Question:
Find the weight of solid rectangular iron piece of size 50 cm × 40 cm × 10cm, if 1 cm3 of iron weighs 8 gm.
Solution:
The dimension of the rectangular piece of iron is $50 \mathrm{~cm} \times 40 \mathrm{~cm} \times 10 \mathrm{~cm}$.
i. e., volume $=50 \mathrm{~cm} \times 40 \mathrm{~cm} \times 10 \mathrm{~cm}=20000 \mathrm{~cm}^{3}$
It is given that the weight of $1 \mathrm{~cm}^{3}$ of iron is $8 \mathrm{gm}$.
$\therefore$ The weight of the given piece of iron $=20000 \times 8 \mathrm{gm}$
$=160000 \mathrm{gm}$
$=160 \times 1000 \mathrm{gm}$
$=160 \mathrm{~kg}(\because 1 \mathrm{~kg}=1000 \mathrm{gm})$