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Ans:
Let a person of mass m is standing on a weighing machine in lift. Actual weight of person = mg. Machine offers a reaction R given by reading of weighing machine. Reaction gives the apparent weight of the person.
In first three cases.
(1) When lift is at rest;
(2) When lift is moving up with uniform velocity and
(3) When the lift is moving downward with uniform velocity, we have a = 0.
Net force on the person =ma=m x 0=0
                                    R -mg =0
or                                 [ R =mg ]
i.e., apparent weight of the person is equal to the actual weight of the person.
(4) When lift is moving upward with acceleration a, then
                                    R -mg =ma
                                    R =mg +ma m ( g+a )
i.e., apparent weight of the person is more than the actual weight.
(5) When lift is moving downward wigh acceleration a, then
                                     mg -R =ma
                                            R =mg-ma =m ( g-a )
i.e., apparent weight of the person is less than the actual weight.
(6) If lift is moving down with a = g, then substituting this value, in above equation, we have
                                       R =m ( g - g ) = 0
i.e, the body will appear weightless.

Ans:
When a deforming force is applied it deforms the shape of the body. Due to deformation in shape, a restoring force is produced in the body which tries to restore back the body to its original position. This internal restoring force acting per unit area is called stress.
Stress = Restoring force/Area
Its S.I unit is Nm^-2 and its dimensions are [ML^-1T^-2]
Types of stress. Normal stress . When deforming and hence restoring force acts normally over the area of the body, we call the restoring force per unit area as normal stress.
There are two types of normal stress.
(i) Tensile stress : Stress is called tensile stress if the increase in length of the body is in the direction of force applied.
(ii) Compressive stress : Stress is called compressive stress if there is a decrease in length of the wire or there is a compression of the body due to force applied.
Tangential stress: When deforming force acting tangent to the surface of the body produces a change in shape of the body, then stress set up in the body is called tangential stress.
Strain: When a deforming force is applied on a a body, there is a change in some configuration. This rate of change in configuration to the original configuration is called strain.
Strain = Change in configuration/Original configuration
Strain has no units or dimensions.
Types of strain There are four types of strain.
(i) Longitudinal strain = Change in length/Original length
(ii) Lateral strain = Change in diameter/Original diameter
(iii) Volumetric strain = Change in volume/Original volume
(iv) Shearing strain or shear is defined as the angle in radian through which a plane perpendicular to the fixed surface of the cubical body gets turned under the effect of tangential force. In figure above shear = q = D l/l = Dl/l

Ans:
Let us first show that First law is contained in the second law. According to Newton's second law of motion F = ma.
    If no external force is applied to the body.
i.e. F = 0 or 0 = ma
or
a = 0 since m =/ 0
                   v-u/t = 0 = u
i.e. if no external force is applied the velocity of the body will remain the same i.e., if u = 0, v= 0 and if u = 5; v = 5ms^-1 which is the first law.
Third law is contained in the second law. Let the two bodies A and B collide. A is applying a force^-> F₁ on B for Dt time and B is applying ^->F₂ on A for Dt.
Since change in momentum = Impluse.
Similarly change in momentum of A = ^->F₂ x Dt.
and change in momentum of B = ^->F₁ x Dt.
Total change in momentum = ^->F₁ Dt + ^->F₂ Dt
Since no external force is acting on the system, therefore the total change of linear momenturn is zero (Newton's second law).
Similarly, ^->F₁ Dt + ^->F₂ Dt = 0
or             ^->F₁ x Dt = - ^->F₂ x Dt
or      Action = - (Reaction)
i.e., action and reaction are equal and opposite which is Newton's third law of motion.
Hence we find that Newton's first and third laws can be deduced from second law of motion and it is the real law of motion.

Ans:
(a) We know that work done during expansion is equal to area between P-V and volume axis. Area below PV diagram depends upon the slope of P - V diagram. More the slope, lesser the area and lesser the work done by the gas.
Slope of isothermal or aadiabatic curve is given by dP/dV
For isothermal operation :           PV = K
                           P.dV + V.dP = 0
or                          V . dP = - P .dV

or          (dP/dV)_iso   = - (P/V)          (i)
For adiabatic operation :              PV^g= K
                PgV^{g -1} .dV + Vg . dP = 0
or                       Vg dP = -g PV^{g -1}dV
or (dP/dV) _ad= -gP-/V                       (ii)
From (i) and (ii) we find that slopes of adiabatic curve is g times isothermal curve and g > 1.
Similarly area below P - V diagram and volume axis in case of adiabatic operation is less than that below isothermal operation.
Hence, more work is done during isothermal operation than in case of adiabatic operation.
(b) Here, T₁ = 288 K ; T₂ = ? ; V₁= x ; V₂ = 8x ; g = 5/3
        T₂V₂^g-1 = (T₁ V₁)^{g -1}
or          T₂ = T₁ (V₁/V₂)^{g -1}    = 288 (x/8x)^5/3 ^-1
or          T₂ = 288/(8) ^2/3
          Log T₂ = log 288 - 2/3 log 8 = 2.4594 - 2/3 (o.9031) = 1.8573
         T₂ = Antilog 1.8573 = 71.99 K = 72 K
   Fall in temperature      = 288 - 72 = 216 K = 216 oC.

