Question

19. The sum of the inner and the outer curved
surfaces of a hollow metallic cylinder is
1056 cm2 and the volume of material in it is 1056
cm3. Find its internal and external radii. Given
that the height of the cylinder is 21 cm.​

Answer 1

AnswEr :

• External radius = R = 5 cm

• Internal radius = r = 3 cm

GivEn :

• Sum of inner and outer CSA = 1056 cm²

• Volume of material = 1056 cm³

• Height (h) = 21 cm

SoluTion :

Let R be the outer radius and r be the inner radius respectively!

According to the given information,

$:\implies \sf \: \: \: Outer \: CSA + Inner \: CSA = 1056$

$:\implies \sf \: \: \: 2\pi Rh + 2\pi rh = 1056$

$\ \ \ \ \ \ \dag \ { \underline{ \sf{On \ taking \ common \ values : - }}}$

$:\implies \sf \: \: \: 2\pi h(R + r)= 1056$

$:\implies \sf \: \: \: 2 \times \dfrac{22}{7} \times 21(R + r)= 1056$

$:\implies \sf \: \: \: (R + r)= \dfrac{1056 \times 7}{21 \times 22 \times 2}$

$:\implies \sf \: \: \: (R + r)= \dfrac{7392}{924}$

$:\implies \sf \: \: \: R + r= 8 \: \: \: \: \: \: \: \: - eqn1$

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As volume is also given,

$:\implies \sf \: Volume \: of \: material \: inside = 1056 \: {cm}^{3}$

$:\implies \sf \: \pi {R}^{2}h - \pi {r}^{2} h = 1056 \: {cm}^{3} \: \: \: \: \: (as \: 2 \: materials \: are \: present)$

$\ \ \ \ \ \ \dag \ { \underline{ \sf{On \ taking \ common \ values : - }}}$

$:\implies \sf \: \pi h( {R}^{2} - {r}^{2} )= 1056 \: {cm}^{3}$

$:\implies \sf \: \frac{22}{7} \times 21 \times ( {R}^{2} - {r}^{2} )= 1056 \: {cm}^{3}$

$:\implies \sf \: ( {R}^{2} - {r}^{2} )= \dfrac{1056 \times 7}{21\times 22}$

$:\implies \sf \: ( {R}^{2} - {r}^{2} )= \dfrac{7392}{462}$

$:\implies \sf \: ( {R}^{2} - {r}^{2} )= 16$

$:\implies \sf \: (R + r)(R - r)= 16 \: \: \: \: \: \: \: { \bf{\bigg({a}^{2} - {b}^{2} =(a+b)(a-b) \bigg)}}$

$:\implies \sf \: R - r = 2 \: \: \: \: \: - eqn2$

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Add eqn 1 and 2 ,

$: \implies \sf \: \: \: R + r + R - r = 8 + 2$

$: \implies \sf \: \: \: 2R = 10$

$: \implies \sf \: \: \: R = \dfrac{10}{2}$

$: \implies \sf \: \: \:{ \underline{ \boxed{ \sf{ \purple{ \: R = 5 \: }}}}} \: \bigstar$

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Substitute the value of R in eqn 2,

$:\implies \sf \: R - r = 2$

$:\implies \sf \: 5 - r = 2$

$:\implies \sf \: r = 5 - 2$

$:\implies \sf \: { \underline{ \boxed{ \sf{ \orange{ \: r = 3 \: }}}}} \: \bigstar$

Therefore,

• External radius = R = 5 cm

• Internal radius = r = 3 cm

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