Question

A hollow iron pipe is 21 cm long and its exterior diameter is 8 cm. If the thickness of the pipe is 1 cm and iron weighs 8 g/cm³, then the weight of the pipe is $[take \pi=\frac{22}{7}]$

• A. 3.696 kg
• B. 3.6 kg
• C. 36 kg
• D. 36.9 kg

Answer 1

Answer:

A. 3.696 kg

The pipe can be assumed as hollow cylinder.

External radius $=\frac{8}{2}=4cm$

Thickness = 1 cm

Internal radius = 4 – 1 = 3 cm.

Volume of the material $=\pi{h}{(R^2-r^2)}$

$=\frac{22}{7}\times{21}\times{(4^2-3^2)}$

$=\frac{22}{7}\times{21}\times{7}$

$=462cm^3$

Now, 1cm3 iron weighs = 8 gm

462 cm3 iron weighs = 462 × 8 gm

$=\frac{462\times{8}}{1000}$

$=3.696kg$