Question

The difference between the outside and inside surfaces of a cylindrical metal pipe $14cm$ long is $44c{m}^2$. If the pipe is made of $99c{m}^3$ of metal, find the sum of the inner and outer radii of the pipe.

• A. $4.5cm$
• B. $3.5cm$
• C. $2cm$
• D. $5.5cm$

Answer 1

The correct option is A $4.5cm$

Let the $r$ and $R$ be the inner and outer radius of the cylindrical metallic pipe, respectively.

Height of the metallic pipe $h=14cm$

Curved surface area of the outer cylinder $-$ Curved surface area of the inner cylinder $=44c{m}^2$

$\therefore 2\pi Rh-2\pi rh=44$

$\Rightarrow$$2\pi (R-r)\times 14=44$

$\Rightarrow$$(R-r)=\frac{44\times 7}{44\times 14}$

$\Rightarrow$$(R-r)=\frac{1}{2}$ ...(i)

$\Rightarrow$Volume of the metal in the pipe $=99c{m}^3$

$\Rightarrow$$\pi R^{2}h-\pi r^{2}h=99$

$\Rightarrow$$\pi (R^2-r^2)\times 14=99$

$\Rightarrow$$\frac{22}{7}\times (R+r)(R-r)\times 14=99$

$\Rightarrow$$\frac{22}{7}\times (R+r)\times \frac{1}{2}\times 14=99$ ....(from (i))

$\Rightarrow$$(R+r)=\frac{99}{22}$$=\frac{9}{2}$ ....(ii)

Adding (i) and (ii) we get,

$\Rightarrow$$R-r+R+r=\frac{1}{2}+\frac{9}{2}$

$\Rightarrow$$2R=\frac{10}{2}$

$\Rightarrow$$2R=5$

$\Rightarrow$$R=\frac{5}{2}$

From (ii) we have,

$\Rightarrow$$\frac{5}{2}+r=\frac{9}{2}$

$\Rightarrow$$r=\frac{9}{2}-\frac{5}{2}$

$\Rightarrow$$r=\frac{4}{2}$

$\Rightarrow$$r=2$

$\therefore$ The inner and outer radii of the pipe is $2cm$ and $\frac{5}{2}cm$.