{"id":200718,"date":"2021-03-03T08:25:50","date_gmt":"2021-03-03T08:25:50","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=200718"},"modified":"2023-09-20T08:16:43","modified_gmt":"2023-09-20T08:16:43","slug":"ncert-solutions-for-11th-class-physics-chapter-9-mechanical-properties-of-solids","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-11th-class-physics-chapter-9-mechanical-properties-of-solids\/","title":{"rendered":"NCERT Solutions for 11th Class Physics: Chapter 9-Mechanical properties of Solids"},"content":{"rendered":"\n

Class 11: Physics Chapter 9 solutions. Complete Class 11 Physics Chapter 9 Notes.<\/p>\n\n\n\n

NCERT Solutions for 11th Class Physics: Chapter 9-Mechanical properties of Solids<\/strong><\/h2>\n\n\n\n

NCERT 11th Physics Chapter 9, class 11 Physics chapter 9 solutions<\/p>\n\n\n\n

Page No: 242<\/p>\n\n\n\n

Excercises<\/h4>\n\n\n\n

9.1. A steel wire of length 4.7 m and cross-sectional area 3.0 \u00d7 10\u20135<\/sup> m2<\/sup> stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 \u00d7 10\u20135<\/sup> m2<\/sup> under a given load. What is the ratio of the Young\u2019s modulus of steel to that of copper?<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Length of the steel wire, L<\/em>1<\/sub> = 4.7 m
Area of cross-section of the steel wire, A<\/em>1<\/sub> = 3.0 \u00d7 10\u20135<\/sup> m2<\/sup>
Length of the copper wire, L<\/em>2<\/sub> = 3.5 m
Area of cross-section of the copper wire, A<\/em>2<\/sub> = 4.0 \u00d7 10\u20135<\/sup> m2<\/sup>
Change in length = \u0394L<\/em>1<\/sub> = \u0394L<\/em>2<\/sub> = \u0394L<\/em>
Force applied in both the cases = F<\/em>
Young\u2019s modulus of the steel wire:
Y<\/em>1<\/sub> = (F<\/em>1<\/sub> \/ A<\/em>1<\/sub>) (L<\/em>1<\/sub> \/ \u0394L1<\/sub>)<\/em>
= (F<\/em> \/ 3 X 10-5<\/sup>) (4.7 \/ \u0394L)<\/em>     ….(i)
Young\u2019s modulus of the copper wire:
Y<\/em>2<\/sub> = (F<\/em>2<\/sub> \/ A<\/em>2<\/sub>) (L<\/em>2<\/sub> \/ \u0394L<\/em>2<\/sub>)
= (F<\/em> \/ 4 \u00d7 10-5<\/sup>) (3.5 \/ \u0394L)<\/em>     ….(ii)
Dividing (i<\/em>) by (ii<\/em>), we get:
Y<\/em>1<\/sub> \/ Y<\/em>2<\/sub>  =  (4.7 \u00d7 4 \u00d7 10-5<\/sup>) \/ (3 \u00d7 10-5<\/sup> \u00d7 3.5)
= 1.79 : 1
The ratio of Young\u2019s modulus of steel to that of copper is 1.79 : 1.<\/p>\n\n\n\n

9.2. Figure 9.11 shows the strain-stress curve for a given material. What are (a) Young\u2019s modulus and (b) approximate yield strength for this material?<\/strong><\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(a) It is clear from the given graph that for stress 150 \u00d7 106<\/sup> N\/m2<\/sup>, strain is 0.002.
\u2234Young\u2019s modulus, Y<\/em> = Stress \/ Strain
= 150 \u00d7 106<\/sup> \/ 0.002  =  7.5 \u00d7 1010<\/sup> Nm-2<\/sup>
Hence, Young\u2019s modulus for the given material is 7.5 \u00d71010<\/sup> N\/m2<\/sup>.<\/p>\n\n\n\n

(b) The yield strength of a material is the maximum stress that the material can sustain without crossing the elastic limit.
It is clear from the given graph that the approximate yield strength of this material is 300 \u00d7 106<\/sup> Nm\/2<\/sup> or 3 \u00d7 108<\/sup> N\/m2<\/sup>.<\/p>\n\n\n\n

