Question

A hollow cylindrical pipe of length 4 cm and thickness 7cm is made from the material of a solid cylinder of
diameter 8 cm and length 21 cm with 88 cm
3
of material extra . Find the internal diameter of the pipe

Answer 1

Given:

The length of the hollow cylindrical pipe = 4 cm

Thickness of the hollow pipe = 7 cm

Diameter of the solid cylinder = 8 cm

length of the solid cylinder = 21 cm

Extra material of the solid cylinder = 88 cm³

To find:

The internal diameter of the pipe

Solution:

For the hollow cylinder:

If we consider "r" as the internal radius then "R = (r + 7)" will be the external radius  

∴ Volume of the hollow cylinder will be,

= $\pi\:h\:[R^2 - r^2]$

= $\pi\:h\:[(r + 7 )^2- r^2]$

= $\frac{22}{7} \times4\times[r^2+14r+49 -r^2]$

= $\frac{22}{7} \times4\times[14r+49]$ ........ (i)

For solid cylinder:

Radius of the solid cylinder = $\frac{diameter}{2} = \frac{8}{2} = 4\:cm$

∴ Volume of the solid cylinder along with the volume of the extra material will be,

= $\pi\:h\:r^2 + 88$

= $[\frac{22}{7}\times\:21\times\: 4^2] + 88$

= $1056 + 88$

= $1144\:cm^3$ ........ (ii)

Since the hollow cylinder is made from the solid cylinder along with the extra material, so their volume will be the same.

So we can say that, eq. (i) will be equal to eq. (ii),

$\frac{22}{7} \times4\times[14r+49]$  = $1144$

⇒  $[14r+49] = \frac{1144 \times7}{22\times4}$

⇒ $[14r+49] = 91$

⇒ $14r = 91 - 49$

⇒ $14r = 42$

⇒ $r = \frac{42}{14}$

⇒ $r = 3\:cm$ ← internal radius of the pipe

∴ Internal diameter of the pipe = 2 × r = 2 × 3 = 6 cm

Thus, the internal diameter of the pipe is 6 cm.

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