Ncert Solution CBSE Class 11 Chapter 8 MECHANICAL PROPERTIES OF SOLIDS - Param Himalaya
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?
Solution :
$$Y= \frac{F/A}{\Delta L/L}$$
Or
$$Y=\frac{FL}{\Delta L A}$$
Here F (Load) and $\Delta L$ (elongation) are the same in the two cases :
For steel :
$$Y_{s}=\frac{F \times4.7}{3.0 \times 10^{-5}\times \Delta L}$$
For copper :
$$Y_{c}=\frac{F \times3.5}{4.0 \times 10^{-5}\times \Delta L}$$
$$\therefore \frac{Y_{s}}{Y_{C}} = \frac{4.7\times 4.0 \times 10^{-5}}{3.0\times 3.0 \times 10^{-5}} = 1.79$$
8.2 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:
(a) From the stress-strain curve in Fig. 8.9 , it is clear that corresponding to stress = $150\times 10^{6} Nm^{-2}$ , the strain is 0.002
$$\therefore Y = \frac{Stress}{Stain} = \frac{150\times 10^{6}}{0.002}$$
$$Y= 7.5 \times 10^{10}Nm^{-2}$$
(b) The approximate yield strength of a material is the maximum stress it can bear without crossing the elastic limit. It is clear from figure 9.01 that approximate yield strength= $300\times 10^{6}Nm^{-2}$
8.3 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:
(a) Young's modulus , y = stress/strain. it is clear from the figure 8.10 that for a given strain, stress for material A is more than that of material B. therefore young's modulus for A is greater than that of B.
(b) The strength of a material is measured by the amount of stress it can be bear fracture. it is clear from the figure 8.10 that material A is stronger than material B.
8.4 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.
Solution:
(a) False , young's modulus , Y = stress/strain . For a given stress, strain produced in a rubber is more than that in steel. Therefore Y of rubber is less than that of Steel and the statement is false.
(b) When a coil spring is stressed, neither its length nor its volume changes. There is only change in its shape. Therefore, stretching of a coil spring is determined by its share modulus.
8.5 Two wires of diameter 0.25 cm, one made of steel and the other made of brass are 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. Given Young's modulus for steel = $2.0 \times 10^{11}$ Pa and that for brass = $0.91 \times 10^{11}$Pa.
Solution :
Steel wire :
Y = $2.0 \times 10^{11}Pa$ ; L = 1.5m ; F = 9.8 (4+6) = 98N;
$$A=\frac{\pi }{4}d^{2} = \frac{\pi }{4}\times(0.25\times10^{-2})^{2} m^{2}$$
Now
$$Y=\frac{FL}{A\Delta L}$$
$$\therefore \Delta L =\frac{FL}{AY}$$
$$\Delta L = \frac{98 \times 1.5 }{\frac{\pi}{4} \times(0.25\times10^{-2})^{2} \times 2.0 \times 10^{11}}$$
$$\Delta L = 1.5\times10^{-4}m$$
Brass wire :
Y = $0.91 \times 10^{11}Pa$ ; L = 1.0m ; F = 6×9.8;
$$A=\frac{\pi }{4}d^{2} = \frac{\pi }{4}\times(0.25\times10^{-2})^{2} m^{2}$$
Now
$$Y=\frac{FL}{A\Delta L}$$
$$\therefore \Delta L =\frac{FL}{AY}$$
$$\Delta L = \frac{6\times 9.8 \times 1.0 }{\frac{\pi}{4} \times(0.25\times10^{-2})^{2} \times 0.91 \times 10^{11}}$$
$$\Delta L = 1.3\times10^{-4}m$$
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?
