Work, Energy and Power

56. An object A moving horizontally with KE 800 J experiences a constant opposing force of 100 N while moving from a place X to Y, where XY = 2m. What is energy of A at Y? What further distance will it move if opposing force continues? [600J, 6m ] Hints: Calculate work done by the opposing force. Work done is equal to the decrease of energy. So, subtract it from the initial KE. Again apply the same relation W = F × d to calculate further distance traveled by the body.
57. A ball of mass 0.1 kg is thrown vertically upward with speed 20 m/s; find PE at the maximum height. Also calculate the KE and PE of the ball half way up. [ 20 J, 10 J, 10J] Hints: The KE of the ball is given by E = . According to conservation of energy, this energy is equal to the PE at the maximum height. At half way up, PE is half of the energy of the maximum height.

58. A 4 kg ball moving with velocity 10 m/s collides with a 16 kg ball moving with velocity 4 m/s (i) in same direction, (ii) in opposite direction. Calculate the velocity of the ball in each case if they coalesce on impact and loss of energy. [5.2m/s, 57.6 J, 1.2 m/s, 313.6 J ]
Hints: Apply the principle of conservation of linear momentum and determine common velocity. i.e. m1u1 + m2u2 = (m1+m2)V. In second case, since the second body is moving in opposite direction, its velocity will be negative. Also, Initial kinetic energy E1 = + Final kinetic energy E2 = and therefore, loss of energy E = E1 – E2
59. A stationary mass explodes into two parts of mass 4 units and 40 units respectively. If the larger mass has KE 100 J, what is the energy of the smaller mass? [1000 J] Hints: Initial momentum of the body before explosion is zero. So, from the principle of conservation of linear momentum, sum of final momentum is also zero i.e. after explosion, each mass possess equal momentum. So, KE of first mass E1 = …………(i) and KE of second mass E2 = …………(ii) Dividing (i) by (ii)
60. A bullet of mass 10 g is fired vertically with a velocity 100 m/s into a block of wood of mass 190 g suspended by a long string. If the bullet comes into rest in the block, through what height does the block move? [1.25m]
Hints: At first use principle of conservation of linear momentum and determine common velocity. i.e. m1u1 + m2u2 = (m1+m2)V Again apply conservation of energy i.e. KE = PE and determine height.
61. A car of mass 1000 kg moves at a constant speed 20 m/s along horizontally where frictional force is 200 N. Calculate the power developed by the engine. If the car now moves up an inclined plane of inclination 1 / 20 at the same speed calculate new power developed.
[4, 14 kw]
Hints: Power developed by engine is given by P = ff . v where ff be the opposing or frictional force. In second case, total opposing force is the sum of frictional force and component of gravitational force along the plane i.e. f + mg Sin. So, power developed by the engine P = (ff + mg Sin). V
62. A train of mass 2 x 105 kg moves with speed 72 km / hr up a straight inclined plane against a frictional force 1.28 x 104 N. The inclination is such that it rises vertically 1.0 m for every 100 m traveled along the inclination. Calculate (i) the rate of increase per second of PE, (ii) the power developed by the train. [ 400 kw, 656 kw] Hints: Rate of increase of PE per second is which reduces into mgv Sin .
63. A stationary nucleus of mass 210 units disintegrates into an alpha particle of mass 4 units and residual nucleus of mass 206 units. If the KE of the alpha particle is E, calculate KE of the residual nucleus. [2E/103] Hints: Same as that of Q 44.

64. A bullet of mass 10 g traveling with velocity 300 m/s, strikes a wood of mass 290 g and embedded into it, then they come into rest after covering distance 15 m. Calculate the coefficient of sliding fraction. [ 1/3 ] Hints: At first, determine common velocity from conservation of linear momentum. From equation of motion, determine retardation and use formula coefficient of friction .
65. An iron block of mass 10 kg, rest on a wooden plane inclined at 300 to the horizon. It is found that the least force parallel to the plane that causes the block to slide up the plane is 100 N. Calculate the coefficient of sliding friction. [ ] Hints: According to question mg Sin + ff = 100. Calculate ff and then use relation to calculate coefficient of friction.

