NCERT Solutions For Class 11 Physics Chapter 9

Q.1 A steel wire of length 4.7 m and cross-sectional area 3.0 × 10^–5 m² stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 × 10^–5 m² under a given load. What is the ratio of the Young’s modulus of steel to that of copper?

Ans.

\begin{array}{l}\text{Here,}\text{length of steel wire,}{\text{l}}_{\text{s}}\text{= 4}\text{.7 m}\\ \text{Area of cross-section}\text{of steel wire,}{\text{a}}_{\text{s}}\text{= 3}{\text{.0×10}}^{\text{–5}}{\text{m}}^{\text{2}}\\ {\text{Length of copper wire, l}}_{\text{c}}\text{= 3}\text{.5 m}\\ \text{Area of cross-section of copper wire,}{\text{a}}_{\text{c}}\text{= 4}{\text{.0 × 10}}^{\text{–5}}{\text{m}}^{\text{2}}\\ {\text{Change in length = Δl}}_{\text{s}}{\text{= Δl}}_{\text{c}}\text{= Δl}\\ \text{Let}\text{force exerted in both the cases = F}\\ \text{Young’s modulus of steel wire}\text{is}\text{given}\text{as:}\\ {\text{Y}}_{\text{s}}\text{=}\frac{{\text{F}}_{\text{s}}}{{\text{a}}_{\text{s}}}\text{×}\frac{{\text{l}}_{\text{s}}}{\text{Δl}}\\ {\text{Y}}_{\text{s}}\text{=}\frac{\text{F × 4}\text{.7 m}}{\text{3}{\text{.0×10}}^{\text{–5}}{\text{m}}^{\text{2}}\text{× Δl}}\mathrm{\dots }\left(\text{i}\right)\\ \text{Young’s modulus of copper wire}\text{is}\text{given}\text{as:}\\ {\text{Y}}_{\text{c}}\text{=}\frac{{\text{F}}_{\text{c}}}{{\text{a}}_{\text{c}}}\text{×}\frac{{\text{l}}_{\text{c}}}{\text{Δl}}\\ {\text{Y}}_{\text{c}}\text{=}\frac{\text{F×3}\text{.5 m}}{\text{4}{\text{.0×10}}^{\text{–5}}{\text{m}}^{\text{2}}\text{×Δl}}\mathrm{\dots }\left(\text{ii}\right)\\ \text{Dividing equation}\left(\text{i}\right)\text{by equation}\left(\text{ii}\right)\text{, we}\text{obtain:}\\ \frac{{\text{Y}}_{\text{s}}}{{\text{Y}}_{\text{c}}}\text{=}\frac{\text{4}\text{.7 m × 4}{\text{.0 × 10}}^{\text{–5}}{\text{m}}^{\text{2}}}{\text{3}{\text{.0 × 10}}^{\text{–5}}{\text{m}}^{\text{2}}\text{× 3}\text{.5 m}}\text{= 1}\text{.79}\\ \therefore \text{Ratio of the}\text{Young’s modulus of steel to that of copper =1}\text{.79}\end{array}

Q.2 Figure 9.11 shows the strain-stress curve for a given material. What are (a) Young’s modulus and (b) approximate yield strength for this material?

Ans.

$\begin{array}{l}{\text{(a)From the given graph, for stress =150×10}}^{\text{6}}{\text{Nm}}^{\text{-2}}\text{,}\hfill \\ \text{Corresponding}\text{strain}\text{=}\text{0}\text{.002}\hfill \\ \therefore \text{Young’s modulus,}\text{Y}\text{=}\frac{\text{Stress}}{\text{Strain}}\text{=}\frac{\text{150}\text{×}{\text{10}}^{\text{6}}{\text{Nm}}^{\text{-2}}}{\text{0}\text{.002}}\hfill \\ \text{Y =}\text{7}\text{.5}\times {\text{10}}^{\text{10}}{\text{Nm}}^{\text{-2}}\hfill \\ \therefore \text{Young’s modulus for the given material}\text{=}\text{7}\text{.5}\text{×}{\text{10}}^{\text{10}}{\text{Nm}}^{\text{-2}}\hfill \\ \text{(b)The approximate}\text{yield strength}\text{of a material is the}\hfill \\ \text{maximum stress it can}\text{bear without crossing the elastic}\hfill \\ \text{limit}\text{.}\hfill \\ \therefore \text{From the given graph, the approximate}\text{yield strength}\hfill \\ \text{of the}\text{given material =}{\text{300 × 10}}^{\text{6}}{\text{Nm}}^{\text{-2}}\text{=}{\text{3×10}}^{\text{8}}{\text{Nm}}^{\text{-2}}\hfill \end{array}$

Q.3 The stress-strain graphs for materials A and B are shown in Fig. 9.12.
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?

Ans.

\begin{array}{l}\left(\text{a}\right)\text{From the given graphs, it is clear that for a given}\\ \text{strain, the stress for material A is more than that for}\\ \text{material B}\text{.}\\ \text{Since,}\text{Young’s modulus,}\text{Y}\text{=}\frac{\text{Stress}}{\text{Strain}}\\ \therefore \text{For a given value}\text{of}\text{strain,}\text{if the stress for a}\\ \text{material is greater,}\text{then Young’s modulus will also}\text{be}\\ \text{greater for that material}\text{.}\end{array}

Hence, Young’s modulus for material A is greater in comparison to it is for material B.

(b) Strength of a material is determined by the amount of stress required for fracturing a material, corresponding to its fracture point. The extreme point in a stress-strain curve is called as the Fracture point.

From the given graphs, it is clear that material A can bear more strain than material B.

Therefore, material A is stronger in comparison to material B.

Q.4 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.

