Question

82. The difference between outside and inside surface of a cylindrical metallic pipe 14 cm long is 44 sq cm. If the pipe is made of 99 cu cm of metal, find the outer and inner radii of the pipe. ​

Answer 1

EXPLANATION.

Difference between outside and inside surface of a cylindrical pipe 14 cm long is 44 cm².

If pipe is made of 99 cm³ of metal.

As we know that,

Formula of :

Curved surface area of cylinder = 2πrh.

Volume of cylinder = πr²h.

Using this formula in the equation, we get.

Outside surface of a cylinder = 2πRh.

Inside surface of a cylinder = 2πrh.

Height = 14 cm.

⇒ 2πRh - 2πrh = 44.

⇒ 2πh(R - r) = 44.

⇒ 2 x 22/7 x 14 (R - r) = 44.

⇒ 2 x 22 x 2 (R - r) = 44.

⇒ 44 x 2(R - r) = 44.

⇒ 2(R - r) = 1.

⇒ (R - r) = 1/2. - - - - - (1).

Pipe is made of 99 cm³.

⇒ πR²h - πr²h = 99.

⇒ πh(R² - r²) = 99.

⇒ 22/7 x 14(R² - r²) = 99.

⇒ 22 x 2 x (R - r)(R + r) = 99.

⇒ 44(R + r)(R - r) = 99.

Put the value of (R - r = 1/2) in the equation, we get.

⇒ 44(R + r)(1/2) = 99.

⇒ (R + r) x 22 = 99.

⇒ (R + r) = 99/22.

⇒ (R + r) = 9/2. - - - - - (2).

From equation (1) and (2), we get.

Adding equation (1) and (2), we get.

⇒ (R - r) = 1/2. - - - - - (1).

⇒ (R + r) = 9/2. - - - - - (2).

We get,

⇒ 2R = 1/2 + 9/2.

⇒ 2R = 10/2.

⇒ 2R = 5.

⇒ R = 2.5 cm

Put the value of R = 2.5 in equation (2), we get.

⇒ R + r = 4.5.

⇒ 2.5 + r = 4.5.

⇒ r = 4.5 - 2.5.

⇒ r = 2 cm.

Outer radii of pipe = R = 2.5 cm.

Inner radii of pipe = r = 2 cm.

Answer 2

Step-by-step explanation:

Let, external radius = R cm and internal radius =r cm.

Then, outside surface

$\red{ \tt=2 \pi R h=\left(2 \times \cfrac{22}{7} \times R \times 14\right) cm ^{2}=(88 \: \: R ) cm ^{2} . }$

Inside surface

$\blue{\tt =2 \pi r h=\left(2 \times \dfrac{22}{7} \times r \times 14\right) cm ^{3}=(88 \: \: r) cm ^{2} }$

$\tt \therefore(88 R-88 r)=44$

$\tt \: \: \: \Rightarrow(R-r)=\dfrac{44}{88}$

$\\ \tt \Rightarrow(R-r)=\frac{1}{2} \qquad \ldots \ldots\ldots (i)$

External volume

$\pink{ \tt=\pi R ^{2} h=\left(\dfrac{22}{7} \times R ^{2} \times 14\right) cm ^{3}=\left(44 \: \: R ^{2}\right) cm ^{3} .}$

Internal volume

$\orange{\tt =\pi r^{2} h=\left(\dfrac{22}{7} \times r^{2} \times 14\right) cm ^{3}=\left(44 \: \: r^{2}\right) cm ^{3} . }$

$\tt \therefore\left(44 R ^{2}-44 r^{2}\right)=99$

$\tt\Rightarrow\left( R ^{2}-r^{2}\right)=\dfrac{99}{44}$

$\tt\Rightarrow\left( R ^{2}-r^{2}\right)=\dfrac{9}{4} \: \: \: \: \: \: \: \: \: \: ...(ii)$

On dividing (ii) by (i), we get:

$\tt ( R +r)=\left(\dfrac{9}{4} \times \dfrac{2}{1}\right)$

$\tt\Rightarrow( R +r)=\dfrac{9}{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: ...(iii)$

Solving (i) and (iii) , we get, R =2.5 and r=2 .

Hence, outer radius =2.5 cm and inner radius =2 cm.