Question

The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if I cm³ of wood has a mass of 0.6 g.

Answer 1

Answer:

Mass of the Pipe=3432gms          $taking \pi=\frac{22}{7}$

Step-by-step explanation:

In this question we have been provided,

Outer diameter, D = 28cm,

Therefore radius, R =$\frac{diameter}{2}=\frac{28}{2}=14cm$

Inner diameter, d = 24cm

Therefore radius, r = $\frac{diameter}{2}=\frac{24}{2}=12cm$

Height of the pipe, h = 35cm

$\textrm{volume of cylinder=}\pi r^{2} h$

$\textrm{Volume of cylinder with r=14cm=}\pi\times 14^{2}\times h=\pi\times 14^{2}\times 35=6860\pi cm^{3}$

$\textrm{Volume of cylinder with r=12cm=}\pi\times 12^{2}\times h=\pi\times 12^{2}\times 35=5040\pi cm^{3}$

$\textrm{Net Volume=}6860\pi-5040\pi=1820\pi cm^{3}$

$\textrm{Mass of the pipe=}1820\pi\times 0.6=1092\pi=1092\pi\times \frac{22}{7}=156\times 22=3432grams.$

Hence Mass of Pipe we are getting is 3432grams.

Answer 2

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

${\boxed{\sf\:{Internal\;Radius}}}$

= r cm

Diameter = 24 cm

$\tt{\rightarrow radius(r)=\dfrac{24}{2}}$

r = 12 cm

Now,

${\boxed{\sf\:{External\;Radius}}}$

$\tt{\rightarrow Radius(R)=\dfrac{28}{2}}$

R = 14 cm

${\boxed{\sf\:{Length\;of\;Pipe}}}$

= 35 cm

$\textbf{\underline{Volume\;of\;wood\;that\;is\;used\;in\;Pipe}}$

= πR²h - πr²h

= π(R² - r²)h

$\tt{\rightarrow\dfrac{22}{7}(14^2 - 12^2)\times 35}$

= 22 × 5(14² - 12²)

= 110(14 - 12)(14 + 12)

= 110 × 2 × 26

= 220 × 26 cm³

Now

${\boxed{\sf\:{Mass\;of\;Wooden\;pipe}}}$

= Volume × Density of wood

= 220 × 26 × 0.6 gm

$\tt{\rightarrow 220\times 26\times\dfrac{6}{10}\times\dfrac{1}{1000}kg}$

$\tt{\rightarrow\dfrac{3432}{1000}}$

= 3.432 kg