MECHANICAL PROPERTIES OF SOLIDS-Exercise
Physics - Exercise
Q1. A steel wire of length 4.7 m and cross-sectional area \(3.0 × 10^{-5}\ m^2\) stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of \(4.0 × 10^{–5}\ m^2\) under a given load. What is the ratio of the Young’s modulus of steel to that of copper?
Solution
Given:
Length of steel wire, \( l_s = 4.7\,\text{m} \)
Cross-sectional area of steel wire, \( A_s = 3.0 \times 10^{-5}\,\text{m}^2 \)
Length of copper wire, \( l_c = 3.5\,\text{m} \)
Cross-sectional area of copper wire, \( A_c = 4.0 \times 10^{-5}\,\text{m}^2 \)
Both wires are subjected to the same load and undergo the same extension \(\Delta l\).
Let the applied load be \(W\). The force acting on each wire is \(F = Wg\).
For the steel wire:
\[ \begin{aligned} Y_s &= \frac{\text{stress}}{\text{strain}} \\ \sigma_s &= \frac{F}{A_s} = \frac{Wg}{3.0 \times 10^{-5}} \\ \varepsilon_s &= \frac{\Delta l}{l_s} = \frac{\Delta l}{4.7} \\ Y_s &= \frac{\sigma_s}{\varepsilon_s} = \frac{Wg}{3.0 \times 10^{-5}} \times \frac{4.7}{\Delta l} \end{aligned} \]
For the copper wire:
\[ \begin{aligned} Y_c &= \frac{\text{stress}}{\text{strain}} \\ \sigma_c &= \frac{F}{A_c} = \frac{Wg}{4.0 \times 10^{-5}} \\ \varepsilon_c &= \frac{\Delta l}{l_c} = \frac{\Delta l}{3.5} \\ Y_c &= \frac{\sigma_c}{\varepsilon_c} = \frac{Wg}{4.0 \times 10^{-5}} \times \frac{3.5}{\Delta l} \end{aligned} \]
Ratio of Young’s moduli:
\[ \begin{aligned} \frac{Y_s}{Y_c} &= \frac{\left(\dfrac{Wg}{3.0 \times 10^{-5}} \times \dfrac{4.7}{\Delta l}\right)} {\left(\dfrac{Wg}{4.0 \times 10^{-5}} \times \dfrac{3.5}{\Delta l}\right)} \\ &= \frac{4.7 \times 4.0}{3.0 \times 3.5} \\ &\approx 1.8 \end{aligned} \]
Result: The ratio of the Young’s modulus of steel to that of copper is approximately 1.8.
Q2. Figure 8.9 shows the strain-stress curve for a given material. What are (a) Young’s modulus and (b) approximate yield strength for this material?
Solution
The strain–stress graph given in Fig. 8.9 is used to determine the elastic properties of the material.
From the linear (elastic) region of the curve, a representative point is taken where the stress and strain are:
\[ \begin{aligned} \sigma &= 150 \times 10^{6}\ \text{N m}^{-2} \\ \varepsilon &= 0.002 \end{aligned} \]
The Young’s modulus \(Y\) of the material is defined as the ratio of stress to strain in the elastic region.
\[ \begin{aligned} Y &= \frac{\sigma}{\varepsilon} \\ &= \frac{150 \times 10^{6}}{0.002} \\ &= 7.5 \times 10^{10}\ \text{N m}^{-2} \end{aligned} \]
Hence, the Young’s modulus of the material is \(7.5 \times 10^{10}\ \text{N m}^{-2}\).
The yield strength is the stress corresponding to the onset of plastic deformation, which is read approximately from the yield point of the strain–stress curve.
\[ \begin{aligned} \text{Yield stress} &\approx 300 \times 10^{6}\ \text{N m}^{-2} \\ &= 3.0 \times 10^{8}\ \text{N m}^{-2} \end{aligned} \]
Thus, the approximate yield strength of the material is \(3.0 \times 10^{8}\ \text{N m}^{-2}\).
Q3. The stress-strain graphs for materials A and B are shown in Fig. 8.10.
The graphs are drawn to the same scale.
(a)
Which of the materials has the greater Young’s modulus?
(b)
Which of the two is the stronger material?