Page No: 243<\/p>\n\n\n\n

9.3. The stress-strain graphs for materials A and B are shown in Fig. 9.12.<\/strong><\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

The graphs are drawn to the same scale.
(a) Which of the materials has the greater Young\u2019s modulus?
(b) Which of the two is the stronger material?<\/p>\n\n\n\n

(a) From the two graphs we note that for a given strain, stress for A<\/strong> is more than that of B<\/strong>. Hence, Young’s modulus (=stress\/strain) is greater for A<\/strong> than that of B<\/strong>.<\/p>\n\n\n\n

(b) A is stronger than B<\/strong>. Strength of a material is measured by the amount of stress required to cause fracture, corresponding to the point of fracture.<\/p>\n\n\n\n

NCERT 11th Physics Chapter 9, class 11 Physics chapter 9 solutions<\/p>\n\n\n\n

9.4. Read the following two statements below carefully and state, with reasons, if it is true or false.(a) The Young\u2019s modulus of rubber is greater than that of steel;
(b) The stretching of a coil is determined by its shear modulus.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(a) False, because for given stress there is more strain in rubber than steel and modulus of elasticity is inversely proportional to strain.<\/p>\n\n\n\n

(b) True, because the stretching of coil simply changes its shape without any change in the length of the wire used in the coil due to which shear modulus of elsticity is involved.<\/p>\n\n\n\n

9.5. Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in Fig. 9.13. 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.<\/strong><\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Elongation of the steel wire = 1.49 \u00d7 10\u20134<\/sup> m
Elongation of the brass wire = 1.3 \u00d7 10\u20134<\/sup> m
Diameter of the wires, d<\/em> = 0.25 m
Hence, the radius of the wires, r = d\/2  =  0.125 cm
Length of the steel wire, L<\/em>1<\/sub> = 1.5 m
Length of the brass wire, L<\/em>2<\/sub> = 1.0 m
Total force exerted on the steel wire:
F<\/em>1<\/sub> = (4 + 6) g = 10 \u00d7 9.8 = 98 N
Young\u2019s modulus for steel:
Y1<\/sub> = (F1<\/sub>\/A1<\/sub>) \/ (\u0394L<\/em>1<\/sub> \/ L<\/em>1<\/sub>)
Where,
\u0394L<\/em>1<\/sub> = Change in the length of the steel wire
A<\/em>1<\/sub> = Area of cross-section of the steel wire = \u03c0r<\/em>1<\/sub>2<\/sup>
Young\u2019s modulus of steel, Y<\/em>1<\/sub> = 2.0 \u00d7 1011<\/sup> Pa
\u2234 \u0394L<\/em>1<\/sub> = F<\/em>1<\/sub>  \u00d7 L<\/em>1<\/sub> \/ (A<\/em>1<\/sub>  \u00d7 Y<\/em>1<\/sub>)
= (98  \u00d7 1.5) \/ [ \u03c0(0.125  \u00d7 10-2<\/sup>)2<\/sup>  \u00d7 2  \u00d7 1011<\/sup>]   =   1.49  \u00d7 10-4<\/sup> m<\/p>\n\n\n\n

Total force on the brass wire:
F<\/em>2<\/sub> = 6 \u00d7 9.8 = 58.8 N
Young\u2019s modulus for brass:
Y<\/em>2<\/sub> = (F<\/em>2<\/sub>\/A<\/em>2<\/sub>) \/ (\u0394L<\/em>2<\/sub> \/ L<\/em>2<\/sub>)
Where,
\u0394L<\/em>2<\/sub> = Change in the length of the brass wire
A<\/em>1<\/sub> = Area of cross-section of the brass wire = \u03c0r1<\/sub>2<\/sup>
\u2234 \u0394L<\/em>2<\/sub> = F<\/em>2<\/sub>  \u00d7 L<\/em>2<\/sub> \/ (A<\/em>2<\/sub>  \u00d7 Y<\/em>2<\/sub>)
= (58.8 X 1) \/ [ (\u03c0  \u00d7 (0.125  \u00d7 10-2<\/sup>)2<\/sup>  \u00d7 (0.91  \u00d7 1011<\/sup>) ] = 1.3  \u00d7 10-4 <\/sup>m
Elongation of the steel wire = 1.49 \u00d7 10\u20134<\/sup> m
Elongation of the brass wire = 1.3 \u00d7 10\u20134<\/sup> m.<\/p>\n\n\n\n

