Chapter 21: Interatomic and Intermolecular Forces PDF

60 2 Elasticity 9.1 Interatomic Forces. E3 The forces between the atoms due to electrostatic interaction between the charges of the atoms are called interatomic forces. These forces are electrical in nature and these are active if the distance between the two atoms is of the order of atomic size i.e...

60 2 Elasticity 9.1 Interatomic Forces. E3 The forces between the atoms due to electrostatic interaction between the charges of the atoms are called interatomic forces. These forces are electrical in nature and these are active if the distance between the two atoms is of the order of atomic size i.e. 10–10 metre. ID (1) Every atom is electrically neutral, the number of electrons (negative charge) orbiting around the nucleus is equal to the number of proton (positive charge) in the nucleus. So if two atoms are placed at a very large distance from each other then there will be a very small (negligible) interatomic force working between them. U (2) When these two atoms are brought close to each other to a distance of the order of 10 –10 m, the distances between their positive nuclei and negative electron clouds get disturbed, and due to this, attractive interatomic force is produced between two atoms. Repulsion D YG (3) This attractive force increases continuously with decrease in r and becomes maximum for one value of r called critical distance, represented by x (as shown in the figure). Beyond this the attractive force starts decreasing F rapidly with further decrease in the value of r. (4) When the distance between the two atoms becomes r0, the x O r r0 Attracti on interatomic force will be zero. This distance r0 is called normal or equilibrium distance. U U (r0 = 0.74 Å for hydrogen). (5) When the distance between the two atoms further decreased, O r r0 ST the interatomic force becomes repulsive in nature and increases very rapidly with decrease in distance between two atoms. (6) The potential energy U is related with the interatomic force F by the following relation. F  dU dr (i) When two atoms are at very large distance, the potential energy is negative and becomes more negative as r is decreased. (ii) When the distance between the two atoms becomes r0, the potential energy of the system of two atoms becomes minimum (i.e. attains maximum negative value). As the state of Elasticity 3 minimum potential energy is the state of equilibrium, hence the two atoms at separation r0 will be in a state of equilibrium. ( U 0  7.2  10 19 Joule for hydrogen). positive value with further decrease in r (as shown in the figure). 9.2 Intermolecular Forces. 60 (iii) When the distance between the two atoms is further decreased (i.e. r < r0) the negative value of potential energy of the system starts decreasing. It becomes zero and then attains E3 The forces between the molecules due to electrostatic interaction between the charges of the molecules are called intermolecular forces. These forces are also called Vander Waal forces and are quite weak as compared to inter-atomic forces. These forces are also electrical in nature ID and these are active if the separation between two molecules is of the order of molecular size i.e.  10–9 m. (1) It is found that the force of attraction between molecules varies inversely as seventh power of the distance between them i.e. 1 r7 Fatt  or a r7 U Fatt  D YG The negative sign indicates that the force is attractive in nature. (2) When the distance between molecules becomes less than r0, the forces becomes repulsive in nature and is found to vary inversely as ninth power of the distance between them i.e. Frep  1 r9 or Frep  b. r9 a b  r7 r9 U Therefore force between two molecules is given by F  Fatt  Frep  The value of constants a and b depend upon the structure and nature of molecules. ST (3) Intermolecular forces between two molecules has the same general nature as shown in the figure for interatomic forces. (4) Potential Energy : Potential energy can be approximately expressed by the formula A B U n  m r r A B represents repulsive contribution and term m represents the attractive n r r contribution. Constants A, B and numbers m and n are different for different molecules. where the term For majority of solids n = 12 and m = 6. So potential energy can be expressed as U  A B  6 12 r r 9.3 Comparison Between Interatomic and Intermolecular Forces. 4 Elasticity (1) Similarities (i) Both the forces are electrical in origin. (ii) Both the forces are active over short distances. (iii) General shape of force-distance graph is similar for both the forces. 60 (iv) Both the forces are attractive up to certain distance between atoms/molecules and become repulsive when the distance between them become less than that value. (2) Dissimilarities E3 (i) Interatomic force depends upon the distance between the two atoms, whereas the intermolecular force depends upon the distance between the two molecules as well as their relative orientation. (ii) Interatomic forces are about 50 to100 times stronger than intermolecular forces. ID (iii) The value of r0 for two atoms is smaller than the corresponding value for the molecules. Therefore one molecule is not restricted to attract only one molecule, but can attract atoms of other molecules. D YG 9.4 States of Matter. U many molecule. It is not so incase of atoms, since the atoms of one molecule cannot bind the The three states of matter differ from each other due to the following two factors. (1) The different magnitudes of the interatomic and intermolecular forces. (2) The extent of random thermal motion of atoms and molecules of a substance (which depends upon temperature). U Comparison Chart of Solid, Liquid and Gaseous States Property Solid Liquid Gas Definite Not definite Not definite Volume Definite Definite Not definite Density Maximum Less than solids but more than gases. Minimum Incompressible Less than gases but more than solids. Compressible Crystalline Non-crystalline Constant Not constant ST Shape Compressibility Crystallinity Interatomic or intermolecular distance Not constant Elasticity 5 K> U K >> U Strongest Less than solids but more than gases. Weakest Freedom of motion Molecules vibrate about their mean position but cannot move freely. Molecules have limited free motion. Effect of temperature Matter