Solution:
Shear modulus,
$\eta = \frac{\frac{F}{A}}{\frac{\Delta x}{L}}$
$\eta = \frac{FL}{A\Delta x}$
$\Delta x = \frac{FL}{A\eta}$
Here , F = 100×9.8 N
L = 10 cm = 0.1 m
$A = L^{2}=(0.1)^{2}m^{2}$
$\eta = 25GPa=25×10^{9}Pa$
$\Delta x= \frac{100 \times 9.8 \times 0.1}{(0.1)^{2} \times 25 \times 10^{9}}$
$\Delta x= 3.92 \times 10^{-7}m$
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.The Young's Modulus of steel is $2.0\times 10^{11}Pa$
Solution :
r=30 cm=0.3m ;
R=60cm=0.6m;
$Y=2.0\times 10^{11}N/m^{2}$
Area of X-section of each column is
$$A=\pi(R^{2}-r^{2})$$
$$A=\pi[(0.6)^{2}-(0.3)^{2}]$$
$$A=0.27\pi m^{2}$$
The compressional force on each column is given by :
$$F= \frac{mg}{4}N$$
$$F= \frac{50000\times9.8}{4}N$$
$$F=12500\times9.8N$$
Now, Young's modulus, $Y = \frac{Stress}{Strain}$
$\therefore$ Compressional strain on each column
$$Strain=\frac{Stress}{Y}$$
$$= \frac{F/A}{Y}= \frac{F}{AY}$$
$$=\frac{12500\times9.8}{0.27\pi\times2\times10^{11}}$$
$$Strain=7.22 \times 10^{-7}m$$
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.Given the shear modulus for copper is $42 \times 10^{9} Nm^{-2}$.
Solution:
Given : $$L = 19.1 mm = 19.1\times 10^{-3}m$$
$$B= 15.2mm = 15.2 \times 10^{-3}m$$
Shear modulus, $$\eta = \frac{FL}{A\Delta L}$$
Here $\frac{\Delta L}{L}$ is the shear strain.
Now , $$\eta = 42 \times 10^{9} Nm^{-2}$$
$$F=44500N$$
$$A=L.B=15.2 \times 19.1 \times 10^{-6}m^{2}$$
$$Shear strain =\frac{\Delta L}{L} = \frac{F}{A\eta}$$
$$= \frac{44500}{(15.2 \times 19.1 \times 10^{-6}m^{2})\times(42\times10^{9})}$$
$$Strain=3.65\times10^{-3}$$
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 $10^{8}Nm^{–2}$, what is the maximum load the cable can support ?
Solution:
The maximum load $(F_{max})$ the cable can support is
$F_{max}$=Maximum stress ×Area of X-Section of cable
$F_{max}= (10^{8}) \times \pi \times (1.5\times10^{-2})^{2} =7.07 \times10^{4}N$
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.Young's modulus for copper and iron are $11\times 10^{10} Nm^{-2}$ and $19 \times 10^{10} Nm^{-2}$ respectively.
Solution:
As the bar is supported symmetrically by the three wires , therefore, extension in each wire must be the same .
Young's modulus ,
$Y = \frac{FL}{A\Delta L}$
As per the condition of the problem , F (tension), L(=2.0m) and $\Delta L$ (extension) for each wire are the same.
$Y \propto \frac{1}{A}$ Or $Y \propto \frac{1}{d^{2}}$
$$\frac{d_{copper}}{d_{iron}} = \sqrt\frac{Y_{iron}}{Y_{copper}}$$
$$\frac{d_{copper}}{d_{iron}} = \sqrt\frac{19\times10^{10}}{11\times10^{10}}$$
$$\frac{d_{copper}}{d_{iron}} = 1.314$$
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 $0.065 cm^{2}$. Calculate the elongation of the wire when the mass is at the lowest point of its path.Given that Young's modulus of steel is $2.0 \times 10^{11}Nm^{-2}$.
Solution:
Here m = 14.5 kg ; A = 0.065 \times 10^{-4}m^{2}; f = 2 r.p.s ; L = 1 m (=r)
When the mass is at the lowest point of its circular path , the streching force is
$F= mg + mr\omega^{2} = 14.5 \times 9.8 + 14.5 \times 1 (2\pi\times 2)^{2}$
$F=2431.85N$
$\therefore Elongation, \Delta L = \frac{FL}{AY}$
$\Delta L = \frac{2431.85 \times 1}{(0.065\times10^{-4})\times (2.0 \times 10^{11})}$
$=1.87 \times 10^{-3}m = 1.87 mm$
8.12 Compute the bulk modulus of water from the following data: Initial volume = 100.0 litre, Pressure increase = 100.0 atm (1 atm = $1.013 \times 10^{5}Pa$ , 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? Given bulk modulus of air at S.T.P = $1.0 \times 10^{-4}Nm^{-2}$.
