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Chapter 9 Mechanical Properties Of Solids Solutions

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 - 11 : - Here, m = 14.5 kg; l = r = 1 m; v = 2 rps; A = 0.065 x 10-4 m2 Totalpulling 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 - 12 : -


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 - 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 x 103 kg m-3?

Answer - 13 : -


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

Answer - 14 : -


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 - 15 : - Here a side of copper cube a = 10 cm, hence volume V = a3 =10-3 m3 , hydraulic pressure applied p = 7.0 x 106 Paand from table we find that bulk modulus of copper B = 140 G Pa = 140 x 109 Pa.

Question - 16 : - How much should be pressure the a litre of water be changed to compress it by 0.10 %? Bulk modulus of elasticity of water = 2.2 x 109 Nm-2.

Answer - 16 : -


Question - 17 : - Anvils made of single crystals of diamond, with the shape as shown in figure 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 of50,000 N. What is the pressure at the tip of the anvil?

Answer - 17 : -


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 mm_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 - 18 : -

For steel wire A, l1=l; Az = 1 mm2;Y1= 2 x 1011 Nm-2
For 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 wireA is connected. Let F1 and F2 be the tensionsin 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 endwhere wire A is connected. Let F1 and F2 be thetension 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. Amass of 100g is suspended from the mid-point of the wire. Calculate the depression at the mid-point.

Answer - 19 : - Let AB be a mild steel wire of length 2L = lm and its cross-section areaA = 0.50 x 10-2 cm2. A mass m = 100 g = 0.1 kg issuspended 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 - 20 : -

Diameter = 6mm;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.

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