ$\longmapsto$ $\bold{\dfrac{4\pi (r_{2})^2}{4\pi (r_{1})^2}}$

ㅤㅤㅤ$\longmapsto$ $\bold{\dfrac{4\pi}{4\pi}\:\times\:\dfrac{(r_{2})^2}{(r_{1})^2}}$

ㅤㅤㅤ$\longmapsto$ $\bold{\cancel{\dfrac{4\pi}{4\pi}}\:\times\:\dfrac{(r_{2})^2}{(r_{1})^2}}$

ㅤㅤㅤ$\longmapsto$ $\bold{\dfrac{(r_{2})^2}{(r_{1})^2}}$

ㅤㅤㅤ$\longmapsto$ $\bold{\dfrac{(x)^2}{(4x)^2}}$

ㅤㅤㅤ$\longmapsto$ $\bold{\dfrac{x^2}{16x^2}}$

ㅤㅤㅤ$\longmapsto$ $\bold{\dfrac{1}{16}\:\times\:\dfrac{x^2}{x^2}}$

ㅤㅤㅤ$\longmapsto$ $\bold{\dfrac{1}{16}\:\times\:\cancel{\dfrac{x^2}{x^2}}}$

ㅤㅤㅤ$\longmapsto$ $\boxed{\bold{\dfrac{1}{16}}}$

$\boxed {\frak {\therefore \purple {Ratio\:of\:their\:surface\:areas\:\leadsto\:1\:\ratio\:16}}}\:\orange\bigstar$

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$\bold{Let \:diameter\:of\:earth\:\leadsto\:16x}$

$\therefore{\bold{Radius(r_{1}) \:of\: earth \:\leadsto\: \dfrac{16x}{2}\:\leadsto\:\cancel{\dfrac{16x}{2}}\:\leadsto\: \boxed{8x}}}$

Now,

$\bold{Diameter\:of\:moon\:is \:\dfrac{1}{4}\: of\: diameter \:of \:earth}$

$\therefore{\bold{Diameter\:of\:moon\:\leadsto\:\dfrac{16x}{4}\:\leadsto\:\cancel{\dfrac{16x}{4}}\:\leadsto\:\boxed{4x}}}$

• $\bold{It's \:radius(r_{2})\:\leadsto\:\dfrac{4x}{2} \:\leadsto\:\cancel{\dfrac{4x}{2}}\:\leadsto\:\boxed{2x}}$

◍ ${ \bold { \underline{Ratio\:of\:their\:volumes : - }}} \:$

ㅤㅤㅤ$\longmapsto$ $\bold{\dfrac{Volume_{(moon)}}{Volume_{(Earth)}}}$

ㅤㅤㅤ$\longmapsto$ $\bold{\dfrac{4/3\pi (r_{2})^3}{4/3\pi (r_{1})^3}}$

ㅤㅤㅤ$\longmapsto$ $\bold{\dfrac{4/3\pi}{4/3\pi}\:\times\:\dfrac{(r_{2})^3}{(r_{1})^3}}$

ㅤㅤㅤ$\longmapsto$ $\bold{\cancel{\dfrac{4/3\pi}{4/3\pi}}\:\times\:\dfrac{(r_{2})^3}{(r_{1})^3}}$

ㅤㅤㅤ$\longmapsto$ $\bold{\dfrac{(r_{2})^3}{(r_{1})^3}}$

ㅤㅤㅤ$\longmapsto$ $\bold{\dfrac{(2x)^3}{(8x)^3}}$

ㅤㅤㅤ$\longmapsto$ $\bold{\dfrac{2x}{8x}\:\times\:\dfrac{2x}{8x}\:\times\:\dfrac{2x}{8x}}$

ㅤㅤㅤ$\longmapsto$ $\bold{\cancel{\dfrac{2x}{8x}}\:\times\:\cancel{\dfrac{2x}{8x}}\:\times\:\cancel{\dfrac{2x}{8x}}}$

ㅤㅤㅤ$\longmapsto$ $\bold{\dfrac{1}{4}\:\times\:\dfrac{1}{4}\:\times\:\dfrac{1}{4}}$

ㅤㅤㅤ$\longmapsto$ $\bold{\dfrac{1\:\times\:1\:\times\:1}{4\:\times\:4\:\times\:4}}$

ㅤㅤㅤ$\longmapsto$ $\boxed{\bold{\dfrac{1}{64}}}$

$\boxed {\frak {\therefore \purple {Ratio\:of\:their\:surface\:areas\:\leadsto\:1\:\ratio\:64}}}\:\orange\bigstar$

$\large\underline{\boxed{\bold\blue{SOLVED}}}$

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${ \bold { \underline{\large{Formulas\: that \:we\:used\:above : - }}}} \:$

$\bold{Surface\:area\:of\:sphere\:=\:4\pi r^2}$

$\bold{Volume\:of\:sphere\:=\:\dfrac{4}{3}\pi r^3}$

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${ \bold { \underline{\large\red{Note : - }}}} \:$

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Answer 2

Refer to the attachment