Ans.

\begin{array}{l}\begin{array}{l}\text{(a) The}\text{given}\text{statement}\text{is}\text{false,}\text{because}\text{for a given}\hfill \\ \text{stress}\text{there}\text{is}\text{more}\text{strain in rubber than steel}\text{.}\hfill \\ \text{Young’s modulus}\text{is}\text{given}\text{by}\text{the}\text{relation:}\hfill \\ \text{Y}\text{=}\frac{\text{Stress}}{\text{Strain}}\hfill \\ \therefore \text{For a constant}\text{value}\text{of}\text{stress:}\text{Y}\propto \frac{\text{1}}{\text{Strain}}\hfill \\ \therefore \text{The}\text{Young’s}\text{modulus}\text{of rubber is less as}\text{compared}\text{to}\hfill \\ \text{it}\text{is}\text{for steel}\text{.}\hfill \end{array}\\ \text{(b) Shear modulus is given by the ratio of the applied stress to}\\ \text{the change in the shape of a body}\text{. The stretching of a coil changes}\\ \text{its shape without any change in the length of the wire used in the}\\ \text{coil}\text{. Therefore, shear modulus of elasticity is involved in it}\text{.}\end{array}

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Q.5 Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in Fig. 9.13. 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.

Ans.

$\begin{array}{l}\text{Here,}\text{diameter of wires,}\text{d}\text{=}\text{0}\text{.25 m}\hfill \\ \therefore \text{Radius of wires,}\text{r =}\frac{\text{d}}{\text{2}}\text{= 0}\text{.125 cm}\hfill \\ \text{Length of steel wire,}{\text{L}}_{\text{1}}\text{= 1}\text{.5 m}\hfill \\ \text{Length of brass wire,}{\text{L}}_{\text{2}}\text{= 1}\text{.0 m}\hfill \\ \text{Total force applied}\text{on the steel wire}\text{is}\text{given}\text{as:}\hfill \\ {\text{F}}_{\text{s}}\text{= (4 kg + 6 kg)}\times \text{g}\hfill \\ {\text{F}}_{\text{s}}\text{= 10 kg × 9}{\text{.8 ms}}^{\text{-2}}\text{= 98 N}\hfill \\ \text{Young’s modulus for steel,}{\text{Y}}_{\text{s}}\text{= 2}{\text{.0 ×10}}^{\text{11}}\text{Pa}\hfill \\ {\text{Let change in the length of the steel wire = ΔL}}_{\text{s}}\hfill \\ \text{Area}\text{of}\text{cross-section}\text{of}\text{steel}{\text{wire = A}}_{\text{s}}{\text{= πr}}^{\text{2}}\hfill \\ \therefore {\text{ΔL}}_{\text{1}}\text{=}\frac{{\text{F}}_{\text{s}}{\text{× L}}_{\text{s}}}{{\text{A}}_{\text{s}}{\text{× Y}}_{\text{s}}}\text{=}\frac{{\text{F}}_{\text{s}}{\text{× L}}_{\text{s}}}{{\text{πr}}^{\text{2}}{\text{× Y}}_{\text{s}}}\hfill \\ {\text{ΔL}}_{\text{1}}\text{=}\frac{\text{98 N × 1}\text{.5 m}}{\text{π(0}{\text{.125 ×10}}^{\text{-2}}{\text{m}}^{\text{2}}{\text{)}}^{\text{2}}{\text{× (2 × 10}}^{\text{11}}\text{Pa)}}\hfill \\ {\text{ΔL}}_{\text{1}}\text{= 1}{\text{.49 ×10}}^{\text{-4}}\text{m}\hfill \\ \text{Total force applied}\text{on brass wire,}{\text{F}}_{\text{b}}\text{= 6 × 9}\text{.8 = 58}\text{.8}\text{N}\hfill \\ {\text{Young’s modulus for brass,Y}}_{\text{b}}\text{= 0}{\text{.91×10}}^{\text{11}}\text{Pa}\hfill \\ {\text{Let change in the length of the brass wire = ΔL}}_{\text{b}}\hfill \\ \text{Area}\text{of}\text{cross-section}\text{of}\text{brass}\text{wire =}{\text{A}}_{\text{b}}{\text{= πr}}^{\text{2}}\hfill \\ \therefore {\text{ΔL}}_{\text{2}}\text{=}\frac{{\text{F}}_{\text{b}}{\text{× L}}_{\text{b}}}{{\text{A}}_{\text{b}}{\text{× Y}}_{\text{b}}}\text{=}\frac{{\text{F}}_{\text{2}}{\text{× L}}_{\text{2}}}{{\text{πr}}_{\text{2}}{}^{\text{2}}{\text{× Y}}_{\text{2}}}\hfill \\ {\text{ΔL}}_{\text{2}}\text{=}\frac{\text{58}\text{.8 N × 1}\text{.0 m}}{\text{π(0}{\text{.125 × 10}}^{\text{-2}}{\text{m)}}^2\text{× (0}{\text{.91 × 10}}^{\text{11}}\text{Pa)}}\text{= 1}{\text{.3×10}}^{\text{-4}}\text{m}\hfill \\ \therefore \text{Elongation}\text{of}\text{steel}{\text{wire = ΔL}}_{\text{s}}\text{= 1}{\text{.49 ×10}}^{\text{-4}}\text{m}\hfill \\ \text{Elongation}\text{of}\text{brass}{\text{wire = ΔL}}_{\text{b}}\text{= 1}{\text{.3 ×10}}^{\text{-4}}\text{m}\hfill \end{array}$

Q.6 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?

Ans.

\begin{array}{l}\text{Here,}\text{edge of aluminium cube,}\text{L}\text{=}\text{10 cm =}\text{0}\text{.1 m}\\ \text{Mass attached}\text{to aluminium cube, m}\text{=}\text{100 kg}\\ \text{Shear modulus of aluminium,}\text{η}\text{=}\text{25 GPa =}\text{25}\text{×}{\text{10}}^{\text{9}}\text{Pa}\\ \text{Shear modulus}\text{is}\text{given}\text{as:}\\ \text{η =}\frac{\text{Shear}\text{stress}}{\text{Shear}\text{strain}}\text{=}\frac{\text{F/A}}{\text{L/ΔL}}\text{(i)}\\ \text{Here, Applied force,}\text{F}\text{=}\text{mg}\\ \text{F = 100 N × 9}{\text{.8 ms}}^{\text{-2}}\text{= 980 N}\\ \text{Area of one of faces of the cube,}\text{A = 0}\text{.1 m × 0}\text{.1m}\\ \text{A = 0}{\text{.01 m}}^{\text{2}}\\ \text{ΔL = Vertical deflection of the cube}\\ \text{From}\text{equation}\text{(i),}\text{we}\text{have}\\ \text{ΔL=}\frac{\text{FL}}{\text{Aη}}\\ \text{ΔL=}\frac{\text{980 N × 0}\text{.1 m}}{{\text{10}}^{\text{–2}}{\text{m}}^{\text{2}}\text{×}\left({\text{25 × 10}}^{\text{9}}\text{Pa}\right)}\text{= 3}{\text{.92 × 10}}^{\text{–7}}\text{m}\\ \therefore \text{Vertical deflection of this face of the cube = 3}{\text{.92 × 10}}^{\text{–7}}\text{m}\end{array}

Q.7 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 cm and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column.

