NCERT Solutions for Class 11 Physics Chapter 8 Mechanical Properties Of Solids

8.1 A steel wire of length 4.7 m and cross-sectional area $3.0\times {10}^{-5}{m}^2$ stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of $4.0\times {10}^{-5}{m}^2$ under a given load. What is the ratio of the Young’s modulus of steel to that of copper?

Answer:

Let the Young's Modulus of steel and copper be Y _S and Y _C respectively.

Length of the steel wire l _S = 4.7 m

Length of the copper wire l _C = 4.7 m

The cross-sectional area of the steel wire A _S = $3.0\times {10}^{-5}{m}^2$

The cross-sectional area of the Copper wire A _C = $4.0\times {10}^{-5}{m}^2$

Let the load and the change in the length be F and $\mathrm{\Delta }l$ respectively

$$\begin{gathered} Y=\frac{Fl}{A\mathrm{\Delta }l} \\ \frac{F}{\mathrm{\Delta }l}=\frac{AY}{l} \end{gathered}$$

Since F and $\mathrm{\Delta }l$ is the same for both wires we have

$$\begin{gathered} \frac{A_S{Y}_S}{l_S}=\frac{A_C{Y}_C}{l_C} \\ \frac{Y_S}{Y_C}=\frac{A_C{l}_S}{A_S{l}_C} \\ \frac{Y_S}{Y_C}=\frac{4.0\times {10}^{-5}\times 4.7}{3.0\times {10}^{-5}\times 3.5} \\ \frac{Y_S}{Y_C}=1.79 \end{gathered}$$

The ratio of Young’s modulus of steel to that of copper is 1.79.

8.2 (a) Figure 8.9 shows the strain-stress curve for a given material. What are (a)

Young’s modulus

Fig. 8.9

Answer:

Young’s modulus is given as the ratio of stress to strain when the body is behaving elastically.

For the given material

$$\begin{gathered} Y=\frac{Stress}{Strain} \\ Y=\frac{150\times {10}^6}{0.002} \\ Y=7.5\times {10}^{2}N{m}^{-2} \end{gathered}$$

8.2(b)What are approximate yield strength for this material?

Answer:

The Yield Strength is approximately $3\times {10}^{8}N{m}^{-2}$ for the given material. We can see above this value of strain, the body stops behaving elastically.

8.3(a) The stress-strain graphs for materials A and B are shown in Fig. 8.10. The graphs are drawn to the same scale. Which of the materials has the greater Young’s modulus?

Fig 8.10

Answer:

As we can see in the given Stress-Strain graphs that the slope is more in the graph corresponding to material A. We conclude that A has a greater Young's Modulus.

8.3(b) Which of the two is the stronger material?
Answer: The material which fractures at higher stress is said to be stronger. As we can see in the given Stress-Strain graphs the stress at which the material fractures is higher in A than that in B we conclude that A is the stronger material.

8.4(a) Read the following two statements below carefully and state, with reasons, if it is true or false.

The Young’s modulus of rubber is greater than that of steel;

Answer:

False: Young's Modulus is defined as the ratio of the stress applied on a material and the corresponding strain that occurs. As for the same amount of pressure applied on a piece of rubber and steel, the elongation will be much lesser in case of steel than that in the case of rubber and therefore the Young's Modulus of rubber is lesser than that of steel.

8.4 (b) Read the following two statements below carefully and state, with reasons, if it is true or false. The stretching of a coil is determined by its shear modulus.

Answer:

True: As the force acts Normal to the parallel planes in which helical parts of the wire lie, the actual length of the wire would not change but it's shape would. Therefore the amount of elongation of the coil taking place for corresponding stress depends upon the Shear Modulus of elasticity.

8.5 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.

