Question

The difference between the Outer and inner surface areas of a hollow cylinder, 14 cm Long is 88 sq. cm. Find out the outer and inner radii of the cylinder, given that volume of metal used is 176 cm3

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 88 cm².


$\sf{=⟩ 2 \pi R h - 2 \pi r h = 88}$

$\sf{=⟩ 2 \pi h (R - r) = 88}$

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

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

$\sf{=⟩ R - r = 1}$....eq i



♠ Also, Volume of metal used to make the cylinder is 176 cm³


$\sf{=⟩ \pi {R}^{2} h - \pi {r}^{2} h = 176}$

$\sf{=⟩ \pi h ({R}^{2} - {r}^{2}) = 176}$

$\sf{=⟩ \frac{22}{7} \times 14 \times ({R}^{2} - {r}^{2}) = 176}$

$\sf{=⟩ ({R}^{2} - {r}^{2}) = 176 \times \frac{1}{14} \times \frac{7}{22}}$

$\sf{=⟩ ({R}^{2} - {r}^{2}) = 4 }$... eq ii

$\sf{=⟩ (R + r)(R - r) = 4}$




♠ Then, from eq i and ii,

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

$\sf{=⟩ (R + r) (1) = 4}$

$\sf{=⟩ R + r = 4 \times 1}$

$\sf{=⟩ R + r = 4}$... eq iii




♠ Adding eq i and iii,


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

$\sf{=⟩ R - r + R + r = 5}$

$\sf{=⟩ 2R = 5}$

$\sf{=⟩ R = 5 \times \frac{1}{2}}$

$\sf{=⟩ R = \frac{5}{2}}$




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


$\sf{\frac{5}{2} + r = 4}$

$\sf{=⟩ r = 4 - \frac{5}{2}}$

$\sf{=⟩ r = \frac{3}{2}}$




♥♥♥ •°• The outer radius is 5/2 cm and inner radius is 3/2 cm. ♥♥♥
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Thank you... (^_-)