Question

Find the weight of a hollow cylindrical lead pipe

28 cm

long and

\frac{1}{2} cm

thick. Its internal diameter is

8 cm

. Weight of

1 {cm}^{3}

of lead is

11.4 g (π = \frac{22}{7})

Accepted Answer

Given,

Height of hollow cylinder lead pipe $= 28 c m$

Inner diameter $= 8 c m$

Inner radius of hollow cylinder lead pipe $= \frac{d}{2} = \frac{8}{2} = 4 c m$

Thichness of hollow cylinder $= \frac{1}{2} c m = 0.5 c m$

Exterior radius of hollow cylinder lead pipe $=$ inner radius $+$ thickness $= 4 + 0.5 = 4.5 c m$

Volume of hollow cylinder lead pipe $=$ Volume of external cylinder $-$ Volume of inner cylinder.

$= {πR}^{2} h - {πr}^{2} h$

$= πh (R^{2} - r^{2})$

$= \frac{22}{7} \times 28 [{(4.5)}^{2} - {(4)}^{2}]$

$= 22 \times 4 (20.25 - 16)$

$= 22 \times 4 \times 4.25$

$= 374 {cm}^{3}$

Weight of $1 c m^{3}$ of lead $= 11.4 g$

Weight of $374 c m^{3} = 374 \times 11.4 = 4263.6 g$

We know, $1 k g = 1000 g \Rightarrow 1 g = \frac{1}{1000} k g$

$4263.6 g = \frac{4263.6}{1000} = 4.2636 k g$

Hence, weight of hollow cylinder is $4.2636 k g$

Find the weight of a hollow cylindrical lead pipe $28 cm$ long and $\frac{1}{2} cm$ thick. Its internal diameter is $8 cm$ . Weight of $1 {cm}^{3}$ of lead is $11.4 g (π = \frac{22}{7})$

Important Questions on Unit Test - 4 (Chapters 16 - 25)

Learn and Prepare for any exam you want !

Find the weight of a hollow cylindrical lead pipe 28 cm long and 12 cm thick. Its internal diameter is 8 cm. Weight of 1 cm3 of lead is 11.4 g π=227

Important Questions on Unit Test - 4 (Chapters 16 - 25)

Learn and Prepare for any exam you want !

Find the weight of a hollow cylindrical lead pipe $28 cm$ long and $\frac{1}{2} cm$ thick. Its internal diameter is $8 cm$ . Weight of $1 {cm}^{3}$ of lead is $11.4 g (π = \frac{22}{7})$