Question

Height of a hollow right circular cylinder, open at both ends, is 2.8 metre. If length of innerdiametre of the cylinder is 4.6 dem and the cylinder is made up of 84.48 cubic dcm of iron,let us calculate the length of outer diameter of the cylinder.​

Answer 1

$\sf\blue{Correct \ question}$

Height of a hollow right circular cylinder, open at both ends, is 2.8 metre. If length of inner diametre of the cylinder is 4.6 cm and the cylinder is made up of 84.48 cubic cm of iron,let us calculate the length of outer diameter of the cylinder.

$\huge\red{\underline{\underline{\pink{Ans}\red{wer:-}}}}$

$\sf{Length \ of \ outer \ diameter \ is \ 4.8 \ cm(approx)}$

$\huge\sf\orange{Given}$

$\sf{For \ a \ hollow \ circular \ cylinder,}$

➡$\sf{Height(h)=2.8 \ m}$

➡$\sf{Inner \ diameter=4.6 \ cm}$

➡$\sf{Volume=84.48 \ cu \ cm}$

$\sf\pink{To \ find:}$

$\sf{Let \ of \ outer \ diameter \ of \ the \ cylinder.}$

$\huge\sf\green{\underline{\underline{Solution:}}}$

$\sf{Let \ outer \ radii \ be \ R \ and \ inner \ radii}$

$\sf{be \ r.}$

$\sf{Inner \ radii(r)=\frac{Inner \ diameter}{2}}$

$\sf{\therefore{r=\frac{4.6}{2}}}$

$\sf{r=2.3 \ cm}$

$\sf{=>Height(h)=2.8 \ m}$

$\sf{1 \ m=100 \ cm}$

$\sf{\therefore{Height(h)=280 \ cm}}$

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$\sf{Volume \ of \ cylinder=\pi×r^{2}×h}$

$\sf{...formula}$

$\sf{\therefore{Volume \ of \ hollow \ cylinder}}$

$\sf{=Volume \ of \ whole \ cylinder \ - }$

$\sf{volume \ of \ hollow \ part.}$

$\sf{\therefore{84.42=(\frac{22}{7}×R^{2}×280)-(\frac{22}{7}×2.3^{2}×280)}}$

$\sf{84.42=\frac{22}{7}×280×(R^{2}-2.3^{2})}$

$\sf{(R^{2}-2.3^{2})=\frac{84.48×7}{22×280}}$

$\sf{R^{2} -2.3^{2}= 0.096}$

$\sf{(R+2.3)(R-2.3)=0.096}$

$\sf{\therefore{R=0.096-2.3 \ or \ 0.096+2.3}}$

$\sf{\therefore{R=-2.204 \ or \ 2.396}}$

$\sf{But \ R \ can't \ be \ negative.}$

$\sf{\therefore{R=2.396 \ cm}}$

$\sf{R=2.396 \ cm(approx) }$

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$\sf{Outer \ diameter=2×Outer \ radii}$

$\sf{\therefore{Outer \ diameter=2×2.396}}$

$\sf{\therefore{Outer \ diametre=4.792\ cm}}$

$\sf{\therefore{Outer \ diameter=4.8 \ cm(approx)}}$

$\bold\purple{\tt{\therefore{Length \ of \ outer \ diameter \ is \ 4.8 \ cm(approx)}}}$

Answer 2

Question -

The height of a hollow right circular cylinder, open at both ends, is 2.8 metres.

If the length of inner diameter of the cylinder is 4.6 centimetres and the cylinder is made up of 84.48 cubic centimetres of iron, calculate the length of outer diameter of the cylinder.​

Solution -

IN the above question, we have the following information given -

This is a hollow right cylinder, open at two ends .

Height = 2.8 m = 280 cm

$Diameter_{1} = 4.6 \ cm.$

Let us assume that the length of the outer diameter of the cylinder is 2k cm.

We know that the volume of a right angled cylinder is $\pi r^2 h$

SO, Volume Of Outer Cylinder =   $\pi {r_{1}} ^2 h$

SO, Volume Of Inner Cylinder =   $\pi {r_{2}} ^2 h$

Volume Of The Hollow Cylinder =

$\pi {r_{1}} ^2 h - \pi {r_{2}} ^2 h \\ \\ = \pi h [ { r_{1 }} + { r_{2 }} ]$

Here,

The length of inner diameter of the cylinder is 4.6 centimetres .

SO, the length of the inner radius is 2.3 centimetres .

The length of the outer diameter of the cylinder is 2k cm.

SO, the length of the outer radius is k centimetres .

The height of the requires hollow right circular cylinder, open at both ends, is 2.8 metres.

Substituting these values into the required formula -

Volume OF the hollow Cylinder =

$\pi \times 2.8 \times [ k + 2.3 ][ k - 2.3 ]$

But, the volume of the cylinder is 88.48 cubic centimetres .

So,

$\pi \times 2.8 \times [  k + 2.3 ][ k - 2.3 ] = 88.48$

Hence,

$\pi \times 2.8 \times [ k + 2.3 ][ k - 2.3 ] = 88.48 \\ \\ \dfrac{22}{\cancel{7}} \times \dfrac{\cancel{28} \ . \ 4 }{10} \times [ k + 2.3 ][ k - 2.3 ] = 88.48$

$[ ( k - 2.3 ) ] [ ( k + 2.3 ) ] \approx 2.5 \\ \\ k^{2} = 2.5 + 5.29 = 7.79 \\ \\ k \approx 2.8 \ cm.$

Outer Diameter = 2k = 5.6 cm.

Hence, the required outer diameter of the cylinder is 5.6 cm.

Answer -

The required outer diameter of the cylinder is 5.6 cm.

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