Question

If the radius of the base of a right circular cylinder is halved keeping the height same,
find the ratio of the volume of the reduced cylinder of that of the reduced cylinder to that of the original cylinder..​

Answer 1

Answer :

• Let the radius of cylinder be r and height of cylinder be h.

Volume of cylinder is calculated by the given below formula :

• Original Volume of cylinder = πr²h

As in the question it is stated that If the radius of the base of a right circular cylinder is halved keeping the height same, mathematically it can be expressed as :

• New radius = ½ × r = r/2
• Height = h

Hence,the reduced volume of cylinder will become as :

• Reduced volume of cylinder = π × (r/2)² × h

Now, let's find the ratio of the volume of the reduced cylinder of that of the original cylinder :

⇒Required ratio = (Reduced Volume of cylinder) ÷ (Original volume of cylinder)

⇒Required ratio = (π[r/2]²h) ÷ (πr²h)

⇒Required ratio = (r/2)² ÷ r²

⇒Required ratio = (r²/4) ÷ r²

⇒Required ratio = r² ÷ (r² × 4)

⇒Required ratio = r² ÷ 4r²

⇒Required ratio = 1 ÷ 4

• Hence,the required ratio of the volume of the reduced cylinder of that of the original cylinder is 1:4.

Answer 2

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

If the radius of the base of right circular cylinder is halved keeping the height same, find the ratio of the volume of the reduced cylinder to that of the original cylinder.

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$\huge{AηsωeR} ✍$

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☆Case :- 1

$\underline{\boxed{\star{\bf{\blue{ Dimension \: of \: original \: cylinder }}}}}$

$\begin{gathered}\begin{gathered}\bf Let = \begin{cases} &\sf{radius \: of \: cylinder \: be \: r \: units} \\ &\sf{height \: of \: cylinder \: be \: h \: units} \\ &\sf{volume \: of \: cylinder \: be \: V_1} \end{cases}\end{gathered}\end{gathered}$

$\bf\implies \:V_1 = \pi \: {r}^{2} h \: \sf \:  ⟼ \: (1)$

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☆Case :- 2

$\underline{\boxed{\star{\bf{\blue{ Dimension \: of \: reduced \: cylinder }}}}}$

$\begin{gathered}\begin{gathered}\bf Let = \begin{cases} &\sf{radius \: of \: cylinder \: be \: \dfrac{r}{2} \: units} \\ &\sf{height \: of \: cylinder \: be \: h \: units} \\ &\sf{volume \: of \: cylinder \: be \: V_2} \end{cases}\end{gathered}\end{gathered}$

$\bf\implies \:V_2 = \pi \: {(\dfrac{r}{2}) }^{2} h$

$\bf\implies \:V_2 = = \pi \: \dfrac{ {r}^{2} }{4} h$

$\bf\implies \:V_2 = \dfrac{1}{4} \pi \: {r}^{2} h$

$\bf\implies \:V_2 = \dfrac{1}{4} V_1$

$\bf\implies \:\dfrac{V_2}{V_1} = \dfrac{1}{4}$

$\bf\implies \:V_2 : V_1 = 1 : 4$

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☆ So, the ratio of the volume of the reduced cylinder to that of the original cylinder is 1 : 4.