66. A body just slides down in an inclined plane when inclination of the plane is 150. If the inclination of the plane in increased to 300, what will be the acceleration of the body? [2.68m/s2] Hints: The condition to just slide is mg Sin = ff and = Tan, where  is the angle of repose. Calculate μ from this relation. In second case, the net downward force F = mg Sin – ff and acceleration a = F/m.
67. A body of mass 0.5 kg just slides down in an inclined plane when inclination of the plane is 150. If the same body is taken above the plane with velocity 2m/s, determine the power. [ 5 Watt ]
Hints: The coefficient of friction is given by = Tan where  is the angle of repose. Power required is given by P = ( Ff + mg Sin). V and Ff can be determined by relation .

Circular Motion

68. An object of mass 10 kg is moving round the horizontal circle of radius 4 m by revolving string inclined to the vertical. If the speed of the object is 5 m / s, calculate (i) the tension on the string, (ii) the angle of inclination of the string to the vertical. [117.9N, 320] Hints: Use the expression of the conical pendulum T Cos = mg and T Sin = . Dividing, the obtained result is Tan = . Substitute the value of  in T Cos = mg to obtain T.
69. A racing car of mass 1000 kg moves round a banked track with speed 108 km / hr. If the radius of the track is 100 m, calculate the angle of inclination of the track and reaction of the wheels. Hints: The angle of inclination Tan = and reaction is given by R Cos = mg.
70. An object of mass 8 kg is whirled round the vertical circle of radius 2 m with a constant speed 6 m / s. Calculate the maximum and minimum tension in the string. [224N, 64N] Hints: The maximum tension at the lowest position is Tmax = and the minimum tension at the maximum height is Tmax =
71. Calculate the mean angular velocity of the earth assuming that it takes 24 hr to rotated about it axis. An object of mass 2 kg is taken from equator to pole of the earth. Calculate the change in its weight. [7.27 x 10-5 rad/s, 0.068N ] Hints: Angular velocity is given by  = , where T is the time period of earth, 24 hr = 24 x 60 x 60 seconds. Weight of a body at pole W1 = mg, since there is no effect of rotation of earth. Weight of a body at equator W2 = mg – = mg – mr2 Therefore, the change in its weight W = W1 – W2
72. An object of mass 0.5 kg is rotated in a horizontal circle by a string 1 m long. The maximum tension in the string before it breaks is 50 N, what is the greatest number of revolution per second of the object? [ 1.58 Hz ] Hints: The tension in horizontal circle is T = and the angular velocity  = 2f.
73. A mass of 0.5 kg is rotated by a string at a constant speed v in a vertical circle of radius 1 m. If the minimum tension of the string is 3 N, calculate (i) v, (ii) the maximum tension, (iii) tension when the string is just horizontal. [ 4m/s, 13N, 8N] Hints: The minimum tension is T = , maximum tension is T = and the tension when the string is horizontal is T =
74. A spaceman in training is rotated in a seat at the end of horizontal rotating arm of length 5m. If he can withstand acceleration up to 9g, what is the maximum number of revolution per second permissible? [0.67 Hz] Hints: Centrifugal force is F = mr 2 and therefore the acceleration is a = F/m = r 2. Take a = 9g and  = 2f and then determine number of revolution per second, i.e. frequency.
Gravity

75. The gravitational force on a mass of 1 kg at the earth’s surface is 10 N. Assuming the earth is a sphere of radius R, calculate the gravitational force on a satellite of mass 100 kg in a circular orbit of radius 2R from the center of the earth.
[ 250 N ]
Hints: For first body, F1 = ………….(i) where, F1 = 10N, m1 = 1 kg, r1 = R and for second body, F2 = ………….(ii) where, m2 = 100 kg, r1 = 2R & F2=? Dividing equation (i) and (ii), F2 can be determined.