Solution
Because both stress–strain graphs are drawn to the same scale, we can compare their slopes and maximum stresses directly.
(a) Greater Young’s modulus
Young’s modulus is given by the slope of the stress–strain curve in the initial linear (elastic) region.
From the figure, material B has a steeper initial slope than material A.
Therefore, material B has the greater Young’s modulus, meaning it is stiffer and undergoes less
strain for the same applied stress.
(b) Stronger material
The strength of a material is determined by the maximum stress it can withstand before failure (ultimate tensile stress).
From the graph, material A reaches a higher maximum stress than material B before failure.
Hence, material A is the stronger material, even though it is less stiff than B.
Answers:
- (a) Material B has the greater Young’s modulus.
- (b) Material A is the stronger material.
Q4. Read the following two statements below carefully and state, with reasons, if it is true
or false.
(a) The Young’s modulus of rubber is greater than that of steel;
(b) The stretching of a coil is determined by its shear modulus.
Solution
(a) False.
Young’s modulus is a measure of the stiffness of a material, that is, its resistance to longitudinal (tensile or compressive) deformation. Steel is much stiffer than rubber and therefore requires a very large stress to produce even a small strain. Rubber, on the other hand, undergoes large strain even under small stress. Hence, the Young’s modulus of steel is far greater than that of rubber, not the other way round.
(b) True.
When a coil or helical spring is stretched by an applied load, the wire of the spring does not undergo simple extension along its length. Instead, it experiences twisting due to shear. Therefore, the deformation of a stretched coil depends on the shear strain in the material of the wire, and hence on its shear modulus.
Q5. Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in Fig. 8.11. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m. Compute the elongations of the steel and the brass wires.
Solution
The diameter of each wire is \(0.25\,\text{cm}\). Hence the radius is \(r = 0.125\,\text{cm} = 0.125 \times 10^{-2}\,\text{m}\). Since both wires have the same diameter, their cross-sectional area is the same.
\[ A = \pi r^2 = \pi \left(0.125 \times 10^{-2}\right)^2 \,\text{m}^2 \]
The unloaded length of the steel wire is \(1.5\,\text{m}\) and that of the brass wire is \(1.0\,\text{m}\).
The mass supported by the steel wire is the total mass hanging below it, which is \(4.0 + 6.0 = 10\,\text{kg}\). Hence the force on the steel wire is
\[ F_s = 10 \times 9.8 = 98\,\text{N} \]
The mass supported by the brass wire is \(6.0\,\text{kg}\), so the force on the brass wire is
\[ F_b = 6 \times 9.8 = 58.8\,\text{N} \]
The Young’s modulus of steel is taken as \(Y_s = 200 \times 10^{9}\,\text{N m}^{-2}\) and that of brass as \(Y_b = 100 \times 10^{9}\,\text{N m}^{-2}\).
Elongation of the steel wire:
\[ \begin{aligned} Y_s &= \frac{F_s L_s}{A\,\Delta L_s} \\ \Delta L_s &= \frac{F_s L_s}{A Y_s} \\ &= \frac{98 \times 1.5}{3.14 \times \left(0.125 \times 10^{-2}\right)^2 \times 200 \times 10^{9}} \\ &\approx 1.5 \times 10^{-4}\,\text{m} \end{aligned} \]
Elongation of the brass wire:
\[ \begin{aligned} Y_b &= \frac{F_b L_b}{A\,\Delta L_b} \\ \Delta L_b &= \frac{F_b L_b}{A Y_b} \\ &= \frac{6 \times 9.8 \times 1.0}{3.14 \times \left(0.125 \times 10^{-2}\right)^2 \times 100 \times 10^{9}} \\ &\approx 1.2 \times 10^{-4}\,\text{m} \end{aligned} \]
Thus, the elongation of the steel wire is approximately \(1.5 \times 10^{-4}\,\text{m}\) and the elongation of the brass wire is approximately \(1.2 \times 10^{-4}\,\text{m}\).
Q6. The edge of an aluminium cube is 10 cm long. One face of the cube is firmly fixed to a vertical wall. A mass of 100 kg is then attached to the opposite face of the cube. The shear modulus of aluminium is 25 GPa. What is the vertical deflection of this face?
