Question

The difference between outside and inside surfaces of a cylindrical metal pipe is 14 cm long is 44 cm². If the pipe is made of 99 cm³ of metal, find the outer and inner radius of the pipe.​

Answer 1

Step-by-step explanation:

$\large \blue{ \underline{✇ \sf \: Question:-}}$

• The difference between outside and inside surfaces of a cylindrical metal pipe is 14 cm is 44 cm²

• Pipe is made of 99 cm³ of metal.

$\large \blue{ \underline{✇ \sf \: Solution:-}}$

$\red{ \sf \Large ❏\: Let: }$

• R be the external radius
• r be the internal radius

We have h = Length of the pipe = 14 cm.

It is given that

$\bf \: Outside \: surface \: area \: - Internal \: surface \: area = 44 \: {cm}^{2} \\ \\ ☞ \sf \: 2 \pi Rh - 2\pi rh = 44 \\☞ \sf \: 2\pi(R - r)h = 44 \\☞ \sf \: 2 \times \frac{22}{7} (R - r) = \frac{44}{14} \\ ☞ \: \sf \: R - r = \cancel \frac{44 \times 7}{14 \times 22 \times 2} \\ \\ ☞ \: \bf R - r = \frac{1}{2} \: \: \: \: \: \bigg\lgroup \: eq. \: 1 \bigg \rgroup$

It is given that the volume of the metal used = 99 cm³

$\bf \: External \: volume - Internal \: volume = 99 \: {cm}^{3} \\ \\ \sf \: ✏ \: \pi {R}^{2}h - \pi {r}^{2} h = 99 \\✏ \sf\pi( {R}^{2} - {r}^{2} )h = 99 \\ ✏ \sf \: \frac{22}{7} (R + r)(R - r) \times 14 = 99 \\ ✏\sf \: \frac{22}{7} \times (R + r) \times \frac{1}{2} \times 14 = 99 \: \: \: \{ using \: (i) \} \\ ✏\sf \: R + r = \cancel \frac{99 \times 2 \times 7}{22 \times 14} \\ \\ ✏ \bf \: R + r = \frac{9}{2} \: \: \: \bigg \lgroup \: eq. \: 2 \bigg \rgroup$

Adding eq. 1 and eq. 2

$\sf R - r = \dfrac{1}{2}$

$\sf R + r = \dfrac {9}{10}$

_____________

$\sf 2R = 5$

$\Large \pink{\frak{R = 2.5 cm}}$

Putting the value of R in eq. 2

$\sf R + r = \frac{9}{2} \\ \Large \purple{\frak{r = 2 cm}}$

$\bf \therefore ❒ Outer \:radius \:and \:Inner\: radius\: are\: 2.5 \:cm \:and\: 2\: cm\: respectively.$

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