Ans.

\begin{array}{l}\text{Here,}\text{mass of the big structure, M}\text{=}\text{50,000 kg}\\ \text{Inner radius of cylinderical}\text{column, r}\text{=}\text{30 cm =}\text{0}\text{.3 m}\\ \text{Outer radius of cylinderical}\text{column, R}\text{=}\text{60 cm=}\text{0}\text{.6 m}\\ \text{Young’s modulus of steel, Y}\text{=}\text{2}\text{×}{\text{10}}^{\text{11}}\text{Pa}\\ \text{Total force applied, F}\text{=}\text{Mg =}\text{50000 kg}\text{×}\text{9}{\text{.8 ms}}^{-2}\\ \text{Stress}\text{=}\text{Force}\text{applied on each}\text{column}\\ \text{=}\frac{\text{50000 × 9}\text{.8 N}}{\text{4}}\text{=}\text{122500 N}\\ \text{Young’s modulus}\text{is}\text{given}\text{as: Y}\text{=}\frac{\text{Stress}}{\text{Strain}}\\ \therefore \text{Strain}\text{=}\frac{\text{F/A}}{\text{Y}}\left(\text{i}\right)\\ \text{Here, Area,}\text{A}\text{=}\pi \left({\text{R}}^{\text{2}}{\text{–r}}^{\text{2}}\right)\text{=}\pi \left[{\left(\text{0}\text{.6 m}\right)}^{\text{2}}\text{–}{\left(\text{0}\text{.3 m}\right)}^{\text{2}}\right]\\ \text{Using}\text{equation}\left(\text{i}\right)\text{,}\text{we}\text{have}\\ \text{Strain}\text{=}\frac{\text{122500 N}}{\pi \left[{\left(\text{0}\text{.6 m}\right)}^{\text{2}}\text{–}{\left(\text{0}\text{.3 m}\right)}^{\text{2}}\right]\text{×}\text{2}\text{×}{\text{10}}^{\text{11}}Pa}\\ \text{Strain = 7}{\text{.22 × 10}}^{\text{–7}}\\ \therefore \text{Compressional}\text{strain}\text{of}\text{each}\text{column}\text{=}\text{7}\text{.22}\text{×}{\text{10}}^{\text{–7}}\end{array}

Q.8 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?

Ans.

\begin{array}{l}\text{Here,}\text{length of the copper}\text{piece,l}\text{=}\text{19}\text{.1 mm =}\text{19}\text{.1}\text{×}{\text{10}}^{\text{–3}}\text{m}\\ \text{Breadth of the copper}\text{piece,b}\text{=}\text{15}\text{.2 mm =}\text{15}\text{.2}\text{×}{\text{10}}^{\text{–3}}\text{m}\\ \text{Area of the piece}\text{of}\text{copper,}\text{A}\text{=}\text{l}\text{×}\text{b}\\ \text{A}\text{= 19}{\text{.1×10}}^{\text{–3}}\text{m ×15}{\text{.2×10}}^{\text{–3}}\text{m = 2}{\text{.9×10}}^{\text{–4}}{\text{m}}^{\text{2}}\\ \text{Tension force}\text{exerted on the copper}\text{piece,}\text{F}\text{=}\text{44500 N}\\ {\text{Modulus of elasticity of copper, η = 42 ×10}}^{\text{9}}{\text{Nm}}^{\text{-2}}\\ \text{Modulus of elasticity}\text{is}\text{given}\text{by}\text{the}\text{relation:}\\ \text{η =}\frac{\text{Stress}}{\text{Strain}}\text{=}\frac{\text{F/A}}{\text{Strain}}\\ \therefore \text{Strain=}\frac{\text{F}}{\text{Aη}}\\ \text{Strain=}\frac{\text{44500 N}}{\text{2}{\text{.9 × 10}}^{\text{–4}}{\text{m}}^{\text{2}}{\text{× 42 × 10}}^{\text{9}}{\text{Nm}}^{\text{2}}}\\ \text{Strain = 3}{\text{.65×10}}^{\text{-3}}\end{array}

Q.9 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 10⁸ Nm^–2, what is the maximum load the cable can support?

Ans.

$\begin{array}{l}\text{Here,}\text{radius of the steel cable,}\text{r = 1}\text{.5 cm = 0}\text{.015 m}\hfill \\ \text{Area}\text{of}\text{cross-section}\text{of the steel cable =}\pi {\text{r}}^{\text{2}}\text{=}\pi {\text{(0}\text{.015 m)}}^{\text{2}}\hfill \\ {\text{Maximum stress = 10}}^{\text{8}}{\text{Nm}}^{\text{–2}}\hfill \\ \text{Maximum stress =}\frac{\text{Maximum force}}{\text{Area}\text{of}\text{cross-section}}\hfill \\ \therefore \text{Maximum force = Maximum stress ×Area of cross-section}\hfill \\ {\text{F = 10}}^{\text{8}}\text{×}\pi \text{(0}{\text{.015 m)}}^{\text{2}}\text{= 7}{\text{.065 × 10}}^{\text{4}}\text{N}\hfill \\ \therefore \text{The steel}\text{cable can carry the maximum load}\text{of 7}{\text{.065 ×10}}^{\text{4}}\text{N}\text{.}\hfill \end{array}$

Q.10 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 ratio of their diameters if each is to have the same tension.