remains in solid form below a certain temperature. Liquids are found at temperatures more than that of solid. Note :  The Molecules are free to move. These are found at temperatures greater than that of solids and liquids. ID Intermolecular force 60 K stress in wire B so the A will break before B (ii) if rB  r , (let) then rA  2r Stress in wire B = mg 3r 2 and Stress in wire A  4 mg 3 (2r) 2  mg 3r 2 i.e. stress in wire A = stress in wire B. It means either A or B may break. U (iii) If rA  2rB then stress in A will be more than B. i.e. A will break before B. Problem 51. A body of mass 10 kg is attached to a wire 0.3 m long. Its breaking stress is 4.8  10 7 N / m 2. ST The area of cross-section of the wire is 10 6 m 2. What is the maximum angular velocity with which it can be rotated in the horizontal circle (a) 1 rad/sec (b) 2 rad/sec (c) 4 rad/sec (d) 8 rad/sec Solution : (c) Breaking force = centrifugal force Breaking stress × area of cross-section = m  2 l 4.8  10 7  10 6  10   2  0.3   2  16    4rad / sec Problem 52. Two block of masses 1 kg and 4 kg are connected by a metal wire going over a smooth pulley as shown in the figure. The breaking stress of the metal is 3.18  1010 N / m 2. The minimum radius of the wire so it will not break is 1kg 28 Elasticity (a) 1  10 5 m (b) 2  10 5 m (c) 3  10 5 m Solution : (d) Tension in the wire T  60 (d) 4  10 5 m 2m 1 m 2 2 1  4 g  T  10  T  16 N m1  m 2 14 Tension in the wire = 3.18  10 10  r 2 16 3.18  10 10  3.14  4  10 5 m. ID 16  3.18  10 10  r 2  r  E3 Breaking force = Breaking stress × Area of cross-section 9.14 Bulk Modulus. U When a solid or fluid (liquid or gas) is subjected to a uniform pressure all over the surface, such that the shape remains the same, then there is a change in volume. D YG Then the ratio of normal stress to the volumetric strain within the elastic limits is called as Bulk modulus. This is denoted by K. K Normal stress volumetric strain K F/A  pV   V / V V (V – V) where p = increase in pressure; V = original volume; V = change in volume U The negative sign shows that with increase in pressure p, the volume decreases by V i.e. if p is positive, V is negative. The reciprocal of bulk modulus is called compressibility. ST C = compressibility = 1 V  K pV S.I. unit of compressibility is N–1m2 and C.G.S. unit is dyne–1 cm2. Gases have two bulk moduli, namely isothermal elasticity E and adiabatic elasticity E. (1) Isothermal elasticity (E) : Elasticity possess by a gas in isothermal condition is defined as isothermal elasticity. For isothermal process, PV = constant (Boyle’s law) Differentiating both sides PdV + VdP = 0  PdV = – VdP P dP stress  E  (dV / V ) strain Elasticity 29  E = P i.e., Isothermal elasticity is equal to pressure. PV  = constant For adiabatic process, (Poisson’s law) 60 (2) Adiabatic elasticity (E) : Elasticity possess by a gas in adiabatic condition is defined as adiabatic elasticity. Differentiating both sides, P  V  1 dV  V  dP  0   PdV  VdP  0 dP stress  E    dV  strain    V  E3 P E =  P  [where   Note :  ID i.e., adiabatic elasticity is equal to  times pressure. Ratio of adiabatic to isothermal elasticity E E  P P  1 Cp Cv ]  E > E U i.e., adiabatic elasticity is always more than isothermal elasticity. 9.15 Density of Compressed Liquid. increases. As density D YG If a liquid of density  , volume V and bulk modulus K is compressed, then its density  m V so But by definition of bulk modulus K   ST or U From (i) and (ii)        1    VP  V         P  K  P   [1  CP] K   V V …..(i) V P  V K …..(ii) [As  =   –  ]  1   As K  C    9.16 Fractional Change in the Radius of Sphere. A solid sphere of radius R made of a material of bulk modulus K is surrounded by a liquid in a cylindrical container. A massless piston of area A floats on the surface of the liquid. Volume of the spherical body V  V R 3 V R 4 3 R 3 m 30 Elasticity R 1 V  R 3 V Bulk modulus K   V …..(i) P V V P mg   V K AK  R 1 mg V from equation (ii) in equation (i) we get  R V 3 AK E3 Substituting the value of mg    As P  A    …..(ii) 60  Sample problems based on Bulk modulus Problem 53. When a pressure of 100 atmosphere is applied on a spherical ball of rubber, then its volume (a) 10  10 12 ID reduces to 0.01%. The bulk modulus of the material of the rubber in dyne/cm2 is (c) 1  10 12 (b) 100  10 12 P 10 7 Dyne   1  10 11 N / m 2  1  10 12. 0.0001 V / V cm 2 D YG K V  0.0001 V U Solution : (c) 1 atm  10 5 N / m 2 100 atm  10 7 N / m 2 and V  0.01 % V  (d) 20  10 12 Problem 54. Coefficient of isothermal elasticity E and coefficient of adiabatic elasticity E are related by (  C p / Cv ) (a) E   E (b) E   E [MP PET 2000] (c) E   / E (d) E   2 E Solution : (b) Adiabatic elasticity    isothermal elasticity  E   E. Problem 55. A uniform cube is subjected to volume compression. If each side is decreased by 1%,then U bulk strain is ST (a) 0.01 Solution : (d) Volume of cube V  L3 3 (1 %)  3 % [EAMCET (Engg.) 1995; DPMT 2000] (b) 0.06 (c) 0.02 (d) 0.03  Percentage change in V  3  (percentage change in L)= V  3 % of V  Volumetric strain  V 3   0.03 V 100 Problem 56. A ball falling in a lake of depth 200 m shows 0.1% decrease in its volume at the bottom. What is the bulk modulus of the material of the ball (a) 19.6  10 8 N / m 2 Solution : (a) K  (b) 19.6  10 10 N / m 2 (c) 19.6  10 10 N / m 2 P hdg 200  10 3  9.8  19.6  10 8 N / m 2   V / V V / V 0.001 (d) 19.6  10 8 N / m 2 Elasticity 31 Problem 57. The ratio of the adiabatic to isothermal elasticities of a triatomic gas is (a) 3 4 (b) 4 3 (c) 1 (d) 5 3 4. 3 60 Solution : (b) For triatomic gas   4 / 3  Ratio of adiabatic to isothermal elasticity   Problem 58. A gas undergoes a change according to the law P  P0 e V. The bulk modulus of the gas is Solution : (b) P  Po e V  dP  Po e V   P dV (c) P (d) PV  E3 (b) PV (a) P [As P  Po e V ] dP dP V  P V   P V dV dV / V   K  P V ID Problem 59. The ratio of two specific heats of gas C p / Cv for argon is1.6 and for hydrogen is 1.4. Adiabatic elasticity of argon at pressure P is E. Adiabatic elasticity of hydrogen will also be equal to E at the pressure 8 P 7 7 P 8 U (a) P (b) (c) (d) 1.4 P D YG Solution : (b) Adiabatic elasticity =  (pressure) For Argon (E ) Ar  1.6 P and for Hydrogen (E )H 2  1.4 P  According to problem (E )H 2  ( ) Ar  1.4 P   1.6 P  P  8 16 P  P. 7 14 Problem 60. The pressure applied from all directions on a cube is P. How much its temperature should be U raised to maintain the original volume ? The volume elasticity of the cube is  and the coefficient of volume expansion is  P P P  (a) (b) (c) (d)    P Solution : (a) Change in volume due to rise in temperature V  V  V    V stress P P     But bulk modulus    strain    ST  volumetric strain  9.17 Modulus of Rigidity. Within limits of proportionality, the ratio of tangential stress to the shearing strain is called modulus of rigidity of the material of the body and is x F A Shearing stress ' R Q Q denoted by , i.e.    