Fig 8.11

Answer:

Tension in the steel wire is F ₁

$$\begin{gathered} {F}_1=(4+6)\times 9.8 \\ {F}_1=98N \end{gathered}$$

Length of steel wire l ₁ = 1.5 m

The diameter of the steel wire, d = 0.25 cm

$$\begin{gathered} A=\pi {\left(\frac{d}{2}\right)}^2 \\ A=\pi \times {\left(\frac{0.25\times {10}^{-2}}{2}\right)}^2 \\ A=4.9\times {10}^{-6}{m}^2 \end{gathered}$$

Area od the steel wire, $A=4.9\times {10}^{-6}{m}^2$

Let the elongation in the steel wire be $\mathrm{\Delta }{l}_1$

Young's Modulus of steel, Y ₁ = $2\times {10}^{11}N{m}^{-2}$

$$\begin{gathered} {Y}_1=\frac{F_1{l}_1}{\mathrm{\Delta }{l}_{1}A} \\ \mathrm{\Delta }{l}_1=\frac{F_1{l}_1}{Y_{1}A} \\ \mathrm{\Delta }{l}_1=\frac{98\times 1.5}{2\times {10}^{11}\times 4.9\times {10}^{-6}} \\ \mathrm{\Delta }{l}_1=1.5\times {10}^{-4}m \end{gathered}$$

Tension in the Brass wire is F ₂

$$\begin{gathered} {F}_2=(6)\times 9.8 \\ {F}_2=58.8N \end{gathered}$$

Length of Brass wire l ₂ = 1.5 m

Area of the brass wire, $A=4.9\times {10}^{-6}{m}^2$

Let the elongation in the steel wire be $\mathrm{\Delta }{l}_2$

Young's Modulus of steel, Y ₂ = $0.91\times {10}^{11}N{m}^{-2}$

$$\begin{gathered} \mathrm{\Delta }{l}_2=\frac{F_2{l}_2}{Y_{2}A} \\ \mathrm{\Delta }{l}_2=\frac{58.8\times 1}{0.91\times {10}^{11}\times 4.9\times {10}^{-6}} \\ \mathrm{\Delta }{l}_2=1.32\times {10}^{-4}m \end{gathered}$$

8.6 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?

Answer:

Edge of the aluminium cube, l = 10 cm = 0.1 m

Area of the face of the Aluminium cube, A = l ² = 0.01 m ²

Tangential Force is F

$$\begin{gathered} F=100\times 9.8 \\ F=980N \end{gathered}$$

Tangential Stress is F/A

$$\begin{gathered} \frac{F}{A}=\frac{980}{0.01} \\ \frac{F}{A}=98000N \end{gathered}$$

Shear modulus of aluminium $\eta =2.5\times {10}^{10}N{m}^{-2}$

$$\begin{gathered} TangentialStrain=\frac{\frac{F}{A}}{\eta } \\ =\frac{98000}{2.5\times {10}^{10}} \\ =3.92\times {10}^{-6} \end{gathered}$$

Let the Vertical deflection be $\mathrm{\Delta }l$

$$\begin{gathered} \mathrm{\Delta }l=TangentialStress\times Side \\ \mathrm{\Delta }l=3.92\times {10}^{-6}\times 0.1 \\ \mathrm{\Delta }l=3.92\times {10}^{-7}m \end{gathered}$$

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

Answer:

Inner radii of each column, r ₁ = 30 cm = 0.3 m

Outer radii of each column, r ₂ = 60 cm = 0.6 m

Mass of the structure, m = 50000 kg

Stress on each column is P

$$\begin{gathered} P=\frac{50000\times 9.8}{4\times \pi \times ({0.6}^2-{0.3}^2)} \\ P=1.444\times {10}^{5}N{m}^{-2} \end{gathered}$$

Youngs Modulus of steel is $Y=2\times {10}^{11}N{m}^{-2}$

$$\begin{gathered} Compressionalstrain=\frac{P}{Y} \\ =\frac{1.444\times {10}^5}{2\times {10}^{11}} \\ =7.22\times {10}^{-7} \end{gathered}$$

8.8 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?

Answer:

Length of the copper piece, l = 19.1 mm

The breadth of the copper piece, b = 15.2 mm

Force acting, F = 44500 N

Modulus of Elasticity of copper, $\eta =42\times {10}^{9}N{m}^{-2}$

$$\begin{gathered} Strain=\frac{F}{A\eta } \\ =\frac{F}{lb\eta } \\ =\frac{44500}{15.2\times {10}^{-3}\times 19.1\times {10}^{-3}\times 42\times {10}^9} \\ =3.65\times {10}^{-3} \end{gathered}$$

8.9 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 ${10}^{8}N{m}^{-2}$ , what is the maximum load the cable can support?

