Class 11th Chapter – 9 Mechanical Properties of Solids |Physics| NCERT Solution | Edugrown

NCERT Exercises

Question 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:
Here, for steel wire,
Length of wire, l₁= 4.7 m
Area of cross-section, A₁ = 3.0 x 10^-5 m²
Stretching, Δl₁= Δl(say)
Stretching force on steel, F₁ = F
For copper wire, length of wire, l₂ = 3.5 m
Area of cross-section, A₂ = 4.0 x 10^_5 m²
Stretching, Δl₂ = Δl (given);
Stretching force on copper, F₂ = F
Let Y₁ and Y₂ be the Young’s modulus of steel and copper wire respectively

Question 2.
Figure shows the stress-strain curve for a given material. What are
(a) Young’s modulus and
(b) approximate yield strength for this material?

Answer:
(a) From graph,
for stress = 150 x 10⁶ Nm², the corresponding strain = 0.002
Young’s modulus, = 7.5 x 10¹⁰ Nm^-2
(b) Approximate yield strength will be equal to the maximum stress it can sustain without crossing the elastic limit .Therefore, the approximate yield strength = 300 x 10⁶ Nm^-2 = 3 x 10^s Nm^-2.
Question 3.
The stress-strain graphs for materials A and B are shown in figure.

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) Material A has greater value of Y. It is because, for producing same strain, more stress is required in case of material A.
(b) Material A is the stronger of the two materials. It is because, it can bear greater stress before the wire of this material breaks.

Question 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.
Answer:
(a) False. This is because if steel and rubber wires of same length and area of cross-section are subjected to same deforming force, then the extension produced in steel is less than the extension produced in rubber. So Y_s> Y_r. In other words, for producing same strain in steel and rubber, more stress is required in case of steel.

(b) True. The reason is that when a coil spring is stretched, there is neither a change in the length of the coil (i.e., length of the wire forming the coil spring) nor a change in its volume. Since the change takes place in the shape of the coil spring, its stretching is determined by its shear modulus

Question 5.
Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in figure. 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.

Answer:

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:

Question 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. The Young’s modulus of steel is
2.0 x 10¹¹ Pa.
Answer:
Here, γ= 2.0 x 10¹¹ Pa,
Inner radius of each column, r₁ = 30 cm = 0.3 m;
Outer radius of each column, r₂ = 60 cm = 0.6 m
Therefore, area of cross-section of the each column,



Question 8.
A piece of copper having a rectangular cross­section of 15.2 mm x 19.1 mm is pulled in tension with 44, 500 N, force producing only elastic deformation. Calculate the resulting strain. Shear modulus of elasticity of copper is 42 x 10⁹ N m^-2.
Answer:
Here, A = 15.2 x 19.2 x 10^-6 m²;
F = 44,500 N; η = 42 x 10⁹ N m^-2

Question 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 10⁸ N nr², what is the maximum load, the cable can support?
Answer:
Here, maximum stress = 10⁸ Nm^-2;
radius of the cable, r = 1.5 cm = 0.015 m.
Therefore, area of cross-section of the cable,
A = Πr² = n x (0.0152)² = 2.25 x 10^-4 m².
The maximum load, the cable can support,
F = maximum stress x area of cross-section
= 10⁸ x 2.25 x 10^-4Π = 7.07 x 10⁴ N

Question 10.
A rigid bar of mass 15 kg is supported symmetrically by three wires each 2.0 m long. These at each end are of copper and the middle one is of iron. Determine the ratio of their diameters if each is to have the same tension.
[Young’s modulus of elasticity for copper and steel are 110 x 10⁹ Nm² and 190 x 10⁹ N m^-2 respectively.]
Answer:
As each wire has same tension F, so each wire has same extension due to mass of rigid bar. As each wire is of same length, hence each wire has same strain. If D is the diameter of wire, then

Question 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 cm². Calculate the elongation of the wire when the mass is at the lowest point of its path.
Answer:
Here, in = 14.5 kg;
A = 0.065 cm² = 0.065 x 10^-4 m²; L = 1 m and ν = 2 r.p.s.
When the mass is at the lowest point of its circular path, the stretching force on the wire,

Question 12.
Compute the bulk modulus of water from the following data : Initial volume = 100.0 liter. Pressure increase = 100.0 atm (1 atm = 1.013 x 10⁵ Pa), Final volume = 100.5 liter. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large.
Answer:
P = 100 atmosphere = 100 x 1.013 x 10⁵ Pa
Initial volume, V₁ = 100 liter = 100 x 10^-3 m³
Final volume, V₂ = 100.5 liter = 100.5 x 10^-3  m³

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 inter molecular distances in case of liquids are very small as compared to the  corresponding distances in the case of gases. Hence there are larger inter atomic 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? Compressibility of water is 45.8 x 10^-11 Pa^-1.
Answer:

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

∴ Fractional change in volume = Δv/v = 2.74 x 10^-5

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, L = 10 cm = 0.1 m
B = bulk modulus of Cu = 140 x 10⁹ Pa
P = 7 x 10⁶ Pa
ΔV = Volume contraction of solid copper cube = ?

Question 16.
How much should the pressure on a liter of water be changed to compress it by 0.10%.
Answer:

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 of 50,000 N. What is the pressure at the tip of the anvil?

Answer:

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 wire A and B are 1.0 mm² and 2.0 mm², 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; A₁ = 1 mm²;
Y₁ = 2 x 10¹¹ N nr²
For aluminium wire B, l₂ = l; A_z = 2 mm²;
Y₂ = 7 x 10¹⁰ N nr²

(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 100 g is suspended from the mid-point of the wire. Calculate the depression at the mid-point.
Answer:


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 rivets is not to exceed 6.9 x 10⁷ Pa? Assume that each rivet is to carry one quarter of the load.
Answer:
When the riveted strip is subjected to a stretching load W, the tensile force (i.e. tension) in each strip (equal to W) provides the shearing force on the four rivets. Since the load is sheared uniformly, i.e. each rivet is under a shearing force equal to W/4
Maximum shearing stress on each rivet = 6.9 x 10⁷ Pa.
Let A = area of each rivet on which the shearing force acts

Question 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:
Here, P = 1.1 x 10⁸ Pa; V = 0.32 m³;
Bulk modulus for steel B = 1.6 x 10¹¹ Pa

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