Solution
The shear modulus of aluminium is given as \(G = 25\,\text{GPa} = 25 \times 10^{9}\,\text{N m}^{-2}\).
The edge length of the aluminium cube is \(10\,\text{cm} = 0.1\,\text{m}\).
The area of the face to which the force is applied is
\[ A = (0.1)^2 = 0.01\,\text{m}^2 \]
The length of the cube perpendicular to the fixed face is
\[ L = 0.1\,\text{m} \]
The mass attached to the opposite face of the cube is \(100\,\text{kg}\). Hence the applied force is
\[ F = mg = 100 \times 9.8 = 980\,\text{N} \]
The relation between shear modulus, force, and shear deformation is
\[ \begin{aligned} G &= \frac{\text{shear stress}}{\text{shear strain}} \\ &= \frac{F/A}{\Delta x/L} \end{aligned} \]
Rearranging the above expression, the vertical deflection \(\Delta x\) is
\[ \begin{aligned} \Delta x &= \frac{F L}{A G} \\ &= \frac{980 \times 0.1}{0.01 \times 25 \times 10^{9}} \\ &= 3.92 \times 10^{-7}\,\text{m} \end{aligned} \]
Thus, the vertical deflection of the free face of the aluminium cube is \(3.92 \times 10^{-7}\,\text{m}\).
Q7. Four identical hollow cylindrical columns of mild steel support a big structure of mass 50,000 kg. The inner and outer radii of each column are 30 and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column.
Solution
The mass of the structure supported by the columns is \(50{,}000\,\text{kg}\). Hence, the total weight of the structure is
\[ W = 50{,}000 \times 9.8 = 4.9 \times 10^{5}\,\text{N} \]
Since the load is uniformly distributed over four identical columns, the compressive force acting on each column is
\[ F = \frac{4.9 \times 10^{5}}{4} = 1.225 \times 10^{5}\,\text{N} \]
The inner radius of each hollow column is \(r = 30\,\text{cm} = 0.30\,\text{m}\) and the outer radius is \(R = 60\,\text{cm} = 0.60\,\text{m}\).
The cross-sectional area of each hollow cylindrical column is therefore
\[ A = \pi\left(R^{2} - r^{2}\right) = \pi\left(0.60^{2} - 0.30^{2}\right) \]
The Young’s modulus of mild steel is taken as \(Y = 200 \times 10^{9}\,\text{N m}^{-2}\).
The compressive stress on each column is
\[ \begin{aligned} \sigma &= \frac{F}{A} \\ &= \frac{1.225 \times 10^{5}}{\pi\left(0.60^{2} - 0.30^{2}\right)} \\ &\approx 1.44 \times 10^{5}\,\text{N m}^{-2} \end{aligned} \]
The compressional strain in each column is obtained using the relation \(Y = \sigma / \varepsilon\).
\[ \begin{aligned} \varepsilon &= \frac{\sigma}{Y} \\ &= \frac{1.44 \times 10^{5}}{200 \times 10^{9}} \\ &\approx 7.2 \times 10^{-7} \end{aligned} \]
Thus, the compressional strain in each mild steel column is approximately \(7.2 \times 10^{-7}\).
Q8. A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain?
Solution
The rectangular cross-section of the copper piece is \(15.2\,\text{mm} \times 19.1\,\text{mm}\), which in SI units becomes \(15.2 \times 10^{-3}\,\text{m} \times 19.1 \times 10^{-3}\,\text{m}\).
The tensile force applied on the copper piece is \(F = 44{,}500\,\text{N}\).
The relation between Young’s modulus, stress, and strain is
\[ \begin{aligned} Y &= \frac{\sigma}{\varepsilon} \\ \varepsilon &= \frac{\sigma}{Y} \end{aligned} \]
The stress produced in the copper piece is
\[ \begin{aligned} \sigma &= \frac{F}{A} \\ &= \frac{44{,}500}{15.2 \times 10^{-3} \times 19.1 \times 10^{-3}} \\ &\approx 1.53 \times 10^{8}\,\text{N m}^{-2} \end{aligned} \]
The Young’s modulus of copper is taken as \(Y = 110 \times 10^{9}\,\text{N m}^{-2}\).
