Question

find the weight of the hollow cylinder lead pipe 28 cm long and 0.5 CM thick its inner diameter is 8 cm weight of 1 cm cube of lead is 11.4g

Answer 1

$internal \: \: volume = 2\pi \: rh \\ \\ 2 \times \frac{22}{7} \times 28 \times 8 \\ 44 \times 32 = 1408 {cm}^{3}$
$external \: volume = 2 \times \frac{22}{7} \times 28 \times \frac{85}{10} \\ \\ \frac{104720}{70} = 1496 {cm}^{3}$

Hence, Volume of lead used = (1496 - 1408) cm³ = 88 cm³




Therefore,
$weight \: of \: lead \: used = (88 \times 11.4) \: grams \\ = 1003.2 \: grams \: \: \: \: \: or \: \: \: \: \: \: 1 \: kg \: 32 \: g$