NCERT 11th Physics Chapter 9, class 11 Physics chapter 9 solutions<\/p>\n\n\n\n

Page No: 244<\/p>\n\n\n\n

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?<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Edge of the aluminium cube, L<\/em> = 10 cm = 0.1 m
The mass attached to the cube, m<\/em> = 100 kg
Shear modulus (\u03b7) of aluminium = 25 GPa = 25 \u00d7 109<\/sup> Pa
Shear modulus, \u03b7 = Shear stress \/ Shear strain  =  (F\/A) \/ (L<\/em>\/\u0394L)<\/em>
Where,
F<\/em> = Applied force = m<\/em>g = 100 \u00d7 9.8 = 980 N
A<\/em> = Area of one of the faces of the cube = 0.1 \u00d7 0.1 = 0.01 m2<\/sup>
\u0394L<\/em> = Vertical deflection of the cube
\u2234 \u0394L<\/em> = FL<\/em> \/ A<\/em>\u03b7
= 980 \u00d7 0.1 \/ [ 10-2<\/sup> \u00d7 (25 \u00d7 109<\/sup>) ]
= 3.92 \u00d7 10\u20137<\/sup> m
The vertical deflection of this face of the cube is 3.92 \u00d710\u20137<\/sup> m.<\/p>\n\n\n\n

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.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Mass of the big structure, M <\/em>= 50,000 kg
Inner radius of the column, r<\/em> = 30 cm = 0.3 m
Outer radius of the column, R<\/em> = 60 cm = 0.6 m
Young\u2019s modulus of steel, Y<\/em> = 2 \u00d7 1011<\/sup> Pa
Total force exerted, F<\/em> = M<\/em>g = 50000 \u00d7 9.8 N
Stress = Force exerted on a single column = 50000 \u00d7 9.8 \/ 4  =  122500 N
Young\u2019s modulus, Y<\/em> = Stress \/ Strain
Strain = (F<\/em>\/A<\/em>) \/ Y<\/em>
Where,
Area, A<\/em> = \u03c0 (R<\/em>2<\/sup> \u2013 r<\/em>2<\/sup>) = \u03c0 ((0.6)2<\/sup> \u2013 (0.3)2<\/sup>)
Strain = 122500 \/ [ \u03c0 ((0.6)2<\/sup> \u2013 (0.3)2<\/sup>) \u00d7 2 \u00d7 1011<\/sup> ]  =  7.22 \u00d7 10-7<\/sup>
Hence, the compressional strain of each column is 7.22 \u00d7 10\u20137<\/sup>.<\/p>\n\n\n\n

9.8. A piece of copper having a rectangular cross-section of 15.2 mm \u00d7 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain?<\/strong><\/p>\n\n\n\n

Answer
<\/strong><\/p>\n\n\n\n

Length of the piece of copper, l<\/em> = 19.1 mm = 19.1 \u00d7 10\u20133<\/sup> m
Breadth of the piece of copper, b<\/em> = 15.2 mm = 15.2 \u00d7 10\u20133<\/sup> m
Area of the copper piece:
A = l \u00d7 b<\/em>
= 19.1 \u00d7 10\u20133<\/sup> \u00d7 15.2 \u00d7 10\u20133<\/sup>
= 2.9 \u00d7 10\u20134<\/sup> m2<\/sup>
Tension force applied on the piece of copper, F<\/em> = 44500 N
Modulus of elasticity of copper, \u03b7 = 42 \u00d7 109<\/sup> N\/m2<\/sup>
Modulus of elasticity, \u03b7 = Stress \/ Strain
= (F<\/em>\/A<\/em>) \/ Strain
\u2234 Strain = F<\/em> \/ A<\/em>\u03b7
= 44500 \/ (2.9 \u00d7 10-4<\/sup> \u00d7 42 \u00d7 109<\/sup>)
= 3.65 \u00d7 10\u20133<\/sup>.<\/p>\n\n\n\n