Shearing strain  L  In this case the shape of a body changes but its volume P Fixed face S 32 Elasticity remains unchanged. Consider a cube of material fixed at its lower face and acted upon by a tangential force F at its upper surface having area A. The shearing stress, then, will be F|| A  F A 60 Shearing stress  So  QQ ' x  PQ L shear stress F / A F   shear strain  A ID Shearing strain    E3 This shearing force causes the consecutive horizontal layers of the cube to be slightly displaced or sheared relative to one another, each line such as PQ or RS in the cube is rotated through an angle  by this shear. The shearing strain is defined as the angle  in radians through which a line normal to a fixed surface has turned. For small values of angle, Only solids can exhibit a shearing as these have definite shape. 9.18 Poisson’s Ratio. U When a long bar is stretched by a force along its length then its length increases and the radius decreases as shown in the figure. D YG Lateral strain : The ratio of change in radius to the original radius is called lateral strain. Longitudinal strain : The ratio of change in length to the original length is called longitudinal strain. The ratio of lateral strain to longitudinal strain is called Poisson’s ratio ( ).   Lateral strain Longitudin al strain   dr / r dL / L L r r – dr L + dL F U i.e. Negative sign indicates that the radius of the bar decreases when it is stretched. ST Poisson’s ratio is a dimensionless and a unitless quantity. 9.19 Relation Between Volumetric Strain, Lateral Strain and Poisson’s Ratio. If a long bar have a length L and radius r then volume V  r 2 L Differentiating both the sides dV  r 2 dL   2rL dr Dividing both the sides by volume of bar  dr dV r 2 dL  2rL dr dL  2   2 2 V L r r L r L Volumetric strain = longitudinal strain + 2(lateral strain) Elasticity 33  dV dL dL dL   2  (1  2 ) V L L L or  dr / r dr dL    As   dL / L  r   L    1 dV  1  2 AdL  bar] Important points dV dL =0  [1  2 ] V L E3 (i) If a material having  = – 0.5 then 60 [where A = cross-section of  Volume = constant or K =  i.e., the material is incompressible. ID (ii) If a material having  = 0, then lateral strain is zero i.e. when a substance is stretched its length increases without any decrease in diameter e.g. cork. In this case change in volume is maximum. (iii) Theoretical value of Poisson’s ratio  1    0.5. U (iv) Practical value of Poisson’s ratio 0    0.5 9.20 Relation between Y, k,  and . D YG Moduli of elasticity are three, viz. Y, K and  while elastic constants are four, viz, Y, K,  and . Poisson’s ratio  is not modulus of elasticity as it is the ratio of two strains and not of stress to strain. Elastic constants are found to depend on each other through the relations : Y  3 K(1  2 ) and Y  2(1   ) Eliminating  or Y between these, we get Y  9 K 3 K  2 and   3K   6 K  2 U Sample problems based on relation between Y, k,  and  Problem 61. Minimum and maximum values of Poisson’s ratio for a metal lies between ST (a) –  to +  (b) 0 to 1 (c) –  to 1 (d) 0 to 0.5 Solution : (d) Problem 62. For a given material, the Young’s modulus is 2.4 times that of rigidity modulus. Its Poisson’s ratio is [EAMCET 1990; RPET 2001] (a) 2.4 Solution : (d) Y  2(1   ) (b) 1.2 (c) 0.4 (d) 0.2  2.4  2(1   )  1.2  1      0. 2 Problem 63. There is no change in the volume of a wire due to change in its length on stretching. The Poisson’s ratio of the material of the wire is (a) + 0.50 (b) – 0.50 (c) + 0.25 (d) – 0.25 34 Elasticity Solution : (b) dV dL dL dL =0  (1  2 )   2 V L L L 1  2  0     [As there is no change in the volume of the wire] 1 2 60 Problem 64. The values of Young’s and bulk modulus of elasticity of a material are 8  1010 N / m 2 and 10  1010 N / m 2 respectively. The value of Poisson’s ratio for the material will be (a) 0.25 (b) – 0.25 (c) 0.37 (d) – 0.37 E3 Solution : (c) Y  3 K(1  2 )  8  10 10  3  10  10 10 (1  2 )    0. 37 Problem 65. The Poisson’s ratio for a metal is 0.25. If lateral strain is 0.0125, the longitudinal strain will be Solution : (b)   (b) 0.05 Lateral strain Longitudin al strain (c) 0.215  Longitudinal strain  Lateral strain (d) 0.0125  ID (a) 0.125  0.0125  0.05 0.25 Problem 66. The ‘’ of a material is 0.20. If a longitudinal strain of 4.0  10 3 is caused, by what (a) 0.48% U percentage will the volume change (b) 0.32% (c) 0.24% (d) 0.50% Solution : (c) Longitudinal strain  4  10 3 or 0.4% D YG Lateral strain    0.4 %  0.2  0.4 %  0.08 %  Volumetric strain = longitudinal strain – 2 lateral strain  0.4  2  (0.08 ) = 0.24%  Volume will change by 0.24%. 9.21 Torsion of Cylinder. If the upper end of a cylinder is clamped and a torque is applied at the lower end the cylinder gets twisted by angle . Simultaneously shearing strain U  is produced in the cylinder. ST (i) The angle of twist  P r l is directly proportional to the distance from the fixed end of the cylinder.  O At fixed end   0 and at free end  = maximum. o Q  A B (ii) The value of angle of shear  is directly proportional to the radius of the cylindrical shell. At the axis of cylinder  = 0 and at the outermost shell  = maximum. (iii) Relation between angle of twist () and angle of shear () AB = r = l   r l Elasticity 35 (iv) Twisting couple per unit twist or torsional rigidity or torque required to produce unit twist. C r 4  C  r4  A2 2l Sample problems based on Torsion E3 Problem 67. Mark the wrong statement r 4  2 1 C 2  2 4l 60 (v) Work done in twisting the cylinder through an angle  is W  (a) Sliding of molecular layer is much easier than compression or expansion (b) Reciprocal of bulk modulus of elasticity is called compressibility ID (c) It is difficult to twist a long rod as compared to small rod (d) Hollow shaft is much stronger than a solid rod of same length and same mass Solution : (c) U Problem 68. A rod of length l and radius r is joined to a rod of length l / 2 and radius r / 2 of same D YG material. The free end of small rod is fixed to a rigid base and the free end of larger rod is given a twist of , the twist angle at the joint will be (a)  / 4 (b)  / 2 (c) 5 / 6 (d) 8 / 9 Solution : (d) If torque  is applied at the free end of larger rod and twist  is given to it then twist at joint is  1 and twist at the upper end (fixed base)  2 4  r (  1 ) 2l r 2 2(l / 2)    1   2   U   4 (   1 )  ( 1  0) 8 [As  2  0] ST   8  8 1   1  91  8   1  8. 