Ques 27. Determine apparent weight of a person in a lift when (1) lift is at rest; (2) lift is moving upward with uniform velocity; (3) lift is moving downward with uniform velocity ; (4) lift is moving up with acceleration a, (5) lift is moving down with acceleration a ; (6) lift is moving down with a = g.
Ans: Let a person of mass m is standing on a weighing machine in lift. Actual weight of person = mg. Machine offers a reaction R given by reading of weighing machine. Reaction gives the apparent weight of the person. In first three cases. (1) When lift is at rest; (2) When lift is moving up with uniform velocity and (3) When the lift is moving downward with uniform velocity, we have a = 0. Net force on the person =ma=m x 0=0 R -mg =0 or [ R =mg ] i.e., apparent weight of the person is equal to the actual weight of the person. (4) When lift is moving upward with acceleration a, then R -mg =ma R =mg +ma m ( g+a ) i.e., apparent weight of the person is more than the actual weight. (5) When lift is moving downward wigh acceleration a, then mg -R =ma R =mg-ma =m ( g-a ) i.e., apparent weight of the person is less than the actual weight. (6) If lift is moving down with a = g, then substituting this value, in above equation, we have R =m ( g - g ) = 0 i.e, the body will appear weightless.

Ques 28. What is stress and strain ? Describe briefly different types of stress and strain. Give units and dimensions of stress and strain.
Ans: When a deforming force is applied it deforms the shape of the body. Due to deformation in shape, a restoring force is produced in the body which tries to restore back the body to its original position. This internal restoring force acting per unit area is called stress. Stress = Restoring force/Area Its S.I unit is Nm^-2 and its dimensions are [ML^-1T^-2] Types of stress. Normal stress . When deforming and hence restoring force acts normally over the area of the body, we call the restoring force per unit area as normal stress. There are two types of normal stress. (i) Tensile stress : Stress is called tensile stress if the increase in length of the body is in the direction of force applied. (ii) Compressive stress : Stress is called compressive stress if there is a decrease in length of the wire or there is a compression of the body due to force applied. Tangential stress: When deforming force acting tangent to the surface of the body produces a change in shape of the body, then stress set up in the body is called tangential stress. Strain: When a deforming force is applied on a a body, there is a change in some configuration. This rate of change in configuration to the original configuration is called strain. Strain = Change in configuration/Original configuration Strain has no units or dimensions. Types of strain There are four types of strain. (i) Longitudinal strain = Change in length/Original length (ii) Lateral strain = Change in diameter/Original diameter (iii) Volumetric strain = Change in volume/Original volume (iv) Shearing strain or shear is defined as the angle in radian through which a plane perpendicular to the fixed surface of the cubical body gets turned under the effect of tangential force. In figure above shear = q = D l/l = Dl/l

Ques 29. Show that the second law is the real law of motion, first and third can be deduced from it.
Ans: Let us first show that First law is contained in the second law. According to Newton's second law of motion F = ma. If no external force is applied to the body. i.e. F = 0 or 0 = ma or a = 0 since m =/ 0 v-u/t = 0 = u i.e. if no external force is applied the velocity of the body will remain the same i.e., if u = 0, v= 0 and if u = 5; v = 5ms^-1 which is the first law. Third law is contained in the second law. Let the two bodies A and B collide. A is applying a force^-> F₁ on B for Dt time and B is applying ^->F₂ on A for Dt. Since change in momentum = Impluse. Similarly change in momentum of A = ^->F₂ x Dt. and change in momentum of B = ^->F₁ x Dt. Total change in momentum = ^->F₁ Dt + ^->F₂ Dt Since no external force is acting on the system, therefore the total change of linear momenturn is zero (Newton's second law). Similarly, ^->F₁ Dt + ^->F₂ Dt = 0 or ^->F₁ x Dt = - ^->F₂ x Dt or Action = - (Reaction) i.e., action and reaction are equal and opposite which is Newton's third law of motion. Hence we find that Newton's first and third laws can be deduced from second law of motion and it is the real law of motion.

Ques 30. (a) A gas is allowed to expand to n times the volume (i) isothermally ; (ii) adiabatically. In which case work done is more ? (b) Calculate the fall in termperature of the baloon in ^oC if its initial temperature is 15^oC when it is suddenly expanded to 8 times its original volume. Given g = 5/3
Ans: (a) We know that work done during expansion is equal to area between P-V and volume axis. Area below PV diagram depends upon the slope of P - V diagram. More the slope, lesser the area and lesser the work done by the gas. Slope of isothermal or aadiabatic curve is given by dP/dV For isothermal operation : PV = K P.dV + V.dP = 0 or V . dP = - P .dV or (dP/dV)_iso = - (P/V) (i) For adiabatic operation : PV^g= K PgV^{g -1} .dV + Vg . dP = 0 or Vg dP = -g PV^{g -1}dV or (dP/dV) _ad= -gP-/V (ii) From (i) and (ii) we find that slopes of adiabatic curve is g times isothermal curve and g > 1. Similarly area below P - V diagram and volume axis in case of adiabatic operation is less than that below isothermal operation. Hence, more work is done during isothermal operation than in case of adiabatic operation. (b) Here, T₁ = 288 K ; T₂ = ? ; V₁= x ; V₂ = 8x ; g = 5/3 T₂V₂^g-1 = (T₁ V₁)^{g -1} or T₂ = T₁ (V₁/V₂)^{g -1} = 288 (x/8x)^5/3 ^-1 or T₂ = 288/(8) ^2/3 Log T₂ = log 288 - 2/3 log 8 = 2.4594 - 2/3 (o.9031) = 1.8573 T₂ = Antilog 1.8573 = 71.99 K = 72 K Fall in temperature = 288 - 72 = 216 K = 216 oC.