Question

What is the change in weight of a hollow cylinder of height 8 cm and radius 3 cm when the air is pumped out of it

Answer 1

This question involves the concepts of density, weight, and volume.

The change in weight of the cylinder will be "2.72 x 10⁻³ N".

The change in weight of the cylinder will be equal to the weight of the air inside the cylinder which is being removed from it:

$\Delta W = W_a$

where,

• $\Delta W$ = change in weight of cylinder = ?
• $W_a$ = Weight of air inside cylinder = $\rho Vg$
• $\rho$ = density of air = 1.225 kg/m³
• V = Volume of air inside the cylinder = πr²h = π(0.03 m)²(0.08 m) = 2.26 x 10⁻⁴ m³
• g = acceleration due to gravity = 9.81 m/s²

Therefore,

$\Delta W = W_a = \rho Vg\\\\\Delta W = (1.225\ kg/m^3)(2.26\ x\ 10^{-4}\ m^3)(9.81\ m/s^2)$

ΔW = 2.72 x 10⁻³ N

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