Ans.

\begin{array}{l}\text{As}\text{each}\text{wire}\text{has}\text{same}\text{tension}\text{force,}\text{so}\text{extension}\text{is}\\ \text{the}\text{same}\text{in}\text{each}\text{case}\text{.}\\ \text{The relation for Young’s modulus is given}\text{as:}\\ \text{Y}\text{=}\frac{\text{Stress}}{\text{Strain}}\text{=}\frac{\text{F/A}}{\text{Strain}}\\ \text{Y}\text{=}\frac{\text{4F/}\pi {\text{d}}^{\text{2}}}{\text{Strain}}\text{(i)}\\ \text{Here,}\text{F}\text{=}\text{Force}\text{of}\text{tension}\\ \text{A=}\text{Cross-sectional}\text{area}\\ \text{d=}\text{Diameter of wire}\\ \text{From}\text{equation}\text{(i),}\text{it}\text{is}\text{clear}\text{that:}\\ \text{Y}\propto \frac{\text{1}}{{\text{d}}^{\text{2}}}\\ {\text{Young’s modulus of iron, Y}}_{\text{iron}}{\text{= 190 × 10}}^{\text{9}}\text{Pa}\\ \text{Young’s modulus of copper,}{\text{Y}}_{\text{cu}}{\text{= 110 × 10}}^{\text{9}}\text{Pa}\\ \text{Let}{\text{diameter of the iron wire = d}}_{\text{iron}}\\ \text{Let}{\text{diameter of the copper wire = d}}_{\text{cu}}\\ \therefore \frac{{\text{d}}_{\text{cu}}}{{\text{d}}_{\text{iron}}}\text{=}\sqrt{\frac{{\text{Y}}_{\text{iron}}}{{\text{Y}}_{\text{cu}}}}\text{=}\sqrt{\frac{{\text{190 × 10}}^{\text{9}}}{{\text{110 × 10}}^{\text{9}}}}\text{=}\sqrt{\frac{\text{19}}{\text{11}}}\\ \frac{{\text{d}}_{\text{cu}}}{{\text{d}}_{\text{iron}}}\text{=1}\text{.31}\end{array}

Q.11 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^-1 at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm². Calculate the elongation of the wire when the mass is at the lowest point of its path.

Ans.

\begin{array}{l}\text{Here,}\text{mass, m = 14}\text{.5 kg}\\ \text{Length of steel wire, l = 1}\text{.0 m}\\ {\text{Angular velocity, ω = 2 revs}}^{\text{-1}}\\ \text{Area of Cross-section}\text{of the wire, a = 0}{\text{.065 cm}}^{\text{2}}\text{= 0}{\text{.065 ×10}}^{\text{-4}}{\text{m}}^{\text{2}}\\ \text{When the mass is at the lowest point of its path:}\\ \text{Let}\text{elongation of the wire = ΔL}\\ \text{Total force exerted}\text{on the}\text{mass,}\text{when it is placed at}\\ \text{the lowest}\text{position of the vertical circleis}\text{given}\text{as:}\\ {\text{F = mg + mlω}}^{\text{2}}\\ \text{F =14}\text{.5 kg × 9}{\text{.8 ms}}^{-2}\text{+14}\text{.5 kg ×1 m×}{\left({\text{2 revs}}^{\text{-1}}\right)}^{\text{2}}\\ \text{F = 200}\text{.1 N}\\ \text{The}\text{relation}\text{for}\text{Young’s modulus is}\text{given}\text{as,}\\ \text{Y =}\frac{\text{Stress}}{\text{Strain}}\text{=}\frac{\frac{\text{F}}{\text{A}}}{\frac{\text{Δl}}{\text{l}}}\text{=}\frac{\text{Fl}}{\text{AΔl}}\\ \therefore \text{Δl =}\frac{\text{Fl}}{\text{AY}}\text{(i)}\\ \text{Young’s modulus}\text{for}\text{Steel =}{\text{2 × 10}}^{\text{11}}\text{Pa}\\ \text{Applying}\text{equation}\text{(i),}\text{we}\text{obtain:}\\ \text{Δl =}\frac{\text{200}\text{.1 N ×1 m}}{\text{0}{\text{.065×10}}^{\text{-4}}{\text{×2×10}}^{\text{11}}\text{Pa}}\\ \text{Δl = 1}{\text{.539× 10}}^{\text{–4}}\text{m}\\ \therefore \text{Elongation of the wire = 1}{\text{.539 × 10}}^{\text{–4}}\text{m}\end{array}

Q.12 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.

Ans.

\begin{array}{l}\text{Here,}\text{initial volume}\text{of}{\text{water,V}}_{\text{1}}\text{= 100}\text{.0}\text{L = 100}{\text{.0 × 10}}^{\text{–3}}{\text{m}}^{\text{3}}\\ \text{Final volume}\text{of}\text{water,}{\text{V}}_{\text{2}}\text{= 100}\text{.5 L = 100}{\text{.5 × 10}}^{\text{–3}}{\text{m}}^{\text{3}}\\ {\text{Rise in volume, ΔV = V}}_{\text{2}}{\text{– V}}_{\text{1}}\text{= 0}{\text{.5 × 10}}^{\text{–3}}{\text{m}}^{\text{3}}\\ \text{Rise in pressure, Δp = 100}\text{.0 atm}\\ \text{Δp = 100 × 1}{\text{.013 × 10}}^{\text{5}}\text{Pa}\\ \text{Bulk}\text{modulus}\text{of}\text{water =}\frac{{\text{ΔpV}}_{\text{1}}}{\text{ΔV}}\\ \text{=}\frac{\text{100 × 1}{\text{.013 × 10}}^{\text{5}}{\text{Pa × 100 × 10}}^{\text{–3}}{\text{m}}^{\text{3}}}{\text{0}{\text{.5 × 10}}^{\text{–3}}{\text{m}}^{\text{3}}}\\ \text{= 2}{\text{.026 ×10}}^{\text{9}}\text{Pa}\\ \text{Bulk}\text{modulus}\text{of}\text{air = 1}{\text{.0×10}}^{\text{5}}\text{Pa}\\ \therefore \frac{\text{Bulk}\text{modulus}\text{of}\text{water}}{\text{Bulk}\text{modulus}\text{of}\text{air}}\text{=}\frac{\text{2}{\text{.026 × 10}}^{\text{9}}\text{Pa}}{\text{1}{\text{.0 ×10}}^{\text{5}}\text{Pa}}\\ \text{= 2}{\text{.026 × 10}}^{\text{4}}\\ \text{This ratio is very large because air is more compressible}\\ \text{than water}\text{.}\end{array}

Q.13 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 ×10³ kg m^–3?