NCERT 11th Physics Chapter 9, class 11 Physics chapter 9 solutions<\/p>\n\n\n\n

9.9. 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 <\/sup>N m\u20132<\/sup>, what is the maximum load the cable can support?<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Radius of the steel cable, r<\/em> = 1.5 cm = 0.015 m
Maximum allowable stress = 108<\/sup> N m\u20132<\/sup>
Maximum stress = Maximum force \/ Area of cross-section
\u2234 Maximum force = Maximum stress \u00d7 Area of cross-section
= 108<\/sup> \u00d7 \u03c0 (0.015)2<\/sup>
= 7.065 \u00d7 104<\/sup> N
Hence, the cable can support the maximum load of 7.065 \u00d7 104<\/sup> N.<\/p>\n\n\n\n

9.10. 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 ratio of their diameters if each is to have the same tension.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

The tension force acting on each wire is the same. Thus, the extension in each case is the same. Since the wires are of the same length, the strain will also be the same.
The relation for Young\u2019s modulus is given as:
Y<\/em> = Stress \/ Strain
= (F<\/em>\/A<\/em>) \/ Strain  =  (4F<\/em>\/\u03c0d<\/em>2<\/sup>) \/ Strain     ….(i)
Where,
F<\/em> = Tension force
A<\/em> = Area of cross-section
d <\/em>= Diameter of the wire
It can be inferred from equation (i<\/em>) that Y <\/em>\u221d (1\/d2<\/sup>)
Young\u2019s modulus for iron, Y<\/em>1<\/sub> = 190 \u00d7 109<\/sup> Pa
Diameter of the iron wire = d<\/em>1<\/sub>
Young\u2019s modulus for copper, Y<\/em>2 <\/sub>= 120 \u00d7 109<\/sup> Pa
Diameter of the copper wire = d<\/em>2<\/sub>
Therefore, the ratio of their diameters is given as:<\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

9.11. 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<\/sup>. Calculate the elongation of the wire when the mass is at the lowest point of its path.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Mass, m<\/em> = 14.5 kg
Length of the steel wire, l<\/em> = 1.0 m
Angular velocity, \u03c9<\/em> = 2 rev\/s = 2 \u00d7 2\u03c0 rad\/s = 12.56 rad\/s
Cross-sectional area of the wire, a<\/em> = 0.065 cm2<\/sup> = 0.065 \u00d7 10-4<\/sup> m2<\/sup>
Let \u0394l<\/em> be the elongation of the wire when the mass is at the lowest point of its path.
When the mass is placed at the position of the vertical circle, the total force on the mass is:
F<\/em> = m<\/em>g + ml\u03c9<\/em>2<\/sup>
= 14.5 \u00d7 9.8 + 14.5 \u00d7 1 \u00d7 (12.56)2<\/sup>
= 2429.53 N
Young’s modulus = Strss \/ Strain
Y = (F\/A) \/ (\u2206l\/l)<\/em>
\u2234 \u2206l<\/em> = F<\/em>l \/ AY<\/em>
Young\u2019s modulus for steel = 2 \u00d7 1011<\/sup> Pa
\u2206l<\/em> = 2429.53 \u00d7 1 \/ (0.065 \u00d7 10-4<\/sup> \u00d7 2 \u00d7 1011<\/sup>)   =   1.87 \u00d7 10-3<\/sup> m
Hence, the elongation of the wire is 1.87 \u00d7 10\u20133<\/sup> m.<\/p>\n\n\n\n

