Question

What will be the mass of a 21 cm long metallic hollow cylinder with internal diameter 8 cm and thickness of metal being 4 mm, if 1 cm3 of the metal weighs 8 gm?

Answer 1

Answer:

1774.1gm

Step-by-step explanation:

mass=density*volume

density=8gm/cm3

volume=πh(R^2-r^2)

=22/7*21*(4.4^2-4^2)=66*8.4*.4=221.76

mass=8*221.76=1774.1

Answer 2

Answer

Volume of the cylinder = $\tt{\pi r^{2} h}$

= $\tt{\pi h (R^{2} - r^{2})}$, R being the larger radius and r being the smaller radius

= $\tt{\frac{22}{7} \times 21(4.4^{2} - 4^{2})}$ [Smaller diameter = 8 cm, smaller radius = 4 cm. Thickness = 0.4 cm; larger radius = 4.4 cm]

= $\tt{22 \times 21 \times 3.36}$

= 221.76 cm^3

Mass of 221.76 cm^3 of metal = 221.76 × 8

= 1774.08