The resulting strain in the copper piece is therefore
\[ \begin{aligned} \varepsilon &= \frac{1.53 \times 10^{8}}{110 \times 10^{9}} \\ &\approx 1.39 \times 10^{-3} \end{aligned} \]
Thus, the strain produced in the copper piece is approximately \(1.4 \times 10^{-3}\).
Q9. A steel cable with a radius of 1.5 cm supports a chairlift at a ski area. If the maximum stress is not to exceed \(\mathrm{108\ N\ m^{–2}}\), what is the maximum load the cable can support ?
Solution
The radius of the steel cable is given as \(r = 1.5\,\text{cm} = 0.015\,\text{m}\).
The maximum permissible stress in the cable is
\[ \sigma = 10^{8}\,\text{N m}^{-2} \]
The relation between stress and force is
\[ \sigma = \frac{F}{A} \]
where \(A\) is the cross-sectional area of the cable.
\[ A = \pi r^{2} = 3.14 \times (0.015)^{2}\,\text{m}^{2} \]
Substituting the given values,
\[ \begin{aligned} 10^{8} &= \frac{F}{3.14 \times (0.015)^{2}} \\ F &= 3.14 \times (0.015)^{2} \times 10^{8} \\ &= 7.07 \times 10^{4}\,\text{N} \end{aligned} \]
Thus, the maximum load that the steel cable can safely support is \(7.07 \times 10^{4}\,\text{N}\).
Q10. A rigid bar of mass 15 kg is supported symmetrically by three wires each 2.0 m long. Those at each end are of copper and the middle one is of iron. Determine the ratios of their diameters if each is to have the same tension.
Solution
The mass of the rigid bar is \(15\,\text{kg}\), so its weight is \(15 \times 9.8\,\text{N}\). The bar is supported symmetrically by three vertical wires, each of length \(2.0\,\text{m}\). The two end wires are of copper and the middle wire is of iron.
Since the bar is rigid and supported symmetrically, all three wires undergo the same extension. The tension in each wire is also the same, as required in the problem.
The Young’s modulus of iron is \(Y_{Fe} = 190 \times 10^{9}\,\text{N m}^{-2}\) and that of copper is \(Y_{Cu} = 110 \times 10^{9}\,\text{N m}^{-2}\).
For a wire under tension, the relation between tension \(T\), Young’s modulus \(Y\), cross-sectional area \(A\), original length \(L\), and extension \(\Delta L\) is
\[ T = Y A \frac{\Delta L}{L} \]
Since all wires have the same length and undergo the same extension, and the tension in each wire is the same, it follows that
\[ Y_{Fe} A_{Fe} = Y_{Cu} A_{Cu} \]
\[ \begin{aligned} \frac{Y_{Fe}}{Y_{Cu}} &= \frac{A_{Cu}}{A_{Fe}} \\ \frac{190 \times 10^{9}}{110 \times 10^{9}} &= \frac{\pi r_{Cu}^{2}}{\pi r_{Fe}^{2}} \\ \frac{r_{Cu}^{2}}{r_{Fe}^{2}} &= \frac{19}{11} \\ \frac{r_{Cu}}{r_{Fe}} &= \sqrt{\frac{19}{11}} \\ &\approx 1.3 \end{aligned} \]
Thus, the ratio of the diameters of the copper wire to the iron wire is approximately \(1.3 : 1\).
Q11. A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle. The cross-sectional area of the wire is 0.065 \(\mathrm{cm^2}\). Calculate the elongation of the wire when the mass is at the lowest point of its path.
Solution
The mass attached to the steel wire is \(m = 14.5\,\text{kg}\) and the unstretched length of the wire is \(L = 1.0\,\text{m}\).
The angular speed of rotation is given as \(2\,\text{rev s}^{-1}\), which in radian measure is
\[ \omega = 2 \times 2\pi = 4\pi\,\text{rad s}^{-1} \]
The cross-sectional area of the steel wire is \(A = 0.065\,\text{cm}^2 = 0.065 \times 10^{-4}\,\text{m}^2\).
