Question

If the difference between the outer and inner surface areas of a hollow cylinder of height 14cm is 44cm square and 88 cmcube metal is used to make the hollow cylinder then find the inner and outer radius of the pipe

Answer 1

$\sf{\boxed{\bold{\tiny{Heya \: mate.\: Solution\: below}}}}$
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$<b><u><font color ="red">Answer - ↓↓↓</font color></u></b>$




♠ Given Height (h) = 14 cm

Let the outer radius be R and inner radius be r.


Then, Outer Curved surface area (C.S.A) = 2πRh

Inner C.S.A = 2πrh

Outer Volume = πR²h

Inner Volume = πr²h



♠ Now, its given that difference between outer C.S.A and inner C.S.A is 44 cm².


$\sf{=⟩ 2 \pi R h - 2 \pi r h = 44}$

$\sf{=⟩ 2 \pi h (R - r) = 44}$

$\sf{=⟩ 2 \times \frac {22}{7} \times 14 (R - r) = 44}$

$\sf{=⟩ R - r = 44 \times \frac{1}{14} \times \frac{1}{2} \times \frac{7}{22}}$

$\sf{=⟩ R - r = \frac{1}{2}}$....eq i



♠ Also, Volume of metal used to make the cylinder is 88 cm³


$\sf{=⟩ \pi {R}^{2} h - \pi {r}^{2} h = 88}$

$\sf{=⟩ \pi h ({R}^{2} - {r}^{2}) = 88}$

$\sf{=⟩ \frac{22}{7} \times 14 \times ({R}^{2} - {r}^{2}) = 88}$

$\sf{=⟩ ({R}^{2} - {r}^{2}) = 88 \times \frac{1}{14} \times \frac{7}{22}}$

$\sf{=⟩ ({R}^{2} - {r}^{2}) = 2 }$... eq ii

$\sf{=⟩ (R + r)(R - r) = 2}$




♠ Then, from eq i and ii,

$\sf{(R + r)(R - r) = 2}$

$\sf{=⟩ (R + r) (\frac{1}{2}) = 2}$

$\sf{=⟩ R + r = 2 \times 2}$

$\sf{=⟩ R + r = 4}$... eq iii




♠ Adding eq i and iii,


$\sf{(R - r) + (R + r) = 4 + \frac{1}{2}}$

$\sf{=⟩ R - r + R + r = \frac{9}{2}}$

$\sf{=⟩ 2R = \frac{9}{2}}$

$\sf{=⟩ R = \frac{9}{2} \times \frac{1}{2}}$

$\sf{=⟩ R = \frac{9}{4}}$




♠ Now, substituting the value of R in eq iii,


$\sf{\frac{9}{4} + r = 4}$

$\sf{=⟩ r = 4 - \frac{9}{4}}$

$\sf{=⟩ r = \frac{7}{4}}$




♥♥♥ •°• The outer radius is 9/4 cm and inner radius is 7/4 cm. ♥♥♥
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Thank you... (^_-)