9 Problem 69. The upper end of a wire of radius 4 mm and length 100 cm is clamped and its other end is twisted through an angle of 30 o. Then angle of shear is (a) 12 o Solution : (b) L   r (b) 0.12 o   (c) 1.2o [NCERT 1990; MP PMT 1996] (d) 0.012 o r 4  10 3  30 o  0.12 o  L 1 Problem 70. Two wires A and B of same length and of the same material have the respective radii r1 and r2. Their one end is fixed with a rigid support, and at the other end equal twisting couple is 36 Elasticity applied. Then the ratio of the angle of twist at the end of A and the angle of twist at the end of B will be [AIIMS 1980] r12 (b) r22 Solution : (c)  1   2   r141 2l1   r24 2 2l2  r22 (c) r12  1  r2     2  r1  r24 (d) r14 r14 r24 4 60 (a) Problem 71. The work done in twisting a steel wire of length 25 cm and radius 2mm through 45 o will be (a) 2.48 J (b) 3.1 J (c) 15.47 J (d) 18.79 J 3. 14  8  10 10  (2  10 3 ) 4  ( / 4 ) 2  r 4  2 1   2.48 J C 2  2 4l 4  25  10  2 9.22 Interatomic Force Constant. ID Solution : (a) W  E3 (  8  1010 N / m 2 ) U Behaviour of solids with respect to external forces is such that if their atoms are connected to springs. When an external force is applied on a solid, this distance between its atoms changes and interatomic force works to restore the original dimension. The ratio of interatomic force to that of change in interatomic distance is defined as the F r D YG interatomic force constant. K  It is also given by K  Y  r0 [Where Y = Young's modulus, r0 = Normal distance between the atoms of wire] Unit of interatomic force constant is N/m and Dimension MT–2 Note :  The number of atoms having interatomic distance r0 in length l of a wire, N U = l/r0.  The number of atoms in area A of wire having interatomic separation r0 is ST N  A / r02. Sample problems based on Interatomic Force Constant Problem 72. The mean distance between the atoms of iron is 3  10 10 m and interatomic force constant for iron is 7 N/m. The Young’s modulus of elasticity for iron is (a) 2.33  10 5 N / m 2 Solution : (d) Y  (b) 23.3  1010 N / m 2 (c) 233  1010 N / m 2 (d) 2.33  1010 N / m 2 k 7   2.33  10 10 N / m 2. 10 ro 3  10 Problem 73. The Young’s modulus for steel is Y  2  1011 N / m 2. If the inter-atomic distance is 3.2Å, the inter atomic force constant in N/ Å will be Elasticity 37 (b) 6.4  10 9 (a) 6.4  10 9 (d) 3.2  10 9 (c) 3.2  10 9 Solution : (b) k  Y  r0  2  10 11  3.2  10 10  6.4  10 1 N / m = 6.4  10 9 N / Å. 9.23 Elastic Hysteresis. 60 When a deforming force is applied on a body then the strain does not change simultaneously with stress rather it lags behind the stress. The lagging of strain behind the stress is defined as elastic hysteresis. This is the reason why the values of strain for same stress are different while increasing the load and while decreasing the load. E3 Hysteresis loop : The area of the stress-strain curve is called the hysteresis loop and it is numerically equal to the work done in loading the material and then unloading it. For B O ID Stress Stress For A O Strai n Strai n D YG U If we have two tyres of rubber having different hysteresis loop then rubber B should be used for making the car tyres. It is because of the reason that area under the curve i.e. work done in case of rubber B is lesser and hence the car tyre will not get excessively heated and rubber A should be used to absorb vibration of the machinery because of the large area of the curve, a large amount of vibrational energy can be dissipated. 9.24 Factors Affecting Elasticity. (1) Hammering and rolling : Crystal grains break up into smaller units by hammering and rolling. This result in increase in the elasticity of material. (2) Annealing : The metals are annealed by heating and then cooling them slowly. U Annealing results in decrease in the elasticity of material. (3) Temperature : Intermolecular forces decreases with rise in temperature. Hence the ST elasticity decreases with rise in temperature but the elasticity of invar steel (alloy) does not change with change of temperature. (4) Impurities : Due to impurities in a material elasticity can increase or decrease. The type of effect depends upon the nature of impurities present in the material. 9.25 Important Facts About Elasticity. (1) The body which requires greater deforming force to produce a certain change in dimension is more elastic. Example : Ivory and steel balls are more elastic than rubber. (2) When equal deforming force is applied on different bodies then the body which shows less deformation is more elastic. 38 Elasticity Example : (i) For same load, more elongation is produced in rubber wire than in steel wire hence steel is more elastic than rubber. (ii) Water is more elastic than air as volume change in water is less for same applied pressure. 60 (iii) Four identical balls of different materials are dropped from the same height then after collision balls rises upto different heights. The order of their height can be given by hivory > hsteel > hrubber > hclay because Yivory > Ysteel > Yrubber > Yclay. E3 (3) The value of moduli of elasticity is independent of the magnitude of the stress and strain. It depends only on the nature of material of the body. (4) For a given material there can be different moduli of elasticity depending on the type of Name of substance Young’s modulus (Y) 1010N/m2 Aluminium 6.9 Brass 9.0 Copper 11.0 ID stress applied and resulting strain. Bulk modulus (K) 1010N/m2 Modulus of rigidity () 1010N/m2 2.6 6.7 3.4 13.0 4.5 19.0 14.0 4.6 20.0 16.0 8.4 36.0 20.0 15.0 83.0 55.0 34.0 – 0.22 – Glycerin – 0.45 – Air – 1.01 – Steel Tungsten Diamond ST U Water D YG Iron U 7.0 (5) The moduli of elasticity has same dimensional formula and units as that of stress since strain is dimensionless.  