At the lowest point of the vertical circle, the tension in the wire is equal to the sum of the weight of the mass and the centripetal force required to keep it in circular motion.
\[ \begin{aligned} F &= mg + \frac{mv^{2}}{r} \\ &= mg + m r \omega^{2} \\ &= (14.5 \times 9.8) + \left(14.5 \times 1 \times (4\pi)^2\right) \\ &\approx 142 + 229 \\ &\approx 371\,\text{N} \end{aligned} \]
The Young’s modulus of steel is taken as \(Y = 200 \times 10^{9}\,\text{N m}^{-2}\).
The stress produced in the wire is
\[ \begin{aligned} \sigma &= \frac{F}{A} \\ &= \frac{371}{0.065 \times 10^{-4}} \end{aligned} \]
The corresponding strain in the wire is
\[ \begin{aligned} \varepsilon &= \frac{\sigma}{Y} \\ &= \frac{371}{0.065 \times 10^{-4} \times 200 \times 10^{9}} \\ &\approx 2.85 \times 10^{-4} \end{aligned} \]
Since strain is the ratio of elongation to original length, the elongation of the wire is
\[ \begin{aligned} \Delta L &= \varepsilon L \\ &= 2.85 \times 10^{-4} \times 1.0 \\ &= 2.85 \times 10^{-4}\,\text{m} \end{aligned} \]
Thus, the elongation of the steel wire when the mass is at the lowest point of its path is approximately \(2.9 \times 10^{-4}\,\text{m}\).
Q12. Compute the bulk modulus of water from the following data: Initial volume = 100.0 litre, Pressure increase = \(\mathrm{100.0\ atm\ (1 atm = 1.013 × 10^55\ Pa)}\), 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.
Solution
The initial volume of water is \(V = 100.0\,\text{litre} = 100 \times 10^{-3}\,\text{m}^3\). The final volume is \(100.5\,\text{litre}\).
The increase in pressure is \(100.0\,\text{atm}\). Using \(1\,\text{atm} = 1.013 \times 10^{5}\,\text{Pa}\), the pressure increase is
\[ P = 100 \times 1.013 \times 10^{5} = 1.013 \times 10^{7}\,\text{Pa} \]
The change in volume of water is
\[ \begin{aligned} \Delta V &= 100.5 - 100.0 \\ &= 0.5\,\text{litre} \\ &= 0.5 \times 10^{-3}\,\text{m}^3 \end{aligned} \]
The bulk modulus \(B\) is defined as the ratio of the increase in pressure to the corresponding fractional decrease in volume.
\[ \begin{aligned} B &= \frac{P}{\Delta V / V} \\ &= \frac{P V}{\Delta V} \\ &= \frac{1.013 \times 10^{7} \times 100 \times 10^{-3}}{0.5 \times 10^{-3}} \\ &= 2.026 \times 10^{9}\,\text{Pa} \end{aligned} \]
Thus, the bulk modulus of water is approximately \(2.0 \times 10^{9}\,\text{Pa}\).
The bulk modulus of air at constant temperature is of the order of \(10^{5}\,\text{Pa}\), which is much smaller than that of water.
This large difference arises because water molecules are already very closely packed, so even a large increase in pressure produces only a very small change in volume. In contrast, air consists of molecules that are far apart, making it much easier to compress. Hence, the bulk modulus of water is several orders of magnitude greater than that of air.
Q13. What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is \(\mathrm{1.03 × 103 kg m^{–3}}\)?
Solution
The density of water at the surface is given as \(\rho_0 = 1.03 \times 10^{3}\,\text{kg m}^{-3}\).
The pressure at the given depth is \(80.0\,\text{atm}\). Taking atmospheric pressure at the surface as \(1\,\text{atm}\), the increase in pressure is
\[ \begin{aligned} \Delta P &= (80 - 1)\,\text{atm} \\ &= 79 \times 1.013 \times 10^{5}\,\text{Pa} \\ &\approx 8.0 \times 10^{6}\,\text{Pa} \end{aligned} \]
The bulk modulus of water is taken as \(B = 2.2 \times 10^{9}\,\text{Pa}\).
