JEE Challenger
More from Mechanical Properties of Solids
PhyMedAdv 2020 • Paper 2

Change in Edge Length of Aluminium Block Under Hydrostatic Pressure

A cubical block of solid aluminium (having bulk modulus B=−VdPdV=70 GPaB = -V \frac{dP}{dV} = 70\text{ GPa}) has an edge length of 1 m1\text{ m} at the surface of the earth. It is placed on the floor of a 5 km5\text{ km} deep ocean. Assuming average density of water to be 103 kg m−310^3\text{ kg m}^{-3} and acceleration due to gravity as 10 ms−210\text{ ms}^{-2}, determine the change in the edge length of the block in mm.

Official Numerical Answer0.24

Topics & Concepts

Mechanical Properties of SolidsElasticity

Step-by-Step Solution

To find the change in the edge length of the cubical block, we follow these steps:

  1. Calculate the gauge pressure at the bottom of the ocean: The increase in hydrostatic pressure (ΔP\Delta P) at a depth of h=5 km=5000 mh = 5\text{ km} = 5000\text{ m} is given by: ΔP=ρgh\Delta P = \rho g h

Given:

  • Density of water, ρ=103 kg m−3\rho = 10^3\text{ kg m}^{-3}
  • Acceleration due to gravity, g=10 ms−2g = 10\text{ ms}^{-2}
  • Depth, h=5000 mh = 5000\text{ m}

Substituting the values: ΔP=103×10×5000=5×107 Pa\Delta P = 10^3 \times 10 \times 5000 = 5 \times 10^7\text{ Pa}

  1. Relate fractional change in volume to fractional change in edge length: The volume of a cube of edge length LL is: V=L3V = L^3

Differentiating both sides with respect to LL: dV=3L2dLdV = 3L^2 dL

Dividing by V=L3V = L^3, we get the relation between volumetric strain and linear strain: ΔVV=3ΔLL\frac{\Delta V}{V} = 3 \frac{\Delta L}{L}

  1. Use the definition of Bulk Modulus (BB): The Bulk Modulus is defined as: B=ΔPΔVVB = \frac{\Delta P}{\frac{\Delta V}{V}}

Rearranging for volumetric strain: ΔVV=ΔPB\frac{\Delta V}{V} = \frac{\Delta P}{B}

Equating the two expressions for volumetric strain: 3ΔLL=ΔPB  ⟹  ΔL=L⋅ΔP3B3 \frac{\Delta L}{L} = \frac{\Delta P}{B} \implies \Delta L = \frac{L \cdot \Delta P}{3B}

  1. Calculate the change in edge length (ΔL\Delta L): Given:
  • Initial edge length, L=1 mL = 1\text{ m}
  • Bulk modulus of aluminium, B=70 GPa=70×109 PaB = 70\text{ GPa} = 70 \times 10^9\text{ Pa}

Substituting the known values into the equation: ΔL=1×5×1073×70×109=5×107210×109=14200 m\Delta L = \frac{1 \times 5 \times 10^7}{3 \times 70 \times 10^9} = \frac{5 \times 10^7}{210 \times 10^9} = \frac{1}{4200}\text{ m}

Converting ΔL\Delta L to millimeters (mm\text{mm}): ΔL=14200×1000 mm=1042 mm=521 mm≈0.238 mm\Delta L = \frac{1}{4200} \times 1000\text{ mm} = \frac{10}{42}\text{ mm} = \frac{5}{21}\text{ mm} \approx 0.238\text{ mm}

Rounding off to two decimal places gives 0.24 mm0.24\text{ mm} (or 0.238 mm0.238\text{ mm}).

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