NCERT 11th Physics Chapter 9, class 11 Physics chapter 9 solutions<\/p>\n\n\n\n

9.12. Compute the bulk modulus of water from the following data: Initial volume = 100.0 litre, Pressure increase = 100.0 atm (1 atm = 1.013 \u00d7 105<\/sup> 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.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Initial volume, V<\/em>1 <\/sub>= 100.0l = 100.0 \u00d7 10 \u20133<\/sup> m3<\/sup>
Final volume, V<\/em>2 <\/sub>= 100.5 l = 100.5 \u00d710 \u20133<\/sup> m3<\/sup>
Increase in volume, \u0394V<\/em> = V<\/em>2<\/sub> \u2013 V<\/em>1 <\/sub>= 0.5 \u00d7 10\u20133<\/sup> m3<\/sup>
Increase in pressure, \u0394p<\/em> = 100.0 atm = 100 \u00d7 1.013 \u00d7 105<\/sup> Pa
Bulk modulus = \u0394p<\/em> \/ (\u0394V<\/em>\/V1<\/sub>)  =  \u0394p<\/em> \u00d7 V1<\/sub> \/ \u0394V<\/em>
= 100 \u00d7 1.013 \u00d7 105<\/sup> \u00d7 100 \u00d7 10-3<\/sup> \/ (0.5 \u00d7 10-3<\/sup>)
= 2.026 \u00d7 109<\/sup> Pa
Bulk modulus of air = 1 \u00d7 105<\/sup> Pa
\u2234 Bulk modulus of water \/ Bulk modulus of air  =  2.026 \u00d7 109<\/sup> \/ (1 \u00d7 105<\/sup>)  =  2.026 \u00d7 104<\/sup>
This ratio is very high because air is more compressible than water.<\/p>\n\n\n\n

NCERT 11th Physics Chapter 9, class 11 Physics chapter 9 solutions<\/p>\n\n\n\n

9.13. 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 \u00d7 103<\/sup> kg m\u20133<\/sup>?<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Let the given depth be h<\/em>.
Pressure at the given depth, p<\/em> = 80.0 atm = 80 \u00d7 1.01 \u00d7 105<\/sup> Pa
Density of water at the surface, \u03c1<\/em>1 <\/sub>= 1.03 \u00d7 103<\/sup> kg m\u20133<\/sup>
Let \u03c1<\/em>2<\/sub> be the density of water at the depth h<\/em>.
Let V<\/em>1<\/sub> be the volume of water of mass m<\/em> at the surface.
Let V<\/em>2<\/sub> be the volume of water of mass m<\/em> at the depth h<\/em>.
Let \u0394V<\/em> be the change in volume.
\u0394V<\/em> = V<\/em>1<\/sub> – V<\/em>2<\/sub>
= m<\/em> [ (1\/\u03c1<\/em>1<\/sub>) – (1\/\u03c1<\/em>2<\/sub>) ]
\u2234 Volumetric strain = \u0394V<\/em> \/ V<\/em>1<\/sub>
= m<\/em> [ (1\/\u03c1<\/em>1<\/sub>) – (1\/\u03c1<\/em>2<\/sub>) ] \u00d7 (\u03c1<\/em>1<\/sub> \/ m<\/em>)
\u0394V<\/em> \/ V<\/em>1<\/sub> = 1 – (\u03c11<\/sub>\/\u03c1<\/em>2<\/sub>)     ……(i)
Bulk modulus, B<\/em> = pV<\/em>1<\/sub> \/ \u0394V<\/em>
\u0394V<\/em> \/ V<\/em>1<\/sub> = p<\/em> \/ B<\/em>
Compressibility of water = (1\/B<\/em>) = 45.8 \u00d7 10-11<\/sup> Pa-1<\/sup>
\u2234 \u0394V<\/em> \/ V<\/em>1<\/sub> = 80 \u00d7 1.013 \u00d7 105<\/sup> \u00d7 45.8 \u00d7 10-11<\/sup>  =  3.71 \u00d7 10-3<\/sup>    ….(ii)
For equations (i<\/em>) and (ii<\/em>), we get:
1 – (\u03c11<\/sub>\/\u03c1<\/em>2<\/sub>)   =   3.71 \u00d7 10-3<\/sup>
\u03c1<\/em>2<\/sub> = 1.03 \u00d7 103<\/sup> \/ [ 1 – (3.71 \u00d7 10-3<\/sup>) ]
= 1.034 \u00d7 103<\/sup> kg m-3<\/sup>
Therefore, the density of water at the given depth (h<\/em>) is 1.034 \u00d7 103<\/sup> kg m\u20133<\/sup>.<\/p>\n\n\n\n