Dimensional formula ML1 T 2 while units dyne/cm2 or Newton/m2. (6) Greater the value of moduli of elasticity more elastic is the material. But as Y  (1/l), K  (1/V) and   (1/) for a constant stress, so smaller change in shape or size for a given stress corresponds to greater elasticity. (7) The moduli of elasticity Y and  exist only for solids as liquids and gases cannot be deformed along one dimension only and also cannot sustain shear strain. However K exist for all states of matter viz. solid, liquid or gas. Elasticity 39 (8) Gases being most compressible are least elastic while solids are most i.e. the bulk modulus of gas is very low while that for liquids and solids is very high. Ksolid > Kliquid > Kgas (9) For a rigid body l, V or  = 0 so Y, K or  will be , i.e. elasticity of a rigid body is infinite. 60 Diamond and carborundum are nearest approach to rigid bodies. (10) In a suspension bridge there is a stretch in the ropes by the load of the bridge. Due to which length of rope changes. Hence Young’s modulus of elasticity is involved. E3 (11) In an automobile tyre as the air is compressed, volume of the air in tyre changes, hence the bulk modulus of elasticity is involved. ID (12) In transmitting power, an automobile shaft is sheared as it rotates, so shearing strain is set up, hence modulus of rigidity is involved. (13) The shape of rubber heels changes under stress, so modulus of rigidity is involved. 9.26 Practical Applications of Elasticity. U (i) The metallic parts of machinery are never subjected to a stress beyond elastic limit, otherwise they will get permanently deformed. D YG (ii) The thickness of the metallic rope used in the crane in order to lift a given load is decided from the knowledge of elastic limit of the material of the rope and the factor of safety. (iii) The bridges are declared unsafe after long use because during its long use, a bridge under goes quick alternating strains continuously. It results in the loss of elastic strength. (iv) Maximum height of a mountain on earth can be estimated from the elastic behaviour of earth. U At the base of the mountain, the pressure is given by P = hg and it must be less than elastic limit (K) of earth’s supporting material. ST K > P > hg  h K g or hmax  K g (v) In designing a beam for its use to support a load (in construction of roofs and bridges), it is advantageous to increase its depth rather than the breadth of the beam because the depression in rectangular beam.   l b d Wl 3 4Ybd 3  W To minimize the depression in the beam, it is designed as Ishaped girder. (vi) For a beam with circular cross-section depression is given by   WL 3 12  r 4 Y 40 Elasticity (vii) A hollow shaft is stronger than a solid shaft made of same mass, length and material. Torque required to produce a unit twist in a solid shaft  solid  r 4 (r24  r14 ) 2l …..(ii) 60 and torque required to produce a unit twist in a hollow shaft  hollow  From (i) and (ii),.…..(i) 2l  hollow r24  r14 (r22  r12 )(r22  r12 )    solid r4 r4 …..(iii) Substituting this value in equation (iii) we get, E3 Since two shafts are made from equal volume  r 2 l   (r22  r12 )l  r 2  r22  r12  hollow r22  r12   1  hollow > solid  solid r2 ST U D YG U ID i.e., the torque required to twist a hollow shaft is greater than the torque necessary to twist a solid shaft of the same mass, length and material through the same angle. Hence, a hollow shaft is stronger than a solid shaft. 60 Elasticity 37 Problems based on Interatomic and Intermolecular forces In solids, inter-atomic forces are E3 1. (a) Totally repulsive (b) Totally attractive (c) Combination of (a) and (b) None of these The potential energy U between two molecules as a function of the distance X between them has been shown in the figure. The two molecules are ID 2. (d) U +B veand C (a) Attracted when x lies between A and B and are repelled when X lies between (c) Attracted when they reach B 3. D YG (d) Repelled when they reach B U 0 and B (b) Attracted when x lies between B and C and are repelled when X lies between A – ve A X B C The nature of molecular forces resembles with the nature of the (a) Gravitational force (b) Nuclear force (c) Electromagnetic force (d) Weak force Problems based on Stress 4. The ratio of radius of two wire of same material is 2 : 1. Stretched by same force, then the ratio of stress is [PET 1991] (a) 2 : 1 5. (d) 4 : 1 U (b) Compressive stress (c) Tangential stress (d) Working stress ST A vertical hanging bar of length l and mass m per unit length carries a load of mass M at the lower end, its upper end is clamped to a rigid support. The tensile force at a distance x from support is (a) Mg + mg(l – x) 7. (c) 1 : 4 If equal and opposite forces applied to a body tend to elongate it, the stress so produced is called (a) Tensile stress 6. (b) 1 : 2 (b) Mg (c) Mg + mgl (d) (M  m )g x l One end of a uniform rod of mass m1 and cross-sectional area A is hung from a ceiling. The other end of the bar is supporting mass m2. The stress at the midpoint is (a) g(m 2  2m1 ) 2A (b) g(m 2  m1 ) 2A (c) g(2m 2  m1 ) 2A m1 m2 38 Elasticity (d) 8. g(m 2  m1 ) A A uniform bar of square cross-section is lying along a frictionless horizontal surface. A horizontal force is applied to pull it from one of its ends then F 60 (a) The bar is under same stress throughout its length (b) The bar is not under any stress because force has been applied only at one end (c) The bar simply moves without any stress in it E3 (d) The stress developed reduces to zero at the end of the bar where no force is applied Problems based on Strain 9. Which one of the following quantities does not have the unit of force per unit area (a) Stress (b) Strain (a) Volume stress 11. (d) Metallic strain (b) Longitudinal (c) Volume (d) Transverse D YG (b) Fluids (c) Solids (d) Liquids The face EFGH of the cube shown in the figure is displaced 2 mm parallel to itself when forces of 5  10 5 N each are applied on the lower and upper faces. The lower face is fixed. The strain produced in the cube is (a) 2 (b) 0.5 (c) 0.05 E F H G C D A U (d) 1. 