Answer:

Let the maximum Load the Cable Can support be T

Maximum Stress Allowed, P = 10 ⁸ N m ^-2

Radius of Cable, r = 1.5 cm

$$\begin{gathered} T=P\pi r^2 \\ T={10}^8\times \pi \times (1.5\times {10}^{-2}{)}^2 \\ T=7.068\times {10}^{4}N \end{gathered}$$

8.10 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 ratios of their diameters if each is to have the same tension.

Answer:

Each wire must support the same load and are of the same length and therefore should undergo the same extension. This, in turn, means they should undergo the same strain.

$$\begin{gathered} Y=\frac{Fl}{\mathrm{\Delta }lA} \\ Y=\frac{4Fl}{\mathrm{\Delta }l\pi d^2} \\ Y{d}^2=k \end{gathered}$$ As F, l and $\mathrm{\Delta }l$ are equal for all wires

$Y_{iron}=1.9\times {10}^{11}N{m}^{-2}$

$Y_{copper}=1.1\times {10}^{11}N{m}^{-2}$

$$\begin{gathered} \frac{d_{iron}}{d_{copper}}=\sqrt{\frac{Y_{copper}}{Y_{iron}}} \\ \frac{d_{iron}}{d_{copper}}=\sqrt{\frac{1.1\times {10}^{11}}{1.9\times {10}^{11}}} \\ \frac{d_{iron}}{d_{copper}}=0.761 \end{gathered}$$

8.11 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 at the bottom of the circle. The cross-sectional area of the wire is $0.065c{m}^2$ . Calculate the elongation of the wire when the mass is at the lowest point of its path.

Answer:

Mass of the body = 14.5 kg

Angular velocity, $\omega$ = 2 rev/s

$\omega =4\pi rad/s$

The radius of the circle, r = 1.0 m

Tension in the wire when the body is at the lowest point is T

$$\begin{gathered} T=mg+m{\omega }^{2}r \\ T=14.5\times 9.8+14.5\times (4\pi {)}^2 \\ T=2431.84N \end{gathered}$$

Cross-Sectional Area of wire, A = 0.065 cm ²

Young's Modulus of steel, $Y=2\times {10}^{11}N{m}^{-2}$

$$\begin{gathered} \mathrm{\Delta }l=\frac{Fl}{AY} \\ \mathrm{\Delta }l=\frac{2431.84\times 1}{0.065\times {10}^{-4}\times 2\times {10}^{11}} \\ \mathrm{\Delta }l=1.87mm \end{gathered}$$

8.12 Compute the bulk modulus of water from the following data: Initial volume = 100.0 litre, Pressure increase = 100.0 atm ( $1atm=1.013\times {10}^{5}Pa$ ), 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.

Answer:

Pressure Increase, P = 100.0 atm

Initial Volume = 100.0 l

Final volume = 100.5 l

Change in Volume = 0.5 l

Let the Bulk Modulus of water be B

$$\begin{gathered} B=\frac{Stress}{VolumetricStrain} \\ B=\frac{P}{\frac{\mathrm{\Delta }V}{V}} \\ B=\frac{100\times 1.013\times {10}^5\times 100\times {10}^{-3}}{0.5\times {10}^{-3}} \\ B=2.026\times {10}^{9}N{m}^{-2} \end{gathered}$$

The bulk modulus of air is $B_a=1.0\times {10}^{5}N{m}^{-2}$

The Ratio of the Bulk Modulus of water to that of air is

$$\begin{gathered} \frac{B}{B_a}=\frac{2.026\times {10}^9}{1.0\times {10}^5} \\ \frac{B}{B_a}=20260 \end{gathered}$$

This ratio is large as for the same pressure difference the strain will be much larger in the air than that in water.

8.13 What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is $1.03\times 103Kg$ ?

Answer:

Water at the surface is under 1 atm pressure.

At the depth, the pressure is 80 atm.

