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    Mechanical Properties of Solids
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    Compute the bulk modulus of water from the following data: Initial volume = 100.0 litre, Pressure increase = 100.0 atm (1 atm = 1.013 × 105 Pa), Final volume = 100.5 litre. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large.
    Initial volume, V1 = 100.0 l = 100.0 × 10 –3 m3

    Final volume, V2 = 100.5 l = 100.5 ×10 –3 m3 

    Increase in volume, ΔV = V2 – V1 = 0.5 × 10–3 m3 

    Increase in pressure, Δp = 100.0 atm = 100 × 1.013 × 105 Pa

    Bulk space modulus space equals space fraction numerator increment straight p over denominator left parenthesis begin display style bevelled fraction numerator increment straight V over denominator straight V subscript 1 end fraction end style right parenthesis end fraction space equals increment straight p cross times fraction numerator straight V subscript 1 over denominator increment straight V end fraction space space space space space space space space space space space space space space space space space space space space space space space space space equals fraction numerator 100 space cross times space 1.013 space cross times space 10 to the power of 5 space end exponent cross times space 100 space cross times space 10 to the power of negative 3 end exponent over denominator left parenthesis 0.5 space cross times space 10 to the power of negative 3 end exponent right parenthesis end fraction space space space space space space space space space space space space space space space space space space space space space space space space space equals 2.026 space cross times space 10 to the power of 9 space Pa Bulk space modulus space of space air space equals space 1 space cross times space 10 to the power of 5 space end exponent Pa Therefore comma space fraction numerator Bulk space modulus space of space water over denominator space Bulk space modulus space of space air end fraction space equals space fraction numerator 2.026 space cross times space 10 to the power of 9 over denominator left parenthesis 1 space cross times space 10 to the power of 5 right parenthesis end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 2.026 space cross times space 10 to the power of 4 space This space ratio space is space very space high space because space air space is space more space compressible space than space water.
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    Answer
    What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is 1.03 × 103kg m–3?
    Let the given depth be h.

    Pressure at the given depth, p = 80.0 atm = 80 × 1.01 × 105Pa

    Density of water at the surface, ρ1 = 1.03 × 103 kg m–3

    Let ρ2 be the density of water at the depth h.

    Let V1 be the volume of water of mass m at the surface.

    Let V2 be the volume of water of mass m at the depth h.

    Let ΔV be the change in volume.

                       ΔV = V1 - V2
    straight m open square brackets open parentheses 1 over straight rho subscript 1 close parentheses minus open parentheses 1 over straight rho subscript 2 close parentheses close square brackets Therefore comma space Volumetric space strain space equals space ΔV over straight V subscript 1 space space space space space space space space space space space space equals straight m open square brackets open parentheses 1 over straight rho subscript 1 close parentheses minus open parentheses 1 over straight rho subscript 2 close parentheses close square brackets cross times open parentheses straight rho subscript 1 over straight m close parentheses space rightwards double arrow space ΔV over straight V subscript 1 space equals space 1 space minus space open parentheses straight rho subscript 1 over straight rho subscript 2 close parentheses space space space space space space space space space space space space space space space space space... space left parenthesis 1 right parenthesis space Bulk space modulus comma space straight B space equals space fraction numerator ρV subscript 1 over denominator increment straight V end fraction fraction numerator ΔV space over denominator space straight V subscript 1 end fraction space equals space straight p over straight B Compressibilty space of space water space equals open parentheses 1 over straight B close parentheses space equals space 45.8 cross times 10 to the power of negative 11 end exponent space Pa to the power of negative 1 end exponent Therefore comma space fraction numerator ΔV space over denominator space straight V subscript 1 end fraction equals space 80 space cross times space 1.013 space cross times space 10 to the power of 5 space cross times space 45.8 space cross times space 10 to the power of negative 11 space end exponent space space space space space space space space space equals space 3.71 space cross times space 10 to the power of negative 3 end exponent space space space space space space space space space space space space space space space space space... space left parenthesis ii right parenthesis From space equations space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis comma space we space get space 1 space minus space open parentheses straight rho subscript 1 over straight rho subscript 2 close parentheses space equals space 3.71 space cross times space 10 to the power of negative 3 end exponent rightwards double arrow space straight rho subscript 2 space equals space fraction numerator 1.03 space cross times space 10 cubed over denominator space 1 space minus space left parenthesis 3.71 space cross times space 10 to the power of negative 3 end exponent right parenthesis end fraction space space space space space space space space space equals space 1.034 space cross times space 10 cubed space kg space straight m to the power of negative 3 end exponent space Therefore comma space the space density space of space water space at space the space given space depth space left parenthesis straight h right parenthesis space is space 1.034 space cross times space 10 to the power of 3 space end exponent kg space straight m to the power of – 3 end exponent.

