Question

A hollow cylindrical iron pipe is of length $28cm$. Its outer and inner diameters are $8cm$ and $6cm$ respectively. Find the volume of the pipe and weight of the pipe if $1cu.cm$ of iron weighs $7gm$. (Take $\pi =\frac{22}{7}$)

Answer 1

Let $r,R$ and $h$ be the inner, outer radii and height of the hollow cylindrical pipe respectively.
Given that $2r=6cm$, $2R=8cm$, $h=28cm$
Now, the volume of the pipe, $V=\pi \times h\times (R+r)(R-r)$
$\therefore$ Volume, $V=616cu.cm$
Weight of $1cu.cm$ of the metal $=7gm$
Weight of the $616cu.cm$ of metal $=7\times 616gm$
Thus, the weight of the pipe $=4.312kg$.