NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids

QUESTIONS FROM TEXTBOOK

Question 9. 1. A steel wire of length 4.7 m and cross-sectional area 3.0 x 10-5 m2 stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 x 10-5 m2 under a given load. What is the ratio of the Young’s modulus of steel to that of copper?
Answer:

Class 11 Physics Mechanical Properties of Solids Notes

Question 9. 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:

Class 11 Physics Mechanical Properties of Solids Notes

(a)  Young’s modulus of the material (Y) is given by
Y =Stress/Strain
=150 x 106/0.002
150 x 106/2 x 10-3
=75 x 109 Nm-2
=75 x 1010 Nm-2
(a)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 .

Class 11 Physics Mechanical Properties of Solids Notes

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 9. 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.

Class 11 Physics Mechanical Properties of Solids Notes
Class 11 Physics Mechanical Properties of Solids Notes

Question 9. 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

Class 11 Physics Mechanical Properties of Solids Notes

Question 9. 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. Young’s modulus, Y = 2.0 x 1011 Pa.
Answer: 

Class 11 Physics Mechanical Properties of Solids Notes
Class 11 Physics Mechanical Properties of Solids Notes
Class 11 Physics Mechanical Properties of Solids Notes
Class 11 Physics Mechanical Properties of Solids Notes
Class 11 Physics Mechanical Properties of Solids Notes

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.

Class 11 Physics Mechanical Properties of Solids Notes
Class 11 Physics Mechanical Properties of Solids Notes
Class 11 Physics Mechanical Properties of Solids Notes
Class 11 Physics Mechanical Properties of Solids Notes
Class 11 Physics Mechanical Properties of Solids Notes
Class 11 Physics Mechanical Properties of Solids Notes
Class 11 Physics Mechanical Properties of Solids Notes
Class 11 Physics Mechanical Properties of Solids Notes

Question 9. 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

Class 11 Physics Mechanical Properties of Solids Notes
Class 11 Physics Mechanical Properties of Solids Notes

Question 9. 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 = 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.

Question 9. 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 x 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?
Answer: 

mechanical properties of solids
mechanical properties of solids

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