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    Answer
    Determine the volume contraction of a solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of 7.0 ×106 Pa.
    Length of an edge of the solid copper cube, l = 10 cm = 0.1 m

    Hydraulic pressure, p = 7.0 × 106 Pa

    Bulk modulus of copper, B = 140 × 109 Pa

    Formula is given by, 

    Bulk space modulus comma space straight B space equals space fraction numerator straight p over denominator open parentheses begin display style bevelled fraction numerator increment straight V over denominator straight V end fraction end style close parentheses end fraction where comma space fraction numerator increment straight V over denominator straight V end fraction equals space Volumetric space strain increment straight V space equals space change space in space volume space equals space pV over straight B straight V space equals space original space volume Original space volume space of space the space cube comma space straight V space equals space straight l cubed therefore space increment straight V space equals space pl cubed over straight B space space space space space space space space space space space space equals fraction numerator space 7 space cross times space 10 to the power of 6 space cross times space left parenthesis 0.1 right parenthesis cubed space over denominator left parenthesis 140 cross times 10 to the power of 9 right parenthesis end fraction space space space space space space space space space space space space equals space 5 space cross times space 10 to the power of negative 8 end exponent space straight m cubed space space space space space space space space space space space space equals space 5 space cross times space 10 to the power of negative 2 end exponent space cm to the power of negative 3 end exponent space Therefore comma space space the space volume space contraction space of space the space solid space copper space cube space is space 5 space cross times space 10 to the power of – 2 end exponent space cm to the power of – 3 end exponent.
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    How much should the pressure on a litre of water be changed to compress it by 0.10%?

    Volume of water, V = 1 L

    It is given that water is to be compressed by 0.10%.

    Therefore,

    Fractional change,     ∆V / V = 0.1 / (100 × 1)  =  10-3

    Bulk modulus, B = ρ / (∆V/V) 

    rightwards double arrow              ρ = B × (∆V/V)
    Bulk modulus of water, B = 2.2 × 109 Nm-2

                                  ρ = 2.2 × 109 × 10-3  =  2.2 × 106 Nm-2 

    Therefore, the pressure on water should be 2.2 ×106 Nm–2. 
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    Compute the fractional change in volume of a glass slab, when subjected to a hydraulic pressure of 10 atm.
    Hydraulic pressure exerted on the glass slab, p = 10 atm = 10 × 1.013 × 105 Pa

    Bulk modulus of glass, B = 37 × 109 Nm–2

    Bulk modulus, B =fraction numerator straight p over denominator begin display style bevelled fraction numerator increment straight V over denominator straight V end fraction end style end fraction

    where, 

    ∆V/V = Fractional change in volume

    ∴ ∆V/V = p / B

              =fraction numerator space 10 space cross times space 1.013 space cross times space 10 to the power of 5 over denominator left parenthesis 37 space cross times space 10 to the power of 9 right parenthesis end fraction
              = 2.73 × 10-5 

    Hence, the fractional change in the volume of the glass slab is 2.73 × 10–5.
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