Deformation of solids

Question

A cast Iron block is used to hold up a beaker of mercury. The face of the block that supports the mercury is 2.5cm2. The blocks length changes by 0.01% (1/10000 the original length). The beaker has a mass of 100g. What volume of mercury is in the beaker?

Answer 1

From Elasticity, we have: F=Y A ΔL/L

Here: ΔL/L = 0.01% = 10^-4

A = 2.5 cm² = 2.5x10^-4 m²

Y = Young's Modulus of Cast Iron. This can vary quite a lot (80-140 GPa) depending on the grade of cast Iron.

F = Gravitational Force = Combined weight of beaker and mercury = (M+m)g, where,

M = Mass of Mercury = ρV

ρ = Density of Mercury = 13600 kg/m³

m = 100g = 0.1 kg

g = 9.81 m/s²

Solving, V = [(YA/g)(ΔL/L)-m]/ρ

Since you haven't given Y, I will use Y = 110 GPa = 1.1x10¹¹ Pa.

So, V = 0.020605 m³