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The difference between the outer and inner curved surface area of hollow right circular cylinder 14 cm long is $88 {cm}^{2}$ . If the volume of metal used in making the cylinder is $176 {cm}^{3}$ , find the outer and inner diameters of the cylinder. (Use $π = \frac{22}{7}$ )

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5 cm and 3 cm

10 cm and 5 cm

8 cm and 3 cm

7 cm and 2 cm

answer is A.

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Detailed Solution

Given,
The difference between the surface areas of the outer and inner curves

= 88 {cm}^{2}

So,
Curved surface area of outer - Curved surface area of inner

= 88 {cm}^{2}

Let r is the cylinder's radius and h are its height, the curved surface area of a cylinder,
CSA=

2 πrh

.
Here, the cylinder of height = 14 cm. Suppose the outer radius is R while inner radius is r. So,

2 πRh - 2 πrh = 88

2 πr (R - r) = 88

R - r = \frac{88}{2 \times \frac{22}{7} \times 14}

R - r = \frac{88 \times 7}{2 \times 22 \times 14}

R - r = 1 \dots \dots . (1)

So, the volume difference between the cylinder's outer and inner halves is reported.
Volume of outer part - Volume of inner part = 176
Thus, the cylinder's radius and h are its height, the volume of a cylinder is

π r^{2} h

.
Here,

π R^{2} h - π r^{2} h = 176

πh (R^{2} - r^{2}) = 176

π h (R - r) (R + r) = 176

π \times 14 (R + r) = 176

R + r = \frac{176 \times 7}{22 \times 14}

R + r = 4 \dots \dots . (2)

Eq. (i) + eq. (ii), we get

R - r + R + r = 5

2 R = 5

R = \frac{5}{2}

cm
So, the inner diameter,

D = \frac{5}{2} \times 2

D = 5

cm
Put the value R in eq. (2)

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

r = 4 - \frac{5}{2}

r = \frac{8 - 5}{2}

r = \frac{3}{2}

cm
So, the inner diameter,

d = 3

cm
Thus, the outer and inner diameters of the cylinder is 5 and 3.
Option 1 is correct.

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