Solution:
Bulk modulus of water ,
$K_{water} = V\frac{\Delta p }{\Delta V}$
Here $V = 100.0 litres = 100.0 \times 10^{-3}m^{3}$
$\Delta p = 100.0 atm = 100.0 \times 1.013 \times 10^{5} Pa$
$\Delta V = 100.5 \times 10^{-3} - 100.0 \times 10^{-3} = 0.5 \times 10^{-3}m^{3}$
$K_{water} = 100.0 \times 10^{-3} \times \frac{100.0 \times 1.013\times 10^{5}}{0.5 \times 10^{-3}}$
$= 2.026 \times 10^{9}Nm^{-2}$
$\frac{K_{water}}{K_{air}} = \frac{2.026 \times 10^{9}}{1.0 \times 10^{-4}} = 2.026 \times 10^{13}$
it is observed that the ratio is very large. It is because water is less compressible as compared to air.
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 $1.03 \times 10^{3} Nm^{-2}$?
Solution:
As per the given conditions mass of $1 m^{3}$ of water at the surface of the ocean is 1030 kg the mass of water remains the same let us calculate the change in volume $\Delta V$ of this water ($V=1m^{3}$) when it is at a pressure of 80 atm.
Now , Bulk modulus,
$$K = -V \frac{\Delta p}{\Delta V}$$
Or $$\Delta V = -\frac{(\Delta p)V}{K}$$
$\Delta p$ = change in pressure = 80 atm - 1 atm ( At the surface , pressure is 1 atm)
$=79 atm = 79 \times 10^{5}N/m^{2}$
$V = 1 m^{3}$
$K= 2.0 \times 10^{9} N/m^{2}$
$\Delta V = - \frac{(79\times 10^{5})\times (1)}{2.0\times10^{9}}=-0.0039 m^{3}$
$\therefore$ Density of water at depth where pressure is 80.0 atm
$= \frac{m}{V+\Delta V} = \frac{1030}{1-0.0039} = 1034 \ kg \ m^{-3}$
Note that even under this huge pressure, water is only compressed to 2% to 3%. Hence solids and liquids are not easily compressed.
8.14 Compute the fractional change in volume of a glass slab, when subjected to a hydraulic pressure of 10 atm. Given that Bulk modulus of glass = $3.7 \times 10^{10}Nm^{-2}$ and 1 atm = $1.013 \times 10^{5}Nm^{-2}$
Solution:
$\Delta p = 10 atm = 10 \times1.013 \times 10^{5}Nm^{-2}$
$K = 3.7 \times 10^{10}Nm^{-2}$
Now , Bulk modulus, $K = V\frac{\Delta p}{\Delta V}$
$\therefore $ Fractional change in the volume of glass is
$$\frac{\Delta V}{V} = \frac{\Delta p}{K}$$
$$\frac{\Delta V}{V} = \frac{10 \times 1.013 \times 10^{5}}{3.7 \times 10^{10}} = 2.74 \times 10^{-5}$$
8.15 Determine the volume contraction of a solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of $7.0 \times 10^{6}$Pa. Bulk modulus of copper = 140GPa.
Solution:
Here ,
$V = 10cm \times 10cm \times 10cm = 1000 cm^{3} = 10^{-3}m^{3}$
$P= 7 \times 10^{6}Pa$
$K= 140 \times 10^{9} Pa$
Now , $K = V\frac{P}{\Delta V}$
$140 \times 10^{9} = 10^{-3} \times \frac{7 \times 10^{6}}{\Delta V}$
$\therefore \Delta V = 5 \times 10^{-8}m^{3} = 50mm^{3}$
8.16 How much should the pressure on a litre of water be changed to compress it by 0.10%? Bulk modulus of water = $2.2 \times 10^{9}Nm^{-2}$.
Solution:
Here , $V = 1 litre = 10^{-3}m^{3}$
$\frac{\Delta V}{V}$ = 0.10% = 0.1/100 =$ 10^{-3}$
Now $K = V\frac{P}{\Delta V}$
$P= K \frac{\Delta V}{V} $
$= 2.2 \times 10^{9} \times 10^{-3} = 2.2 \times 10^{6} Pa$