9.14. Compute the fractional change in volume of a glass slab, when subjected to a hydraulic pressure of 10 atm.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Hydraulic pressure exerted on the glass slab, p <\/em>= 10 atm = 10 \u00d7 1.013 \u00d7 105<\/sup> Pa
Bulk modulus of glass, B<\/em> = 37 \u00d7 109<\/sup> Nm\u20132<\/sup>
Bulk modulus, B<\/em> = p \/ (\u2206V<\/em>\/V<\/em>)
Where,
\u2206V<\/em>\/V<\/em> = Fractional change in volume
\u2234 \u2206V<\/em>\/V<\/em> = p<\/em> \/ B<\/em>
= 10 \u00d7 1.013 \u00d7 105<\/sup> \/ (37 \u00d7 109<\/sup>)
= 2.73 \u00d7 10-5<\/sup>
Hence, the fractional change in the volume of the glass slab is 2.73 \u00d7 10\u20135<\/sup>.<\/p>\n\n\n\n

NCERT 11th Physics Chapter 9, class 11 Physics chapter 9 solutions<\/p>\n\n\n\n

9.15. Determine the volume contraction of a solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of 7.0 \u00d7106<\/sup> Pa.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Length of an edge of the solid copper cube, l<\/em> = 10 cm = 0.1 m
Hydraulic pressure, p<\/em> = 7.0 \u00d7 106<\/sup> Pa
Bulk modulus of copper, B<\/em> = 140 \u00d7 109<\/sup> Pa
Bulk modulus, B<\/em> = p<\/em> \/ (\u2206V<\/em>\/V<\/em>)
Where,
\u2206V<\/em>\/V<\/em> = Volumetric strain
\u0394V<\/em> = Change in volume
V<\/em> = Original volume.
\u0394V<\/em> = pV<\/em> \/ B<\/em>
Original volume of the cube, V<\/em> = l<\/em>3<\/sup>
\u2234 \u0394V<\/em> = pl<\/em>3<\/sup> \/ B<\/em>
= 7 \u00d7 106<\/sup> \u00d7 (0.1)3<\/sup> \/ (140 \u00d7 109<\/sup>)
= 5 \u00d7 10-8<\/sup> m3<\/sup>  =  5 \u00d7 10-2<\/sup> cm-3<\/sup>
Therefore, the volume contraction of the solid copper cube is 5 \u00d7 10\u20132<\/sup> cm\u20133<\/sup>.<\/p>\n\n\n\n

NCERT 11th Physics Chapter 9, class 11 Physics chapter 9 solutions<\/p>\n\n\n\n

9.16. How much should the pressure on a litre of water be changed to compress it by 0.10%?<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Volume of water, V <\/em>= 1 L
It is given that water is to be compressed by 0.10%.
\u2234 Fractional change, \u2206V<\/em> \/ V<\/em> = 0.1 \/ (100 \u00d7 1)  =  10-3<\/sup>
Bulk modulus, B<\/em> = \u03c1 \/ (\u2206V<\/em>\/V<\/em>)
\u03c1<\/em> = B<\/em> \u00d7 (\u2206V<\/em>\/V<\/em>)
Bulk modulus of water, B<\/em> = 2.2 \u00d7 109<\/sup> Nm-2<\/sup>
\u03c1<\/em> = 2.2 \u00d7 109<\/sup> \u00d7 10-3<\/sup>  =  2.2 \u00d7 106<\/sup> Nm-2<\/sup>
Therefore, the pressure on water should be 2.2 \u00d7106<\/sup> Nm\u20132<\/sup>.<\/p>\n\n\n\n

NCERT 11th Physics Chapter 9, class 11 Physics chapter 9 solutions<\/p>\n\n\n\n

Additional Excercises<\/strong><\/p>\n\n\n\n

9.17. Anvils made of single crystals of diamond, with the shape as shown in Fig. 9.14, 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?<\/strong><\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

Diameter of the cones at the narrow ends, d<\/em> = 0.50 mm = 0.5 \u00d7 10\u20133<\/sup> m
Radius, r<\/em> = d<\/em>\/2  =  0.25 \u00d7 10-3<\/sup> m
Compressional force, F<\/em> = 50000 N
Pressure at the tip of the anvil:
P<\/em> = Force \/ Area  =  50000 \/ \u03c0(0.25 \u00d7 10-3<\/sup>)2<\/sup>
= 2.55  \u00d7 1011<\/sup> Pa
Therefore, the pressure at the tip of the anvil is 2.55 \u00d7 1011<\/sup> Pa.<\/p>\n\n\n\n