For small changes, the relation between density and pressure using bulk modulus is
\[ \begin{aligned} \frac{\Delta \rho}{\rho_0} &= \frac{\Delta P}{B} \end{aligned} \]
Substituting the given values,
\[ \begin{aligned} \frac{\Delta \rho}{1.03 \times 10^{3}} &= \frac{8.0 \times 10^{6}}{2.2 \times 10^{9}} \\ \Delta \rho &= 1.03 \times 10^{3} \times \frac{8.0 \times 10^{6}}{2.2 \times 10^{9}} \\ &\approx 3.75\,\text{kg m}^{-3} \end{aligned} \]
The density of water at the given depth is therefore
\[ \begin{aligned} \rho &= \rho_0 + \Delta \rho \\ &= 1.03 \times 10^{3} + 3.75 \\ &\approx 1.034 \times 10^{3}\,\text{kg m}^{-3} \end{aligned} \]
Thus, the density of water at a depth where the pressure is \(80.0\,\text{atm}\) is approximately \(1.034 \times 10^{3}\,\text{kg m}^{-3}\).
Q14. Compute the fractional change in volume of a glass slab, when subjected to a hydraulic pressure of 10 atm.
Solution
The hydraulic pressure applied on the glass slab is \(10\,\text{atm}\). Using \(1\,\text{atm} = 1.013 \times 10^{5}\,\text{Pa}\), the pressure in SI units is
\[ P = 10 \times 1.013 \times 10^{5} = 1.013 \times 10^{6}\,\text{Pa} \]
The bulk modulus of glass is given as \(B = 37 \times 10^{9}\,\text{Pa}\).
The bulk modulus is defined as
\[ \begin{aligned} B &= \frac{P}{\Delta V / V} \end{aligned} \]
Rearranging the above expression, the fractional change in volume is
\[ \begin{aligned} \frac{\Delta V}{V} &= \frac{P}{B} \\ &= \frac{1.013 \times 10^{6}}{37 \times 10^{9}} \\ &\approx 2.7 \times 10^{-5} \end{aligned} \]
Thus, the fractional change in volume of the glass slab is approximately \(2.7 \times 10^{-5}\).
Q15. Determine the volume contraction of a solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of \(\mathrm{7.0 × 10^6\ Pa}\).
Solution
The hydraulic pressure applied on the copper cube is \(P = 7.0 \times 10^{6}\,\text{Pa}\).
The edge length of the cube is \(10\,\text{cm} = 0.1\,\text{m}\).
The volume of the copper cube is
\[ V = 0.1 \times 0.1 \times 0.1 = 0.001\,\text{m}^3 \]
The bulk modulus of copper is given as \(B = 140 \times 10^{9}\,\text{N m}^{-2}\).
The bulk modulus is defined as
\[ \begin{aligned} B &= \frac{P}{\Delta V / V} \end{aligned} \]
Rearranging the expression to find the change in volume,
\[ \begin{aligned} \Delta V &= \frac{P}{B} \times V \\ &= \frac{7.0 \times 10^{6}}{140 \times 10^{9}} \times 0.001 \\ &= \frac{7.0 \times 10^{6} \times 0.001}{140 \times 10^{9}} \\ &= 5.0 \times 10^{-8}\,\text{m}^3 \end{aligned} \]
Thus, the volume contraction of the copper cube under the given hydraulic pressure is \(5.0 \times 10^{-8}\,\text{m}^3\).
Q16. How much should the pressure on a litre of water be changed to compress it by 0.10%? carry one quarter of the load.
Solution
The initial volume of water is \(1\,\text{litre} = 1 \times 10^{-3}\,\text{m}^3\).
The given compression is \(0.10\%\), which corresponds to a fractional change in volume of
\[ \frac{\Delta V}{V} = \frac{0.10}{100} = 1.0 \times 10^{-3} \]
The bulk modulus of water is taken as \(B = 2.2 \times 10^{9}\,\text{Pa}\).
The relation between bulk modulus, pressure change, and fractional change in volume is
\[ \begin{aligned} B &= \frac{\Delta P}{\Delta V / V} \end{aligned} \]
Rearranging the above expression to calculate the required change in pressure,
\[ \begin{aligned} \Delta P &= B \left(\frac{\Delta V}{V}\right) \\ &= 2.2 \times 10^{9} \times 1.0 \times 10^{-3} \\ &= 2.2 \times 10^{6}\,\text{Pa} \end{aligned} \]
Thus, the pressure on one litre of water must be increased by approximately \(2.2 \times 10^{6}\,\text{Pa}\) to compress it by \(0.10\%\).
