Chapter – 8
Appendices
Q1. 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 Young’s modulus of steel to that of copper?
Answer:
Length of the steel wire, l1 = 4.7 m Cross-sectional area of the steel wire, a1 = 3.0 × 10–5 m 2 Length of the copper wire, l2 = 3.5 m Cross-section area of the copper wire, a2 = 4.0 × 10–5 m 2 Change in length = Δl1 = Δl2 = Δl Force applied in both cases = FYoung’s modulus of the steel wire ——(1)Young’s modulus of the copper wire ——–(2)Dividing (1) by (2), we get
Question 2. Figure shows the strain-stress curve for a given material. What are
(a) Young’s modulus and
(b) approximate yield strength for this material?
Answer:
(a) Young’s modulus of the material (Y) is given byY =Stress/Strain=150 x 106/0.002150 x 106/2 x 10-3=75 x 109 Nm-2=75 x 1010 Nm-2
(b)Yield strength of a material is defined as the maximum stress it can sustain. From graph, the approximate yield strength of the given material= 300 x 106 Nm-2= 3 x 108 Nm-2 .
Question 3. The stress-strain graphs for materials A and B are shown in figur. 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?
Answer:
(a) From the two graphs we note that for a given strain, stress for A is more than that of B. Hence Young’s modulus =(Stress /Strain) is greater for A than that of B.
b) Strength of a material is determined by the amount of stress required to cause fracture. This stress corresponds to the point of fracture. The stress corresponding to the point of fracture in A is more than for B. So, material A is stronger than material B.
Question 4. Read the ‘allowing 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.
Answer: (a) False. The-Young’s modulus is defined as the ratio of stress to the strain within elastic limit. For a given stretching force elongation is more in rubber and quite less in steel. Hence, rubber is less elastic than steel.
(b) True. Stretching of a coil is determined by its shear modulus. When equal and opposite forces are applied at opposite ends of a coil, the distance, as well as shape of helicals of the coil change and it, involves shear modulus.
Question 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:
Here, side of cube, L = 10 cm =10/100= 0.1 m.•. Area of each face, A = (0.1)2 = 0.01 m2
Question 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:
Since each wire is to have same tension therefore, each wire has same extension. Moreover, each wire has the same initial length.So, strain is same for each wire
Question 11. A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1 m, is whirled in a vertical circle with an angular velocity of 2 rero./s at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm2. Calculate the elongation of the zvire when the mass is at the lowest point of its path. Ysteel = 2 x 1011 Nm-2.
Answer .
Here, m = 14.5 kg; l = r = 1 m; v = 2 rps; A = 0.065 x 10-4 m2 Total pulling force on mass, when it is at the lowest position of the vertical circle is F = mg + mr w2 = mg + mr 4,π2 v2
Question 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 x 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.
Answer:
The ratio is too large. This is due to the fact that the strain for air is much larger than for water at the same temperature. In other words, the intermolecular distances in case of liquids are very small as compared to the corresponding distances in the case of gases. Hence there are larger interatomic forces in liquids than in gases.
Question 15. Determine the volume contraction of a solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of 7.0 x 106 Pa.
Answer:
Here a side of copper cube a = 10 cm, hence volume V = a3 = 10-3 m3 , hydraulic pressure applied p = 7.0 x 106 Pa and from table we find that bulk modulus of copper B = 140 G Pa = 140 x 109 Pa.
Question 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 figure. The cross-sectional areas of wires A and B are 1.0 mm2 and 2.0 m m 2, 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.
Answer:
For steel wire A, l1=l; Az = 1 mm2; Y1= 2 x 1011 Nm-2For aluminium wire B, l2 = l; A2 = 2mm2; Y2 = 7 x 1010 Nm-2
(a) Let mass m be suspended from the rod at distance x from the end where wire A is connected. Let F1 and F2 be the tensions in two wires and there is equal stress in two wires, then
(b) Let mass m be suspended from the rod at distance x from the end where wire A is connected. Let F1 and F2 be the tension in the wires and there is equal strain in the two wires i.e.,
Question 19. A mild steel wire of length 1.0 m and cross-sectional area 0.50 x 10-2 cm2 is stretched, well within its elastic limit, horizontally between two pillars. A mass of 100g is suspended from the mid-point of the wire. Calculate the depression at the mid-point.
Answer:
Let AB be a mild steel wire of length 2L = lm and its cross-section area A = 0.50 x 10-2 cm2. A mass m = 100 g = 0.1 kg is suspended at mid-point C of wire as shown in figure. Let x be the depression at mid-point i.e., CD = x
Question 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 x 107 Pa? Assume that each rivet is to carry one-quarter of the load.
Answer:
Diameter = 6 mm; Radius, r = 3 x 10-3 m;Maximum stress = 6.9 x 107 Pa Maximum load on a rivet= Maximum stress x cross-sectional area= 6.9 x 107 x 22/7 (3 x 10-3)2 N = 1952 N Maximum tension = 4 x 1951.7 N = 7.8 x 103 N.