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

Answer:

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} . }=2πRh=(2×722×R×14)cm2=(88R)cm2.

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} }=2πrh=(2×722×r×14)cm3=(88r)cm2

\tt \therefore(88 R-88 r)=44∴(88R−88r)=44

\tt \: \: \: \Rightarrow(R-r)=\dfrac{44}{88}⇒(R−r)=8844

\begin{gathered} \\ \tt \Rightarrow(R-r)=\frac{1}{2} \qquad \ldots \ldots\ldots (i)\end{gathered}⇒(R−r)=21………(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} .}=πR2h=(722×R2×14)cm3=(44R2)cm3.

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} . }=πr2h=(722×r2×14)cm3=(44r2)cm3.

\tt \therefore\left(44 R ^{2}-44 r^{2}\right)=99∴(44R2−44r2)=99

\tt\Rightarrow\left( R ^{2}-r^{2}\right)=\dfrac{99}{44}⇒(R2−r2)=4499

\tt\Rightarrow\left( R ^{2}-r^{2}\right)=\dfrac{9}{4} \: \: \: \: \: \: \: \: \: \: ...(ii)⇒(R2−r2)=49...(ii)

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

\tt ( R +r)=\left(\dfrac{9}{4} \times \dfrac{2}{1}\right)(R+r)=(49×12)

\tt\Rightarrow( R +r)=\dfrac{9}{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: ...(iii)⇒(R+r)=29...(iii)

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