2  10 8 14. (c) Longitudinal strain The longitudinal strain is only possible in (a) Gases 13. (b) Shearing strain [EAMCET 1980] When a spiral spring is stretched by suspending a load on it, the strain produced is called (a) Shearing 12. Pressure The reason for the change in shape of a regular body is U 10. (d) ID (c) Young’s modulus of elasticity B 4 cm Forces of 10 5 N each are applied in opposite direction on the upper and lower faces of a cube of side 10 cm, shifting the upper face parallel to itself by 0.5 cm. If the side of the cube were 20 cm, the displacement would ST be E F D C (a) 1 cm (b) 0.5 cm A B (c) 0.25 cm (d) 0.125 cm Problems based on Stress strain curve The stress versus strain graphs for wires of two materials A and B are as shown in the figure. If YA and YB are the Young’s modulii of the materials, then Y Stress 15. O A 60o 30o Strain B X Elasticity 39 (a) YB  2Y A (b) Y A  YB (c) YB  3Y A (d) Y A  3YB The graph is drawn between the applied force F and the strain (x) for a thin uniform wire. The wire behaves as a liquid in the part 60 16. [CPMT 1988] F d E3 (b) bc c (c) cd O (d) oa X x The diagram shows stress v/s strain curve for the materials A and B. From the curves we infer that ID 17. b a (a) ab (c) Both A and B are ductile The figure shows the stress-strain graph of a certain substance. Over which region of the graph is Hooke’s law obeyed (a) AB (b) BC (c) CD (d) ED A D C B Strain E Which one of the following is the Young’s modulus (in N/m2) for the wire having the stress-strain curve shown U 19. Strain Stress 18. D YG (d) Both A and B are brittle B U (b) A is ductile and B is brittle Stress A (a) A is brittle but B is ductile 10 8 6 4 2 Stress (10 N/m2) ST 7 in the figure (a) 24  10 11 (b) 8. 0  10 11 (c) 10  10 11 O 2 4 (d) 2.0  10 11 6 8  10– 4 Strai n Problems based on Young's Modulus The adjacent graph shows the extension (l) of a wire of length 1m suspended from the top of a roof at one end with a load W connected to the other end. If the cross sectional area of the wire is 10–6m2, calculate the young’s modulus of the material of the wire –4 ) [IIT-JEE (Screening) 2003] 4 3 2 l(10 m 20. 1 20 40 60 80 W(N) 40 Elasticity (a) 2  10 11 N / m 2 (b) 2  10 11 N / m 2 (c) 3  10 12 N / m 2 (d) 2  10 13 N / m 2 In the Young’s experiment, if length of wire and radius both are doubled then the value of Y will become[RPET 2003] (a) 2 times 22. (b) 4 times (c) Remains same 2 60 21. (d) Half A rubber cord catapult has cross-sectional area 25mm and initial length of rubber cord is 10cm. It is stretched 23. missile is [CPMT 2002] (a) 20 ms–1 (b) 100 ms–1 E3 to 5cm. and then released to project a missile of mass 5gm. Taking Yrubber  5  10 8 N / m 2 velocity of projected (c) 250 ms–1 Consider the following statements (d) 200 ms–1 Assertion (A) : Stress is the internal force per unit area of a body. ID Reason (R) : Rubber is more elastic than steel. Of these statements [AIIMS 2002] (a) Both A and R are true and the R is a correct explanation of the A (b) Both A and R are true but the R is not a correct explanation of the A U (c) A is true but the R is false (d) Both A and R are false (e) A is false but the R is true The area of cross-section of a steel wire (Y  2.0  10 11 N / m 2 ) is 0.1 cm2. The force required to double its length will be (a) 2  10 N 12 25. (c) 2  10 N (d) 2  10 N 6 U (b) Y A  t (c) YL t A (d) Y AL t (b) Glass (c) Steel (d) Copper ST There are two wires of same material and same length while the diameter of second wire is 2 times the diameter of first wire, then ratio of extension produced in the wires by applying same load will be (a) 1 : 1 28. [MP PET 2002] 10 Which one of the following substances possesses the highest elasticity [MP PMT 1992; RPMT 1999; RPET 2000; MH CET (Me (a) Rubber 27. (b) 2  10 N 11 A metal bar of length L and area of cross-section A is clamped between two rigid supports. For the material of the rod, its Young’s modulus is Y and coefficient of linear expansion is . If the temperature of the rod is increased by t o C , the force exerted by the rod on the supports is (a) Y AL t 26. D YG 24. (b) 2 : 1 (c) 1 : 2 (d) 4 : 1 Consider the following statements Assertion (A) : Rubber is more elastic than glass. Reason (R) : The rubber has higher modulus of elasticity than glass. Of these statements (a) Both A and R are true and the R is a correct explanation of the A (b) Both A and R are true but the R is not a correct explanation of the A (c) A is true but the R is false (d) Both A and R are false (e) A is false but the R is true [AIIMS 2000] Elasticity 41 (a) Thin block of any cross section (b) Thick block of any cross section (c) Long thin wire (d) Short thin wire In suspended type moving coil galvanometer, quartz suspension is used because (a) It is good conductor of electricity (b) Elastic after effects are negligible (c) Young’s modulus is greater (b) B (d) All have the same elasticity (c) 2n times (d) None of the above ID (b) n times A wire of radius r, Young’s modulus Y and length l is hung from a fixed point and supports a heavy metal cylinder of volume V at its lower end. The change in length of wire when cylinder is immersed in a liquid of density  is in fact (a) Decrease by (c) Decrease by (d) V g Y Vlg Y r 2 Vrg Y l 2 D YG (b) Increase by Vg Y r If the ratio of lengths, radii and Young’s modulii of steel and brass wires in the figure are a, b and c respectively. Then the corresponding ratio of increase in their lengths would be (a) (b) 3a 2b 2 c 2 ac b2 ST (c) 2a 2c b U 34. (c) C The ratio of diameters of two wires of same material is n : 1. The length of wires are 4 m each. On applying the same load, the increase in length of thin wire will be (a) n2 times 33. There is no elastic limit You are given three wires A, B and C of the same length and cross section. They are each stretched by applying the same force to the ends. The wire A is stretched least and comes back to its original length when the stretching force is removed. The wire B is stretched more than A and also comes back to its original length when the stretching force is removed. The wire C is stretched most and remains stretched even when stretching force is removed. The greatest Young’s modulus of elasticity is possessed by the material of wire (a) A 32. (d) U 31. 