Ans.

$\begin{array}{l}\text{Pressure at the given depth,}\text{p = 80}\text{.0 atm}\hfill \\ \text{p = 80 × 1}{\text{.013 × 10}}^{\text{5}}\text{Pa}\hfill \\ \text{Density of water at surface,}{\text{ρ}}_{\text{1}}\text{= 1}{\text{.03 × 10}}^{\text{3}}{\text{kgm}}^{\text{–3}}\hfill \\ \text{Let}{\text{ρ}}_{\text{2}}\text{= density of water at the given}\text{depth}\text{.}\hfill \\ {\text{Let V}}_{\text{1}}\text{and}{\text{V}}_{\text{2}}\text{be}\text{the volume of water of mass m at the}\hfill \\ \text{surface}\text{and}\text{at the}\text{given depth}\text{respectively}\text{.}\hfill \\ \therefore \text{Change in volume,}{\text{ΔV = V}}_{\text{2}}{\text{– V}}_{\text{1}}\hfill \\ \text{ΔV = m (}\frac{\text{1}}{{\text{ρ}}_{\text{1}}}\text{–}\frac{\text{1}}{{\text{ρ}}_{\text{2}}}\text{)}\hfill \\ \therefore \text{Volumetric}\text{strain}\text{=}\frac{\text{ΔV}}{{\text{V}}_{\text{1}}}\text{=}\text{m(}\frac{\text{1}}{{\text{ρ}}_{\text{1}}}\text{–}\frac{\text{1}}{{\text{ρ}}_{\text{2}}}\text{)}\text{×}\frac{{\text{ρ}}_{\text{1}}}{\text{m}}\hfill \\ \therefore \frac{\text{ΔV}}{{\text{V}}_{\text{1}}}\text{=1}\text{–}\frac{{\text{ρ}}_{\text{1}}}{{\text{ρ}}_{\text{2}}}\text{(i)}\hfill \\ \text{Bulk}\text{modulus}\text{is}\text{given}\text{as:}\text{B}\text{=}\frac{{\text{pV}}_{\text{1}}}{\text{ΔV}}\hfill \\ \therefore \frac{\text{ΔV}}{{\text{V}}_{\text{1}}}\text{=}\frac{\text{p}}{\text{B}}\hfill \\ \text{Since}\text{compressibility}\text{of}\text{water}\text{=}\frac{\text{1}}{\text{B}}\text{= 45}{\text{.8×10}}^{\text{-11}}{\text{Pa}}^{\text{-1}}\hfill \\ \therefore \frac{\text{ΔV}}{{\text{V}}_{\text{1}}}\text{= 80 atm ×1}{\text{.013 × 10}}^{\text{5}}{\text{kgm}}^{\text{–3}}\text{×45}{\text{.8×10}}^{\text{-11}}{\text{Pa}}^{\text{-1}}\hfill \\ \frac{\text{ΔV}}{{\text{V}}_{\text{1}}}\text{= 3}{\text{.712×10}}^{\text{-3}}\text{(ii)}\hfill \\ \text{From}\text{equation}\text{(i)}\text{and}\text{(ii),}\text{we}\text{obtain:}\hfill \\ \text{1-}\frac{{\text{ρ}}_{\text{1}}}{{\text{ρ}}_{\text{2}}}\text{= 3}{\text{.712×10}}^{\text{-3}}\hfill \\ {\text{ρ}}_{\text{2}}\text{=}\frac{\text{1}{\text{.03 × 10}}^{\text{3}}}{\text{1-(3}{\text{.712×10}}^{\text{-3}}\text{)}}\text{=1}{\text{.034 × 10}}^{\text{3}}{\text{kgm}}^{\text{-3}}\hfill \\ \therefore \text{Density}\text{of}\text{water}\text{at}\text{the}\text{given}\text{depth}\text{= 1}{\text{.034 ×10}}^{\text{3}}{\text{kgm}}^{\text{-3}}\hfill \end{array}$

Q.14 Compute the fractional change in volume of a glass slab, when subjected to a hydraulic pressure of 10 atm.

Ans.

\begin{array}{l}\text{Given,}\text{hydraulic pressure applied on the glass slab,}\\ \text{p = 10 atm = 10 × 1}{\text{.013 × 10}}^{\text{5}}\text{Pa}\\ \text{Bulk modulus of glass,}{\text{B = 37 × 10}}^{\text{9}}{\text{Nm}}^{\text{–2}}\\ \text{Bulk}\text{modulus}\text{is}\text{given}\text{by}\text{the}\text{relation:}\\ \text{B =}\frac{\text{p}}{\frac{\text{ΔV}}{\text{V}}}\\ \text{Here,}\frac{\text{ΔV}}{\text{V}}\text{= Fractional change in volume}\\ \therefore \frac{\text{ΔV}}{\text{V}}\text{=}\frac{\text{p}}{\text{B}}\text{=}\frac{\text{10 × 1}{\text{.013 × 10}}^{\text{5}}\text{Pa}}{{\text{37 × 10}}^{\text{9}}\text{Pa}}\\ \frac{\text{ΔV}}{\text{V}}\text{= 2}{\text{.73× 10}}^{\text{-5}}\\ \therefore \text{Fractional change in the volume of the glass slab = 2}{\text{.73 × 10}}^{\text{–5}}\end{array}

Q.15 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.

Ans.

Q.16 How much should the pressure on a litre of water be changed to compress it by 0.10%?