Page No: 245<\/p>\n\n\n\n

9.18. 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 Fig. 9.15. The cross-sectional areas of wires A and B are 1.0 mm2<\/sup> and 2.0 mm2<\/sup>, respectively. At what point along the rod should a mass m <\/em>be suspended in order to produce (a) equal stresses and (b) equal strains in both steel and aluminium wires.<\/strong><\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Cross-sectional area of wire A<\/strong>, a<\/em>1<\/sub> = 1.0 mm2 <\/sup>= 1.0 \u00d7 10\u20136<\/sup> m2<\/sup>
Cross-sectional area of wire B<\/strong>, a<\/em>2<\/sub> = 2.0 mm2 <\/sup>= 2.0 \u00d7 10\u20136<\/sup> m2<\/sup>
Young\u2019s modulus for steel, Y<\/em>1<\/sub> = 2 \u00d7 1011 <\/sup>Nm\u20132<\/sup>
Young\u2019s modulus for aluminium, Y<\/em>2<\/sub> = 7.0 \u00d71010 <\/sup>Nm\u20132<\/sup>
(a)Let a small mass m<\/em> be suspended to the rod at a distance y<\/em> from the end where wire A<\/strong> is attached.
Stress in the wire = Force \/ Area  =  F<\/em> \/ a<\/em>
If the two wires have equal stresses, then:
F<\/em>1<\/sub> \/ a<\/em>1<\/sub>  =  F<\/em>2<\/sub> \/ a<\/em>2<\/sub>
Where,
F<\/em>1<\/sub> = Force exerted on the steel wire
F<\/em>2<\/sub> = Force exerted on the aluminum wire
F<\/em>1<\/sub> \/ F<\/em>2<\/sub> = a<\/em>1<\/sub> \/ a<\/em>2<\/sub>  =  1 \/ 2    ….(i)<\/strong>
The situation is shown in the following figure:<\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

Taking torque about the point of suspension, we have:
F<\/em>1<\/sub>y<\/em> = F<\/em>2<\/sub> (1.05 – y<\/em>)
F<\/em>1<\/sub> \/ F<\/em>2<\/sub> = (1.05 – y<\/em>) \/ y<\/em>    ……(ii)<\/strong>
Using equations (i)<\/strong> and (ii)<\/strong>, we can write:
(1.05 – y<\/em>) \/ y<\/em>  = 1 \/ 2
2(1.05 – y<\/em>)  =  y<\/em>
y<\/em> = 0.7 m
In order to produce an equal stress in the two wires, the mass should be suspended at a distance of 0.7 m from the end where wire A<\/strong> is attached.<\/p>\n\n\n\n

(b) Young’s modulus = Stress \/ Strain
Strain = Stress \/ Young’s modulus  =  (F<\/em>\/a<\/em>) \/ Y<\/em>
If the strain in the two wires is equal, then:
(F<\/em>1<\/sub>\/a<\/em>1<\/sub>) \/ Y<\/em>1<\/sub>  =  (F<\/em>2<\/sub>\/a<\/em>2<\/sub>) \/ Y<\/em>2<\/sub>
F<\/em>1<\/sub> \/ F<\/em>2<\/sub> = a<\/em>1<\/sub>Y<\/em>1<\/sub> \/ a<\/em>2<\/sub>Y<\/em>2<\/sub>
a<\/em>1<\/sub> \/ a<\/em>2<\/sub> = 1\/2
F<\/em>1<\/sub> \/ F<\/em>2<\/sub> = (1 \/ 2) (2 \u00d7 1011<\/sup> \/ 7 \u00d7 1010<\/sup>)  =  10 \/ 7      …….(iii)<\/strong>
Taking torque about the point where mass m<\/em>, is suspended at a distance y<\/em>1<\/sub> from the side where wire A<\/strong> attached, we get:
F<\/em>1<\/sub>y<\/em>1 <\/sub>= F<\/em>2<\/sub> (1.05 \u2013 y<\/em>1<\/sub>)
F<\/em>1<\/sub> \/ F<\/em>2<\/sub>  =  (1.05 – y<\/em>1<\/sub>) \/ y1<\/sub>    ….(iii)<\/strong>
Using equations (iii)<\/strong> and (iv)<\/strong>, we get:
(1.05 – y<\/em>1<\/sub>) \/ y1<\/sub>  =  10 \/ 7
7(1.05 – y1<\/sub>)  =  10y<\/em>1<\/sub>
y<\/em>1<\/sub> = 0.432 m
In order to produce an equal strain in the two wires, the mass should be suspended at a distance of 0.432 m from the end where wire A<\/strong> is attached.<\/p>\n\n\n\n