60 30. The longitudinal extension of any elastic material is very small. In order to have an appreciable change, the material must be in the form of E3 29. Steel M Brass 2M 3c (d) 2 ab 2 35. A uniform heavy rod of weight W, cross sectional area A and length L is hung from a fixed support. Young’s modulus of the material of the rod is Y. If lateral contraction is neglected, the elongation of the rod under its own weight is 2WL WL WL (a) (b) (c) (d) Zero AY AY 2 AY 36. A constant force F0 is applied on a uniform elastic string placed over a smooth horizontal surface as shown in figure. Young’s modulus of string is Y and area of cross-section is S. The strain produced in the string in the direction of force is (a) F0 Y S F0 42 Elasticity F0 SY (c) F0 2 SY (d) F0 Y 2S A uniform rod of length L has a mass per unit length  and area of cross section A. The elongation in the rod is l 60 37. (b) due to its own weight if it is suspended from the ceiling of a room. The Young’s modulus of the rod is (a) (b) gL2 (c) 2 Al 2 gL Al (d) gl 2 AL AB is an iron wire and CD is a copper wire of same length and same cross-section. BD is a rod of length 0.8 m. A load G = 2kg-wt is suspended from the rod. At what distance x from point B should the load be suspended for E3 38. 2 gL2 Al the rod to remain in a horizontal position (YCu  11.8  10 10 N / m 2 , Y Fe  19.6  10 10 N / m 2 ) C A T1 G U (d) 0.7 m (a) FL r12 Y D YG A slightly conical wire of length L and end radii r1 and r2 is stretched by two forces F, F applied parallel to length in opposite directions and normal to end faces. If Y denotes the Young’s modulus, then extension produced is (b) FL r1Y (c) FL r1r2 Y (b) FKA (c) F K 3 (d) FK AL Y U 1 3 Force (b) ST 1 (c) 2 (d) O 3 2 30o X Displacemen t The force constant of a wire does not depend on (a) Nature of the material 43. FLY r1r2 The value of force constant between the applied elastic force F and displacement will be (a) 42. (d) The force constant of wire is K and its area of cross-section is A. If the force F is applied on it, then the increase in its length will be (a) KA 41. D x (c) 0.5 m 40. T2 B (b) 0.3 m 39. O ID (a) 0.1 m (b) Radius of the wire (c) Length of the wire (d) A metal wire of length L, area of cross-section A and Young’s modulus Y behaves as a spring. The equivalent spring constant will be (a) 44. Y AL (b) YA L (c) YL A (d) L AY A highly rigid cubical block A of small mass M and side L is fixed rigidly onto another cubical block B of the same dimensions and modulus of rigidity  such that the lower face of A completely covers the upper face of B. The lower face of B is rigidly held on a horizontal surface. A small force is applied perpendicular to one of the Elasticity 43 sides faces of A. After the force is withdrawn, block A execute small oscillations the time period of which is given by (a) 2 ML (b) 2 M L (c) 2 ML (d) 2  M L 45. A wire of length L and cross-sectional area A is made of a material of Young’s modulus Y. It is stretched by an amount x. The work done is YxA 2L (b) Yx 2 A L (c) Yx 2 A 2L Two wires of same diameter of the same material having the length l and 2l. If the force F is applied on each, the ratio of the work done in the two wires will be (a) 1 : 2 (b) 1 : 4 ID (b) 5V (c) V/5 [CPMT 1976] (d) 25V The strain energy stored in a body of volume V due to shear S and shear modulus  is (a) S 2V 2 (b) SV 2 2 (c) S 2V  (d) 1 S 2 V 2 K is the force constant of a spring. The work done in increasing its extension from l1 to l2 will be (a) K(l2  l1 ) D YG 49. (d) 1 : 1 If the potential energy of a spring is V on stretching it by 2 cm, then its potential energy when it is stretched by 10 cm will be (a) V/25 48. (c) 2 : 1 U 47. 2 Yx 2 A L (d) E3 (a) 46. 60 Problems based on Stretching a wire (b) K (l2  l1 ) 2 (c) K (l22  l12 ) (d) K 2 2 (l2  l1 ) 2 Problems based on Breaking of wire 50. The breaking stress of a wire depends upon (a) Length of the wire 51. (b) Radius of the wire [AIIMS 2002] (c) Material of the wire (d) Shape of the cross section An aluminium rod has a breaking strain of 0.2%. The minimum cross sectional area of the rod, in m2, in order U to support a load of 104 N is (Y  7  10 9 N / m 2 ) (a) 1.4  10 4 (c) 1.4  10 3 (d) 7.1  10 5 A cable is replaced by another one of the same length and material but of twice the diameter. The maximum load that the new wire can support without exceeding the elastic limit, as compared to the load that the ST 52. (b) 7.1  10 4 original wire could support, is (a) Half 53. (b) Double (d) One-fourth A heavy mass is attached to a thin wire and is whirled in a vertical circle. The wire is most likely to break (a) When the mass is at the highest point (c) When the wire is horizontal the upward vertical 54. (c) Four times (b) When the mass is at the lowest point (d) At an angle of cos–1 (1/3) from A heavy uniform rod is hanging vertically from a fixed support. It is stretched by its own weight. The diameter of the rod is (a) Smallest at the top and gradually increases down the rod (b) Largest at the top and gradually decreases down the rod (c) Uniform everywhere 44 Elasticity (d) Maximum in the middle Problems based on Bulk modulus The isothermal bulk modulus of a gas at atmospheric pressure is (a) 1 mm of Hg (b) C p / Cv (b) 2  10 8 (b) 4  10 5 cc (c) 0.025 cc (d) 0.004 cc An ideal gas of mass m, volume V, pressure p and temperature T undergoes a small change in state at constant Cp temperature. Its adiabatic exponent i.e., is . The bulk modulus of the gas at the constant temperature Cv U m p pV (d) T T An ideal gas of mass m, volume V, pressure p and temperature T undergoes a small change under a condition that heat can neither enter into it from outside nor can it leave the system. Such a process is called adiabatic process. (b)  p (c) D YG (a) p  The bulk modulus of the gas     Cp   is C v  m p pV (d) T T An ideal gas whose adiabatic exponent is  is expanded according to the law p= V where  is a constant. For this process the bulk modulus of the gas is p (a) p (b) (c) p (d) (l – )p (a) p 61. (d) 2  10 9 (c) 10 9 process called isothermal process is 60. (d) 1 / C p Cv The compressibility of water is 4  10–5 per unit atmospheric pressure. The decrease in volume of 100 cubic centimetre of water under a pressure of 100 atmosphere will be (a) 0.4 cc 59. (c) C p Cv If a rubber ball is taken at the depth of 200 m in a pool. Its volume decreases by 0.1%. If the density of the water is 1  10 3 kg / m 3 and g = 10 m/s2, then the volume elasticity in N/m2 will be (a) 10 8 58. 