Ans.

\begin{array}{l}\text{Here,}\text{volume of water,}\text{V = 1 L}\\ \text{Bulk}\text{modulus}\text{of}\text{water,}\text{B = 2}{\text{.2×10}}^{\text{9}}{\text{Nm}}^{\text{-2}}\\ \text{Given that, water is to be compressed by 0}\text{.10\%}\text{.}\\ \therefore \text{Fractional}\text{change,}\frac{\text{ΔV}}{\text{V}}\text{=}\frac{\text{0}\text{.1 L}}{\text{100×1 L}}{\text{= 10}}^{\text{-3}}\\ \text{Bulk}\text{modulus}\text{is}\text{given}\text{by}\text{the}\text{relation:}\\ \text{B =}\frac{\text{p}}{\frac{\text{ΔV}}{\text{V}}}\\ \therefore \text{p = B ×}\frac{\text{ΔV}}{\text{V}}\\ \therefore \text{p =}\left(\text{2}{\text{.2×10}}^{\text{9}}{\text{Nm}}^{\text{-2}}\right){\text{×10}}^{\text{-3}}\text{= 2}{\text{.2 ×10}}^{\text{6}}\text{Pa}\\ \therefore \text{The}\text{required}\text{pressure}\text{on}\text{water}\text{=}\text{2}{\text{.2 × 10}}^{\text{6}}\text{Pa}\end{array}

Q.17 Anvils made of single crystals of diamond, with the shape as shown in Fig. 9.14, are used to investigate behaviour of materials under very high pressures. Flat faces at the narrow end of the anvil have a diameter of 0.50 mm, and the wide ends are subjected to a compressional force of 50,000 N. What is the pressure at the tip of the anvil?

Ans.

\begin{array}{l}\text{Here,}\text{diameter of cones at narrow ends,}\text{d}\text{=}\text{0}\text{.50 mm}\\ \text{d =}\text{0}\text{.5}\text{×}{\text{10}}^{\text{–3}}\text{m}\\ \therefore \text{Radius, r}\text{=}\frac{\text{d}}{\text{2}}\text{=}\frac{\text{0}{\text{.50× 10}}^{\text{–3}}}{\text{2}}\text{= 0}{\text{.25 × 10}}^{\text{-3}}\text{m}\\ \text{Compressional force}\text{at}\text{wide}\text{ends,}\text{F = 50000 N}\\ \text{Pressure at the tip of the anvil}\text{is}\text{given}\text{as:}\\ \text{P =}\frac{\text{Force}}{\text{Area}}\text{=}\frac{\text{F}}{\pi {\text{r}}^{\text{2}}}\\ \text{P =}\frac{\text{50000 N}}{\pi {\left(\text{0}{\text{.25 × 10}}^{\text{-3}}\text{m}\right)}^{\text{2}}}\text{= 2}{\text{.55×10}}^{\text{11}}\text{Pa}\\ \therefore \text{Pressure at the tip of the anvil}\text{=}\text{2}\text{.55}\text{×}{\text{10}}^{\text{11}}\text{Pa}\text{.}\end{array}

Q.18 A rod of length 1.05 m having negligible mass is supported at its ends by two wires of steel (wire A) and aluminium (wire B) of equal lengths as shown in Fig. 9.15. The cross-sectional areas of wires A and B are 1.0 mm² and 2.0 mm² respectively. At what point along the rod should a mass m be suspended in order to produce (a) equal stresses and (b) equal strains in both steel and aluminium wires.

Ans.

\begin{array}{l}\text{Here,}\text{area of cross-section}\text{of}\text{wire}\text{A,}{\text{a}}_{\text{A}}\text{= 1}{\text{.0 mm}}^{\text{2}}\\ \text{= 1}{\text{.0 × 10}}^{\text{–6}}{\text{m}}^{\text{2}}\\ \text{Area of cross-section of wire B,}{\text{a}}_{\text{B}}\text{= 2}{\text{.0 mm}}^{\text{2}}\\ \text{= 2}{\text{.0 × 10}}^{\text{–6}}{\text{m}}^{\text{2}}\\ {\text{Young’s modulus of steel,Y}}_{\text{s}}{\text{= 2 × 10}}^{\text{11}}{\text{Nm}}^{\text{–2}}\\ {\text{Young’s modulus of aluminium,Y}}_{\text{a}}\text{= 7}{\text{.0 × 10}}^{\text{10}}{\text{Nm}}^{\text{–2}}\\ \left(\text{a}\right)\text{Let mass m be suspended to the rod at a distance y}\\ \text{from the end where wire A is suspended}\text{.}\\ \text{Stress =}\frac{\text{Force}}{\text{Area}}\text{=}\frac{\text{F}}{\text{a}}\\ \text{If stress}\text{is}\text{equal}\text{in}\text{the}\text{two}\text{wires,}\text{then:}\\ \frac{{\text{F}}_{\text{1}}}{{\text{a}}_{\text{A}}}\text{=}\frac{{\text{F}}_{\text{2}}}{{\text{a}}_{\text{B}}}\\ \text{Here,}\\ {\text{F}}_{\text{1}}\text{= Force on the steel wire}\\ {\text{F}}_{\text{2}}\text{= Force on the aluminum wire}\\ \frac{{\text{F}}_{\text{1}}}{{\text{a}}_{\text{A}}}\text{=}\frac{{\text{F}}_{\text{2}}}{{\text{a}}_{\text{B}}}\text{=}\frac{\text{1}}{\text{2}}\text{(i)}\\ \text{The situation of}\text{the}\text{problem}\text{is shown in the figure}\text{(a)}\text{.}\\ \text{Taking torque about the point of suspension}\text{of}\text{mass}\\ \text{from}\text{the}\text{rod,}\text{we obtain:}\\ {\text{F}}_{\text{1}}{\text{y = F}}_{\text{2}}\left(\text{1}\text{.05 – y}\right)\\ \frac{{\text{F}}_{\text{1}}}{{\text{F}}_{\text{2}}}\text{=}\frac{\left(\text{1}\text{.05 – y}\right)}{\text{y}}\text{(ii)}\\ \text{From equations}\left(\text{i}\right)\text{and}\left(\text{ii}\right)\text{, we have:}\\ \frac{\left(\text{1}\text{.05 – y}\right)}{\text{y}}\text{=}\frac{\text{1}}{\text{2}}\\ \text{2}\left(\text{1}\text{.05 – y}\right)\text{= y}\\ \text{2}\text{.1 – 2y = y}\\ \text{3y = 2}\text{.1}\\ \therefore \text{y = 0}\text{.7}\text{m}\\ \therefore \text{To produce an equal stress in both}\text{of}\text{the wires, the}\\ \text{mass must be suspended at a distance of 0}\text{.7 m from}\\ \text{the}\text{end}\text{where}\text{A}\text{is}\text{attached}\text{.}\\ \text{(b)}\text{The}\text{relation}\text{for}\text{Young’s}\text{modulus}\text{is}\text{given}\text{as:}\\ \text{Young’s}\text{modulus}\text{=}\frac{\text{Stress}}{\text{Strain}}\\ \therefore \text{Strain}\text{=}\frac{\text{Stress}}{\text{Young’s}\text{modulus}}\text{=}\frac{\text{a}}{\text{Y}}\\ \text{If}\text{the}\text{strain}\text{in}\text{both}\text{of}\text{the}\text{wires}\text{is}\text{equal,}\text{then}\\ \frac{\frac{{\text{F}}_{\text{1}}}{{\text{a}}_{\text{A}}}}{{\text{Y}}_{\text{s}}}\text{=}\frac{\frac{{\text{F}}_{\text{2}}}{{\text{a}}_{\text{B}}}}{{\text{Y}}_{\text{a}}}\\ \frac{{\text{F}}_{\text{1}}}{{\text{F}}_{\text{2}}}\text{=}\frac{{\text{a}}_{\text{A}}}{{\text{a}}_{\text{B}}}\frac{{\text{Y}}_{\text{s}}}{{\text{Y}}_{\text{a}}}\text{=}\frac{\text{1}}{\text{2}}\text{×}\frac{{\text{2 × 10}}^{\text{11}}{\text{Nm}}^{\text{–2}}}{{\text{7 × 10}}^{\text{10}}{\text{Nm}}^{\text{–2}}}\text{=}\frac{\text{10}}{\text{7}}\text{(iii)}\\ \text{Taking torque about the point of suspension}\text{of}\text{mass}\text{m,}\\ \text{we obtain:}\\ {\text{F}}_{\text{1}}{\text{y}}_{\text{s}}{\text{= F}}_{\text{2}}\left(\text{1}{\text{.05 – y}}_{\text{s}}\right)\\ \frac{{\text{F}}_{\text{1}}}{{\text{F}}_{\text{2}}}\text{=}\frac{\left(\text{1}{\text{.05 – y}}_{\text{s}}\right)}{{\text{y}}_{\text{s}}}\left(\text{iv}\right)\\ \text{From equations}\left(\text{iii}\right)\text{and}\left(\text{iv}\right)\text{, we obtain:}\\ \frac{\left(\text{1}{\text{.05 – y}}_{\text{s}}\right)}{{\text{y}}_{\text{s}}}\text{=}\frac{\text{10}}{\text{7}}\\ \text{7}\left(\text{1}{\text{.05 – y}}_{\text{s}}\right)\text{= 10}{\text{y}}_{\text{s}}\\ \text{17}{\text{y}}_{\text{s}}\text{= 7}\text{.35}\\ \therefore {\text{y}}_{\text{s}}\text{= 0}\text{.432}\\ \text{In order to produce an equal strain in both}\text{of}\text{the wires,}\\ \text{the mass must be hanged at a distance of 0}\text{.432 m}\\ \text{from the}\text{end where wire A attached}\text{.}\end{array}

