Question

The difference between outer and inner clowed surface areas ofa hollow right circular cylinder, 14 cm long is 88cm2. if the volume of the metal used in making the cylinder is 176cm3. find the outer and inner diameters of the cylinder.

Answer 1

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Answer 2

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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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