76. A satellite X moves round the earth in a circular orbit of radius R . Another satellite Y of the same mass moves round the earth in the circular orbit of radius 4R. Show that (i) speed of X is twice that of Y, (ii) K E of X is greater than that of Y, (iii) P E of X is less than that of Y.
Hints: For satellite X, ………….(i) For satellite Y, ………….(ii) Dividing equation (i) by (ii), we get vx = 2 vy .
Similarly, the gravitational PE of satellite X is Ux = …………..(iii) and, gravitational PE of satellite Y is Uy = ………….(iv)Divide to get result.

77. Find the period of revolution of a satellite moving in the circular orbit round the earth at a height of 3.6 x 106 m above the surface. Assume earth is a sphere of radius 6.4 x 106 m and mass 6 x 1024 Kg.
Hints: Start from the relation where  is angular velocity equal to  = 2/T. Hence finally we get T = 2
78. The acceleration of free fall at the earth’s surface is 9.8 m / s2 and the radius of the earth is 6400 km, calculate the mass of the earth.
Hints: Use the relation

79. Two binary stars, masses 1020 kg and 2 x 1020 rotate about there common center of mass with angular velocity  . Calculate  if distance between them is 106 m. [ 1.4×10-4 rad/s]
Hints: Since both stars are rotating about common center of mass, both experience equal centrifugal force. Equate the centrifugal force and at first find their distances from the CM , i.e. m1r12 = m2r22 and obtain r1 = 2/3 x106m and r2 = 1/3 x 106 m.
Then equate centrifugal force with gravitational force, and determine angular velocity.

80. Explorer – 38 satellite of mass 200 kg circles the earth in the orbit of radius 3R/2. If gravitational pull of 1 kg mass at earth surface is 10 N, calculate the pull on the satellite. [889N]
Hints: For 1 kg mass at earth surface, the gravitational force, F1 = …………..(i) For satellite of mass 200 kg, the gravitational force, F2 = …………..(ii)
Divide them to obtain result.
81. Earth is elliptical with polar and equilateral radii 6357 km and 6378 km respectively. Determine the difference in g at that places. Given, mass of earth = 5.957 x 1024 kg. [0.0646]
Hints: Use the relation of acceleration due to gravity,

82. Obtain the value of ‘g’ from the motion of the moon that its time period of revolution is 27 days 8 hrs and its radius of orbit is 60.1 times the radius of the earth. [9.77 m/s2]
. Hints: The time period of satellite T = 2 = 2 . Substitute r = 60.1 R.
83. The orbit of moon is approximately a circle of radius 60 times the radius of the earth Calculate the time taken for the moon to complete one orbit. Given g = 9.8 m / s2, R = 6.4 x 106 m, 1 day = 8.6 x 104 sec. [27.4 days]
Hints: The time period of satellite T = 2 = 2 . Substitute r = 60 R.

84. A communication satellite would revolve round the earth on the equatorial plane. Find the radius of the orbit if the radius of the earth is 6400 km. [ 42626 km]
Hints: The time period of communication satellite is exactly equal to that of earth i.e. 24 hour. Using formula, T = 2 . 24 x 60 x 60 = 2 . Determine ‘r’.
85. Calculate the total energy required to raise a satellite of mass 2000 kg to a height of 800 km above the ground and to set it into circular orbit at that altitude. Given, g = 10 m/s2 and R = 6400 km. [7.11x1010J] Hints: Given R = 6400 km = 6.4 x 106 r = R + h = 6400 km + 800 km = 7.2 x 106 m. PE at earth surface is – and the PE of the satellite at the orbit is – . Hence increase in PE = = m g h and KE = . Total energy U = KE + increase in PE = + m g h = + m g h
86. Determine the orbital and escape velocity of earth if its mass is 6.0 x 1024 kg and radius is 6.4 x 106 m. [7.9 km/s, 11.18km/s]
Hints: For orbital velocity, gravitational pull = centripetal force at earth surface. So, = . Or, v =
Similarly, for escape velocity, KE = increase in PE of a body. So, = ,Or, v =