Q.19 A mild steel wire of length 1.0 m and cross-sectional area 0.50 × 10^–2 cm² is stretched, well within its elastic limit, horizontally between two pillars. A mass of 100 g is suspended from the mid-point of the wire. Calculate the depression at the midpoint.

Ans.

\begin{array}{l}\text{Here,}\text{length of the steel wire = 1}\text{.0 m}\\ \text{Cross-sectional}\text{area, A = 0}{\text{.50 × 10}}^{\text{–2}}{\text{cm}}^{\text{2}}\\ \text{=0}{\text{.50 × 10}}^{\text{–6}}{\text{m}}^{\text{2}}\\ \text{A mass 100 g is suspended from midpoint}\text{of}\text{the}\text{wire}\text{.}\\ \text{m= 100 g = 0}\text{.1 kg}\\ \therefore \text{The wire get}\text{depressed as shown in the given figure}\text{.}\\ \text{Original length of}\text{wire = AC}\\ \text{Depression in}\text{the}\text{wire = l}\\ \text{The length of}\text{the}\text{wire}\text{after mass m is attached to the}\\ \text{wire = AO + OC}\\ \text{Increase in length of the}\text{wire,}\text{Δl =}\left(\text{AO + OC}\right)\text{–AC}\\ \text{Here,}\\ \text{AO = OC =}{\left[{\left(\text{0}\text{.5}\right)}^{\text{2}}\text{+}{\left(\text{l}\right)}^{\text{2}}\right]}^{\frac{\text{1}}{\text{2}}}\\ \therefore \text{Δl =}{\left[{\left(\text{0}\text{.5 m}\right)}^{\text{2}}\text{+}{\left(\text{l}\right)}^{\text{2}}\right]}^{\frac{\text{1}}{\text{2}}}\text{– 1}\text{.0}\\ \text{Expanding}\text{the}\text{expression}\text{and}\text{neglecting}\text{higher}\text{terms,}\\ \text{we}\text{obtain:}\\ \text{Δl =}\frac{{\text{l}}^{\text{2}}}{\text{0}\text{.5}}\\ \text{Strain=}\frac{\text{Increase}\text{in}\text{length}}{\text{Original}\text{length}}\\ \text{Let T = Tension in the wire}\text{.}\\ \therefore \text{mg = 2Tcosθ}\\ \text{From the given}\text{figure,we}\text{obtain:}\\ \text{cosθ}\text{=}\frac{\text{l}}{{\left[{\left(\text{0}\text{.5 m}\right)}^{\text{2}}\text{+}{\left(\text{l}\right)}^{\text{2}}\right]}^{\frac{\text{1}}{\text{2}}}}\text{=}\frac{\text{l}}{\left(\text{0}\text{.5 m}\right){\left[\text{1 +}{\left(\frac{\text{l}}{\text{0}\text{.5 m}}\right)}^{\text{2}}\right]}^{\frac{\text{1}}{\text{2}}}}\\ \text{Expanding the}\text{above}\text{expression and neglecting higher}\\ \text{terms,}\text{we}\text{obtain:}\\ \text{cos}\text{θ}\text{=}\frac{\text{l}}{\left(\text{0}\text{.5 m}\right)\left[\text{1}\text{+}\frac{{\text{l}}^{\text{2}}}{\text{2}{\left(\text{0}\text{.5 m}\right)}^{\text{2}}}\right]}\\ \left[\text{1}\text{+}\frac{{\text{l}}^{\text{2}}}{\text{2}{\left(\text{0}\text{.5 m}\right)}^{\text{2}}}\right]\text{;}\text{1}\text{for}\text{small}\text{value}\text{of}\text{l}\\ \therefore \text{cos}\text{θ}\text{=}\frac{\text{l}}{\text{0}\text{.5}}\\ \therefore \text{T}\text{=}\frac{\text{mg}}{\text{2}\left(\frac{\text{l}}{\text{0}\text{.5}}\right)}\text{=}\frac{\text{mg}}{\text{4l}}\\ \text{Stress}\text{=}\frac{\text{Tension}}{\text{Area}}\text{=}\frac{\text{mg}}{\text{4l}\text{×}\text{A}}\\ \text{Young’s modulus}\text{is}\text{given}\text{as:}\text{Y}\text{=}\frac{\text{Stress}}{\text{Strain}}\\ \therefore \text{Y =}\frac{\text{mg × 0}\text{.5}}{{\text{4l × A×l}}^{\text{2}}}\\ \therefore \text{l =}\sqrt[\text{3}]{\frac{\text{mg × 0}\text{.5}}{\text{4YA}}}\\ {\text{Young’s modulus of steel, Y = 2×10}}^{\text{11}}\text{Pa}\\ \therefore \text{l =}\sqrt[\text{3}]{\frac{\text{0}\text{.1 kg × 9}{\text{.8 ms}}^{-2}\text{×0}\text{.5 m}}{{\text{4 × 2 ×10}}^{\text{11}}\text{Pa × 0}{\text{.50 ×10}}^{\text{-6}}{m}^2}}\text{= 0}\text{.0106}\text{m}\\ \therefore \text{The depression at the midpoint = 0}\text{.0106 m}\end{array}

