Question

help me guys...... ​

Answer 1

$\large\underline{\textsf{\textbf{\purple{ \mapsto GiveN:-}}}}$

The radius of a sphere is reduced to one third .

$\large\underline{\textsf{\textbf{\purple{ \mapsto To\:FinD:-}}}}$

How many times its volume will decrease.

$\large\underline{\textsf{\textbf{\purple{ \mapsto AnsweR:-}}}}$.

Given that the radius of sphere is reduced by one third . So ,let the initial radius of sphere be r , then the new radius of sphere will be r/3 .

Let the initial Volume be V , and the final Volume be V' . Now ,we know the Volume of sphere as ;

$\large\qquad\boxed{\red{\bf \blue{\dashrightarrow}\: Volume_{sphere}\:\:=\dfrac{4}{3}\pi r^3 }}$

$\underline{\blue{\sf Case\:1:- Original\: Volume\: of\:sphere:-}}$

$\tt:\implies V = \dfrac{4}{3}\pi (radius)^3$

$\underline{\boxed{\red{\tt\longmapsto V\:\:=\:\:\dfrac{4}{3}\pi r^3}}}$

$\rule{200}2$

$\underline{\blue{\sf Case\:2:- New\: Volume\: of\:sphere:-}}$

$\tt:\implies V' = \dfrac{4}{3}\pi (radius)^3$

$\tt:\implies V' = \dfrac{4}{3}\pi \bigg(\dfrac{1r}{3}\bigg)^3$

$\tt:\implies V' = \dfrac{4}{3}\pi \dfrac{1}{27} r^3$

$\underline{\boxed{\red{\tt\longmapsto V'\:\:=\:\:\dfrac{4}{81}\pi r^3}}}$

$\rule{200}2$

$\underline{\pink{\sf Required\: Ratios\: of\:sphere:-}}$

$\tt:\implies V : V' = \dfrac{4}{3}\pi r^3:\dfrac{4}{81}\pi r^3$

$\tt:\implies V:V' = \dfrac{\cancel{\dfrac{4}{3}\pi r^3}}{\cancel{\dfrac{4}{81}\pi r^3}}$

$\tt:\implies V : V' = 27 : 1$

$\tt:\implies \dfrac{V}{V'}=\dfrac{27}{1}$

$\underline{\boxed{\purple{\bf\orange{\leadsto} V' = \dfrac{V}{27}}}}$

$\large\underline{\underline{\green{\sf \orange{\dag}\:Hence\;the\: Volume\:becomes\:\frac{1}{27}\:of\: Initial\: Volume.}}}$

$\rule{200}4$

Answer 2

$\large\underline{\textsf{\textbf{\purple{ \mapsto GiveN:-}}}}$

The radius of a sphere is reduced to one third .

$\large\underline{\textsf{\textbf{\purple{ \mapsto To\:FinD:-}}}}$

How many times its volume will decrease.

$\large\underline{\textsf{\textbf{\purple{ \mapsto AnsweR:-}}}}$.

Given that the radius of sphere is reduced by one third . So ,let the initial radius of sphere be r , then the new radius of sphere will be r/3 .

Let the initial Volume be V , and the final Volume be V' . Now ,we know the Volume of sphere as ;

$\large\qquad\boxed{\red{\bf \blue{\dashrightarrow}\: Volume_{sphere}\:\:=\dfrac{4}{3}\pi r^3 }}$

$\underline{\blue{\sf Case\:1:- Original\: Volume\: of\:sphere:-}}$

$\tt:\implies V = \dfrac{4}{3}\pi (radius)^3$

$\underline{\boxed{\red{\tt\longmapsto V\:\:=\:\:\dfrac{4}{3}\pi r^3}}}$

$\rule{200}2$

$\underline{\blue{\sf Case\:2:- New\: Volume\: of\:sphere:-}}$

$\tt:\implies V' = \dfrac{4}{3}\pi (radius)^3$

$\tt:\implies V' = \dfrac{4}{3}\pi \bigg(\dfrac{1r}{3}\bigg)^3$

$\tt:\implies V' = \dfrac{4}{3}\pi \dfrac{1}{27} r^3$

$\underline{\boxed{\red{\tt\longmapsto V'\:\:=\:\:\dfrac{4}{81}\pi r^3}}}$

$\rule{200}2$

$\underline{\pink{\sf Required\: Ratios\: of\:sphere:-}}$

$\tt:\implies V : V' = \dfrac{4}{3}\pi r^3:\dfrac{4}{81}\pi r^3$

$\tt:\implies V:V' = \dfrac{\cancel{\dfrac{4}{3}\pi r^3}}{\cancel{\dfrac{4}{81}\pi r^3}}$

$\tt:\implies V : V' = 27 : 1$

$\tt:\implies \dfrac{V}{V'}=\dfrac{27}{1}$

$\underline{\boxed{\purple{\bf\orange{\leadsto} V' = \dfrac{V}{27}}}}$

$\large\underline{\underline{\green{\sf \orange{\dag}\:Hence\;the\: Volume\:becomes\:\frac{1}{27}\:of\: Initial\: Volume.}}}$

$\rule{200}4$