9.19. A mild steel wire of length 1.0 m and cross-sectional area 0.50 \u00d7 10\u20132<\/sup> cm2 <\/sup>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 midpoint.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

From the above figure,<\/p>\n\n\n\n

Let x<\/em> be the depression at the mid point i.e. CD = x.<\/p>\n\n\n\n

In fig.,<\/p>\n\n\n\n

AC= CB = l<\/em> = 0.5 m ;<\/p>\n\n\n\n

m<\/em> = 100 g = 0.100 Kg<\/p>\n\n\n\n

AD= BD = (l<\/em>2<\/sup> + x<\/em>2<\/sup>)1\/2<\/sup><\/p>\n\n\n\n

Increase in length, \u2206l = AD + DB – AB = 2AD – AB<\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n
\"NCERT<\/figure>\n\n\n\n

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 \u00d7 107<\/sup> Pa? Assume that each rivet is to carry one quarter of the load.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Diameter of the metal strip, d<\/em> = 6.0 mm = 6.0 \u00d7 10\u20133<\/sup> m
Radius, r = d\/2 = 3 \u00d7 10-3<\/sup> m
Maximum shearing stress = 6.9 \u00d7 107<\/sup> Pa
Maximum stress = Manimum load or force \/ Area
Maximum force = Maximum stress \u00d7 Area
= 6.9 \u00d7 107<\/sup> \u00d7 \u03c0 \u00d7 (r<\/em>) 2<\/sup>
= 6.9 \u00d7 107<\/sup> \u00d7 \u03c0 \u00d7 (3 \u00d710\u20133<\/sup>)2<\/sup>
= 1949.94 N
Each rivet carries one quarter of the load.
\u2234 Maximum tension on each rivet = 4 \u00d7 1949.94 = 7799.76 N.<\/p>\n\n\n\n

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 \u00d7 108<\/sup> Pa. A steel ball of initial volume 0.32 m3<\/sup> 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?<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Water pressure at the bottom, p <\/em>= 1.1 \u00d7 108<\/sup> Pa
Initial volume of the steel ball, V<\/em> = 0.32 m3<\/sup>
Bulk modulus of steel, B<\/em> = 1.6 \u00d7 1011 <\/sup>Nm\u20132<\/sup>
The ball falls at the bottom of the Pacific Ocean, which is 11 km beneath the surface.
Let the change in the volume of the ball on reaching the bottom of the trench be \u0394V<\/em>.
Bulk modulus, B<\/em> = p<\/em> \/ (\u2206V<\/em>\/V<\/em>)
\u2206V<\/em>  =  B<\/em> \/ pV<\/em>
= 1.1 \u00d7 108<\/sup> \u00d7 0.32 \/ (1.6 \u00d7 1011<\/sup> )  =  2.2 \u00d7 10-4<\/sup> m3<\/sup>
Therefore, the change in volume of the ball on reaching the bottom of the trench is 2.2 \u00d7 10\u20134<\/sup> m3<\/sup>.<\/p>\n\n\n\n

NCERT Solutions for 11th Class Physics: Chapter 9: Download PDF<\/strong><\/h2>\n\n\n\n

NCERT Solutions for 11th Class Physics: Chapter 9-Mechanical properties of Solids<\/p>\n\n\n\n

Download PDF<\/strong>: NCERT Solutions for 11th Class Physics: Chapter 9-Mechanical properties of Solids PDF<\/a><\/p>\n\n\n\n

Chapterwise NCERT Solutions for Class 11 Physics<\/strong>:<\/h4>\n\n\n\n