Change in pressure is $\mathrm{\Delta }P=79atm$

Bulk Modulus of water is $B=2.2\times {10}^{9}N{m}^{-2}$

$$\begin{gathered} B=-\frac{P}{\frac{\mathrm{\Delta }V}{V}} \\ \frac{\mathrm{\Delta }V}{V}=-\frac{P}{B} \\ \frac{\mathrm{\Delta }V}{V}=-\frac{79\times 1.013\times {10}^5}{2.2\times {10}^9} \\ \frac{\mathrm{\Delta }V}{V}=-3.638\times {10}^{-3} \end{gathered}$$

The negative sign signifies that for the same given mass the Volume has decreased

The density of water at the surface $\rho =1.03\times {10}^{3}kg{m}^{-3}$

Let the density at the given depth be ${\rho }^{\prime }$

Let a certain mass occupy V volume at the surface

$\rho =\frac{m}{V}$

${\rho }^{\prime }=\frac{m}{V+\mathrm{\Delta }V}$

Dividing the numerator and denominator of RHS by V we get

$$\begin{gathered} {\rho }^{\prime }=\frac{\frac{m}{V}}{1+\frac{\mathrm{\Delta }V}{V}} \\ {\rho }^{\prime }=\frac{\rho }{1+\frac{\mathrm{\Delta }V}{V}} \\ {\rho }^{\prime }=\frac{1.03\times {10}^3}{1+(-3.638\times {10}^{-3})} \\ {\rho }^{\prime }=1.034\times {10}^{3}kg{m}^{-3} \end{gathered}$$

The density of water at a depth where pressure is 80.0 atm is $1.034\times {10}^{3}kg{m}^{-3}$ .

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

Answer:

Bulk's Modulus of Glass is $B_G=3.7\times {10}^{10}N{m}^{-2}$

Pressure is P = 10 atm.

The fractional change in Volume would be given as

$$\begin{gathered} \frac{\mathrm{\Delta }V}{V}=\frac{P}{B} \\ \frac{\mathrm{\Delta }V}{V}=\frac{10\times 1.013\times {10}^5}{3.7\times {10}^{10}} \\ \frac{\mathrm{\Delta }V}{V}=2.737\times {10}^{-5} \end{gathered}$$

The fractional change in Volume is $2.737\times {10}^{-5}$

8.15 Determine the volume contraction of a solid copper cube, 10 cm on edge, when subjected to a hydraulic pressure of $7.0\times {10}^{6}Pa$

Answer:

The bulk modulus of copper is $B_C=140GPa=1.4\times {10}^{11}N{m}^{-2}$

Edge of copper cube is s = 10 cm = 0.1 m

Volume Of copper cube is V = s ³

V = (0.1) ³

V = 0.001 m ³

Hydraulic Pressure applies is $P=7.0\times {10}^{6}Pa$

From the definition of bulk modulus

$$\begin{gathered} {B}_C=\frac{P}{\frac{\mathrm{\Delta }V}{V}} \\ \frac{\mathrm{\Delta }V}{V}=\frac{P}{B_C} \\ \frac{\mathrm{\Delta }V}{V}=\frac{7\times {10}^6}{1.4\times {10}^{11}}\frac{\mathrm{\Delta }V}{V}=5\times {10}^{-5} \end{gathered}$$

The volumetric strain is $5\times {10}^{-5}$

Volume contraction will be

$$\begin{gathered} \mathrm{\Delta }V=VolumetricStrain\times InitialVolume \\ \mathrm{\Delta }V=5\times {10}^{-5}\times {10}^{-3} \\ \mathrm{\Delta }V=5\times {10}^{-8}{m}^3 \\ \mathrm{\Delta }V=5\times {10}^{-2}c{m}^3 \end{gathered}$$

The volume contraction has such a small value even under high pressure because of the extremely large value of the bulk modulus of copper.

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

Answer:

Change in volume is $\mathrm{\Delta }V=0.10\mathrm{\%}$

$$\begin{gathered} VolumetricStrain=\frac{0.1}{100} \\ \frac{\mathrm{\Delta }V}{V}=0.001 \end{gathered}$$

Bulk modulus of water is $B_w=2.2\times {10}^{9}N{m}^{-2}$

$$\begin{gathered} {B}_w=\frac{P}{\frac{\mathrm{\Delta }V}{V}} \\ P=\frac{\mathrm{\Delta }V}{V}\times B_w \\ P=0.001\times 2.2\times {10}^9 \\ P=2.2\times {10}^{6}Pa \end{gathered}$$

A pressure of $2.2\times {10}^{6}Pa$ is to be applied so that a litre of water compresses by 0.1%.

Note: The answer is independent of the volume of water taken into consideration. It only depends on the percentage change.