Question

If the radius of a sphere is reduced to one-third , by how many times does it"s volume.​

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

Given:- The radius of a sphere is reduced to one-third.

To Find:- by how many times does it"s volume decrease .

Solution:- We know the Volume of sphere as ,

$\boxed{\pink{\sf Volume_{sphere}=\dfrac{4}{3}\pi r^3 }}$

Let Initial Volume be V and final be V¹ .Let the initial radius be r , so new radius will be r/3 .So ,

Volume of larger sphere :-

=> V = 4/3 π r³ .

Volume of smaller Sphere :-

=> V¹ = 4/3 π × (r/3)³.

=> V¹ = 4/3 π × r³/27 .

On dividing them ,

=> V / V¹ = 27/1

=> V¹ = V/27 .

Hence the volume becomes 1/27 times the initial Volume.