60 The specific heat at constant pressure and at constant volume for an ideal gas are Cp and Cv and its adiabatic and isothermal elasticities are E and E respectively. The ratio of E to E is (a) Cv / C p 57. (d) 2.026  10 5 N / m 2 E3 56. (c) 1.013  10 5 N / m 2 (b) 13.6 mm of Hg ID 55. (b)  p (c)  1 c.c. of water is taken from the top to the bottom of a 200 m deep lake. What will be the change in its volume if U 62. K of water is 2. 2  10 9 N / m 2 (b) 8.8  10 2 c.c. ST (a) 8.8  10 6 c.c. 63. (d) 8.8  10 1 c.c Problems based on Modulus of rigidity Modulus of rigidity of a liquid (a) Non zero constant 64. (c) 8.8  10 4 c.c. (b) Infinite (c) Zero The Young’s modulus of the material of a wire is 6  10 12 N /m 2 (d) Cannot be predicted and there is no transverse strain in it, then its modulus of rigidity will be (a) 3  10 12 N / m 2 (b) 2  10 12 N / m 2 (c) 10 12 N / m 2 (d) None of the above Problems based on relation between Y, , K and  65. The value of Poisson’s ratio lies between 2002] [AIIMS 1985; MP PET 1986; DPMT Elasticity 45 (a) –1 to 70. 71. 1 to 1 2 (d) 1 to 2 Which of the following will be  if Y = 2.4 (a) –1 (b) 0.2 (c) 0.1 (d) – 0.25 Which is correct relation (a) Y <  (b) Y >  (c) Y =  (d)  = +1 The relationship between Young’s modulus Y, bulk modulus K and modulus of rigidity  is 9K   3K (a) Y  69. (c)  (b)   9 yK Y  3K The Poisson’s ratio cannot have the value (a) 0.7 (b) 0.2 Which of the following relations is true 9Y (a) 3Y = K(1 – ) (b) K  Y  Shearing stress Shearing strain [RPET 2001] 3K 9  K (d) Y  [EAMCET 1989] (c) 0.1 (d) 0.5 (c)  = (6K + ) Y The wrong relation for modulus of rigidity ( ) is (a)   9K 3  K (c) Y  [RPET 2001] 60 68. 1 3 to  4 2 E3 67. (b)  (b) Unit of  is N / m 2 (c)   Y 2(1   ) ID 66. 1 2 (d)   (d)   [CPMT 1984] 0.5 Y    Y 2(1   ) A rod of 2m length and radius 1 cm is twisted at one end by 0.8 rad with respect to other end being clamped. The shear strain developed in its rod will be (a) 0.002 73. (b) 0.004 D YG 72. U Problems based on Torsion (c) 0.008 (d) 0.016 The upper end of a wire 1 metre long and 2 mm in radius is clamped. The lower end is twisted through an angle of 45 o. The angle of shear is (a) 0.09 o 74. (b) 1. 6  10 3 U ST 15  16 (c) 16  10 3 (d) 16  10 6 (b) 16  15 (c) 16  17 (d) 17  16 Problems based on Interatomic force constant If the interatomic spacing in a steel wire is 3.0Å and Y steel  20  10 10 N / m 2 , then force constant is (a) 6  10 2 N / Å 77. (d) 90 o Two cylinders A and B of the same material have same length, their radii being in the ratio of 1 : 2 respectively. The two are joined in series. The upper end of A is rigidly fixed. The lower end of B is twisted through an angle , the angle of twist of the cylinder A is fig. (a) 76. (c) 9 o The end of a wire of length 0.5m and radius 10–3m is twisted through 0.80 radian. The shearing strain at the surface of wire will be (a) 1. 6  10 3 75. (b) 0. 9 o (b) 6  10 9 N / Å (c) 4  10 5 N / Å (d) 6  10 5 N / Å The Young’s modulus of a metal is 1. 2  10 11 N / m 2 and the inter-atomic force constant is 3.6  10 9 N / Å. The mean distance between the atoms of the metal is (a) 2Å 78. (b) 3 Å The interatomic distance for a metal is 3  10 (c) 4.5 Å 10 (d) 5 Å m. If the interatomic force constant is 3.6  10 9 N / Å , then the Young’s modulus in N / m 2 will be (a) 1. 2  10 11 (b) 4.2  10 11 (c) 10.8  10 19 (d) 2.4  10 10 46 Elasticity Miscellaneous problems (a) About x = 0, with   k / m (b) About x = 0,with   k / m (c) About x = F0/k with   k / m (d) About x = F0/k with   k / m The extension in a string obeying Hooke’s law is x. The speed of sound in the stretched string is v. If the extension in the string is increased to 1.5x, the speed of sound will be (a) 1.22 v 81. 60 80. A particle of mass m is under the influence of a force F which varies with the displacement x according to the relation F  kx  F0 in which k and F0 are constants. The particle when disturbed will oscillate (b) 0.61 v (c) 1.50 v Railway lines and girders for buildings, are I shaped, because (d) 0.75 v E3 79. (a) The bending of a girder is inversely proportional to depth, hence high girder bends less (b) The coefficient of rigidity increases by this shape (c) Less volume strain is caused If Young’s modulus for a material is zero, then the state of material should be (a) Solid 83. (b) Solid but powder The elasticity of invar (c) Does not depend on temperature (d) None of the above D YG (b)  : 1 (c) 3 :  (d) 1 :  m  2 L2 AY (b) m  2 L2 2 AY (c) m  2 L2 3 AY (d) 2m  2 L2 AY U A steel wire is suspended vertically from a rigid support. When loaded with a weight in air, it extends by la and when the weight is immersed completely in water, the extension is reduced to lw. Then the relative density of the material of the weight is (a) la lw (b) la la  lw (c) la la  lw (d) lw la The twisting couple per unit twist for a solid cylinder of radius 4.9 cm is 0.1 N-m. The twisting couple per unit twist for a hollow cylinder of same material with outer and inner radii of 5 cm and 4 cm respectively, will be ST 87. (b) Decreases with temperature rise A uniform rod of mass m, length L, area of cross-section A is rotated about an axis passing through one of its ends and perpendicular to its length with constant angular velocity  in a horizontal plane. If Y is the Young’s modulus of the material of rod, the increase in its length due to rotation of rod is (a) 86. (d) None of the above For the same cross-sectional area and for a given load, the ratio of depressions for the beam of square crosssection and circular cross-section is (a)  : 3 85. (c) Gas U (a) Increases with temperature rise 84. ID (d) This keeps the surface smooth 82. (a) 0.64 N-m (b) 0.64  10–1 N-m (c) 0.64  10–2 N-m (d) 0.64  10–3 N-m 46 46 Elasticity 2. 3. 4. 5. 6. 7. c b c c a a c 11. 12. 13. 14. 15. 16. a c c c d b 21. 22. 23. 24. 25. 26. c c c d b c 31. 32. 33. 34. 35. 36. a a a b c 41. 42. 43. 44. b d b d 51. 52. 53. 54. b c b a 62. 63. c a 71. c 81. 10. b b b 19. 20. b d d a 27. 28. 29. 30. d d c b 37. 38. 39. 40. c b b c c 45. 46. 47. 48. 49. 50. c a d d d c 55. 56. 57. 58. 59. 60. c b d a a b 64. 65. 66. 67. 68. 69. 70. c a a b b a a d 72. 73. 74. 75. 76. 77. 78. 79. 80. b a a c b b a c a 82. 83. 84. 85. 86. 87. b c c c b b U ID 18. ST U a 9. 17. D YG 61. 8. E3 1. 60 Answer Sheet (Practice problems)