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A solid iron pole consists of a solid cylinder of height $200$ cm and base diameter $28$ cm, which is surmounted by another cylinder of height $50$ cm and radius $7$ cm. Find the mass of the pole, given that $1$ cm$^3$ of iron has approximately $8$ g mass.
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Sol.
Radius of lower cylinder = $14$ cm
Volume of pole = $\frac{22}{7} \times 14 \times 14 \times 200 + \frac{22}{7} \times 7 \times 7 \times 50$
$= 130900$ cm$^3$
Mass of the pole= $8 \times 130900$
$=1047200$gm or $1047.2$ kg
OR
Radius of lower cylinder = $14$ cm
Volume of pole = $\frac{22}{7} \times 14 \times 14 \times 200 + \frac{22}{7} \times 7 \times 7 \times 50$
$= 130900$ cm$^3$
Mass of the pole= $8 \times 130900$
$=1047200$gm or $1047.2$ kg
OR