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Exercises · 8.15

Q.Determine the volume contraction of a solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of 7.0×106 Pa7.0 \times 10^{6}\ \text{Pa}.

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The volume contraction is found using the bulk modulus relation ΔV=−PVB\Delta V = -\frac{P V}{B}. For copper, B=1.4×1011 PaB = 1.4 \times 10^{11}\ \text{Pa}, V=10−3 m3V = 10^{-3}\ \text{m}^3, and P=7.0×106 PaP = 7.0 \times 10^{6}\ \text{Pa}, giving ΔV=−5.0×10−8 m3\Delta V = -5.0 \times 10^{-8}\ \text{m}^3 or 0.05 cm30.05\ \text{cm}^3.

When a solid is squeezed uniformly from all sides — that’s hydraulic pressure — the volume shrinks. The question is: by how much? The key idea is that the resistance to uniform compression is measured by the bulk modulus BB. For a given pressure PP, the fractional change in volume is −PB-\frac{P}{B}, so the absolute change is ΔV=−PVB\Delta V = -\frac{P V}{B}. The negative sign simply means volume decreases.

Copper’s bulk modulus is a standard value you should remember or be given: B=1.4×1011 PaB = 1.4 \times 10^{11}\ \text{Pa}. The cube’s edge is 10 cm, so its volume is straightforward. Let’s walk through it.

  1. Find the original volume in SI units.

    Edge length a=10 cm=0.1 ma = 10\ \text{cm} = 0.1\ \text{m}.

    Volume V=a3=(0.1)3=1.0×10−3 m3V = a^3 = (0.1)^3 = 1.0 \times 10^{-3}\ \text{m}^3.

  2. Recall the bulk modulus formula.

    B=−PΔV/VB = -\frac{P}{\Delta V / V}

    Rearranged: ΔV=−PVB\Delta V = -\frac{P V}{B}.

    The minus sign ensures that a positive pressure gives a negative ΔV\Delta V (contraction). We only care about the magnitude of contraction, but we’ll keep the sign for correctness.

  3. Plug in the numbers.

    P=7.0×106 PaP = 7.0 \times 10^{6}\ \text{Pa}, V=1.0×10−3 m3V = 1.0 \times 10^{-3}\ \text{m}^3, B=1.4×1011 PaB = 1.4 \times 10^{11}\ \text{Pa}.

ΔV=−(7.0×106)×(1.0×10−3)1.4×1011=−7.0×1031.4×1011\Delta V = -\frac{(7.0 \times 10^{6}) \times (1.0 \times 10^{-3})}{1.4 \times 10^{11}} = -\frac{7.0 \times 10^{3}}{1.4 \times 10^{11}}

  1. Simplify the fraction.

7.01.4=5.0,1031011=10−8\frac{7.0}{1.4} = 5.0, \quad \frac{10^{3}}{10^{11}} = 10^{-8}

So ΔV=−5.0×10−8 m3\Delta V = -5.0 \times 10^{-8}\ \text{m}^3. …

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