Find the mass of a metallic hollow cylindrical pipe 24 cm long with internal diameter 10 cm and made of 5 mm thick metal, if 1 cm³ of the metal weighs 7.5 grams.

Internal radius of the pipe, r = $\dfrac{\text{diameter}}{2} = \dfrac{10}{2} = 5$ cm

Given,

Thickness = 5 mm = 0.5 cm

Thickness = External radius (R) - Internal radius (r)

⇒ R = r + thickness

⇒ R = 5.5 cm

Length of the pipe (h) = 24 cm

External volume = πR²h

$= \dfrac{22}{7} \times (5.5)^2 \times 24 \\[1em] = \dfrac{22}{7} \times 30.25 \times 24 \\[1em] = \dfrac{15972}{7}$

Internal volume = πr²h

$= \dfrac{22}{7} \times (5)^2 \times 24 \\[1em] = \dfrac{22}{7} \times 25 \times 24 \\[1em] = \dfrac{13200}{7}$

Volume of metal = External volume - Internal volume

$= \dfrac{15972}{7} - \dfrac{13200}{7} \\[1em] = \dfrac{15972 - 13200}{7} \\[1em] = \dfrac{2772}{7} \\[1em] = 396 \text{ cm}^3.$

Given, 1 cm³ of the metal weighs 7.5 grams.

(1g = $\dfrac{1}{1000} \text{kg}$)

Total weight of the pipe = $396 \times 7.5 \times \dfrac{1}{1000} = \dfrac{2970}{1000} = 2.97$ kg

Hence, the mass of a metallic hollow cylindrical pipe is 2.97 kg.