Q.20 Two strips of metal are riveted together at their ends by four rivets, each of diameter 6.0 mm. What is the maximum tension that can be exerted by the riveted strip if the shearing stress on the rivet is not to exceed 6.9 ×10⁷ Pa? Assume that each rivet is to carry one quarter of the load.

Ans.

$\begin{array}{l}\text{Here,}\text{diameter of the metal strip, d}\text{=}\text{6}\text{.0 mm}\hfill \\ \text{d = 6}\text{.0}\text{×}{\text{10}}^{\text{–3}}\text{m}\hfill \\ \therefore \text{Radius}\text{of the metal strip,}\text{r}\text{=}\frac{\text{d}}{\text{2}}\text{= 3}{\text{.0 × 10}}^{\text{–3}}\text{m}\hfill \\ \text{Maximum shearing stress}\text{on}\text{the}\text{riveted}\text{strip = 6}\text{.9}\text{×}{\text{10}}^{\text{7}}\text{Pa}\hfill \\ \text{Maximum}\text{stress}\text{=}\frac{\text{Maximum}\text{load}\text{on}\text{a}\text{rivet}}{\text{Area}}\hfill \\ \therefore \text{Maximum load}\text{on}\text{a}\text{rivet}\text{=}\text{Maximum stress × Area}\hfill \\ \begin{array}{l}\text{= 6}{\text{.9 × 10}}^{\text{7}}\text{Pa ×}\pi {\text{×(r)}}^{\text{2}}\\ \text{= 6}{\text{.9 × 10}}^{\text{7}}\text{Pa ×}\pi {\text{× (3 ×10}}^{\text{–3}}{\text{m)}}^{\text{2}}\end{array}\hfill \\ \text{= 1949}\text{.94 N}\hfill \\ \text{Since}\text{each rivet supports one quarter of the load,}\hfill \\ \therefore \text{Maximum tension}\text{on}\text{each}\text{rivet = 4 × 1949}\text{.94 N}\hfill \\ \text{T = 7799}\text{.76 N}\hfill \end{array}$

Q.21 The Marina trench is located in the Pacific Ocean, and at one place it is nearly eleven km beneath the surface of water. The water pressure at the bottom of the trench is about 1.1 × 10⁸ Pa. A steel ball of initial volume 0.32 m³ is dropped into the ocean and falls to the bottom of the trench. What is the change in the volume of the ball when it reaches to the bottom?

Ans.

$\begin{array}{l}\text{Here, water pressure at the bottom}\text{of the trench,}\hfill \\ \text{P =}\text{1}{\text{.1 × 10}}^{\text{8}}\text{Pa}\hfill \\ \text{Initial volume of the}\text{ball,}\text{V = 0}{\text{.32 m}}^{\text{3}}\hfill \\ \text{Bulk modulus for steel,}\text{B = 1}{\text{.6 × 10}}^{\text{11}}{\text{Nm}}^{\text{–2}}\hfill \\ \text{The ball is}\text{located in the Pacific Ocean,}\text{nearly}\text{11 km below}\hfill \\ \text{the surface}\text{.}\hfill \\ \text{Let the change in the volume of the ball on arriving}\text{at}\text{the}\hfill \\ \text{bottom of the trench = ΔV}\hfill \\ \text{Bulk}\text{modulus}\text{is}\text{given}\text{by}\text{the}\text{relation:}\hfill \\ \text{B =}\frac{\text{p}}{\frac{\text{ΔV}}{\text{V}}}\hfill \\ \therefore \text{ΔV =}\frac{\text{B}}{\text{pV}}\text{=}\frac{\text{1}{\text{.1 × 10}}^{\text{8}}\text{Pa × 0}{\text{.32 m}}^{\text{3}}}{\text{1}{\text{.6 × 10}}^{\text{11}}{\text{Nm}}^{\text{–2}}}\hfill \\ \text{ΔV = 2}{\text{.2 × 10}}^{\text{-4}}{\text{m}}^{\text{3}}\hfill \\ \therefore \text{Change in volume of the ball on reaching the bottom}\hfill \\ \text{of}\text{the trench =}\text{2}\text{.2}\text{×}{\text{10}}^{\text{–4}}{\text{m}}^{\text{3}}\hfill \end{array}$