S H M

87. An object moving with SHM has an amplitude of 0.02 m and a frequency of 20 Hz. Calculate (i) the period of oscillation, (ii) acceleration at the middle and at the end of the oscillation, (iii) the velocities at the corresponding instants. [ 0.05 s ; 0 and 32 2 m/s2 ; 0.8  m/s, 0 ] Hints: (i) Time period T = 1/f, (ii) Acceleration at any position a = 2y. At middle point y = 0 and at end point y = r, (iii) Velocity at any position is given by v = 
88. A steel strip clamped at one end vibrates with frequency 50 Hz and amplitude of 8 mm. Find (i) Velocity at zero position, (ii) acceleration at the maximum displacement. [2.5 m/s, 790 m/s2 ] Hints: (i) The instantaneous velocity is v =  . At zero or mean position, y = 0 and the maximum velocity of the strip is v = r.. (ii) Similarly, the instantaneous acceleration is given by a = 2y . At the maximum displacement, i.e. at end position, y = r. So a = 2r.
89. A spring of force constant 5 N/m is placed horizontally on a smooth table and a mass X of 0.2 kg is attached to the free end which is displayed at a distance of 4 mm along the table and then released. Calculate the (i) the period, (ii) the maximum acceleration, (iii) the maximum kinetic energy, (iv) maximum PE [ 1.26 s; 0.1 m/s2 ; 4×10-5 J ] Hints: (i) The time period of the oscillation is given by T = 2 (ii) Acceleration is maximum at end position, which is a = 2r. (iii) The maximum KE = ½ kx2. (iv) The KE is equal to the PE.
90. A vertical spring is extended 10 mm when a small weight is attached to its free end. The weight is now pulled down slightly and released. Calculate the period of oscillation. [ 0.2 s ] Hints: (i) The maximum acceleration of the body is a = 2r which is equal to the acceleration due to gravity. So, g = 2r.
91. A simple pendulum has a period of 4.2 second. When the pendulum is shortened by 1 m the period will be 3.7 s. Find value of g and original length of the pendulum. If the pendulum is taken from the earth to the moon, where acceleration is g /6, what is the relative change in T ? [ 10 m/s2 ; 4.5 m ; 2.45 : 1 ] Hints: (i) Time period of simple pendulum is given by T1 = …………….(i) When it is shortened by 1m, the new time period, T2 = ……….….(ii). Divide these two expressions to obtained result.
92. A simple pendulum has length 1.8 m with a bob of mass 2.2 kg is pulled aside a horizontal distance 20 cm and then released. What are the values of (i) Time period (ii) KE (ii) velocity of the bob at the lowest point of the swing? [ 2.7 s ; 0.24 J; 0.47 m/s] Hints: (i) Time period of pendulum T = (ii) Determine the height raised by pendulum bob and calculate PE= mgh, which is equal to the kinetic energy of th bob. (iii) Use formula KE = ½ mv2 to determine velocity.
93. Displacement of a particle executing in SHM is given by y = 20 sin 10t, where y is in mm and t is in second. What is (i) the amplitude, (ii) the period, (iii) the velocity at t = 0.
[ 20 mm ; 0.2 s; 628 mm / s; 0.08 J]
Hints: Compare with the standard form of SHM y = r Sin t, where r is amplitude,  is angular velocity and y is displacement. The derivative of y with respect to time, i.e. gives the velocity.
94. A small mass rests on a horizontal floor vibrates vertically in SHM with period 0.5 s. Find the maximum amplitude of the motion which allow the mass to remain on contact with the floor. [ 6.3 cm ] Hints: To remain in contact with the floor, the maximum acceleration should be ‘g’, the acceleration due to gravity. Apply relation a = 2y. Or, g = r and determine r.
95. A particle executes a SHM of time period 4 s. Find time taken by the particle to go directly from its mean position to half the amplitude. [ 1/3 s] Hints: The general formula of SHM is y = r Sin t. Take y = r/2 and determine ‘t’.
96. Assume that a narrow tunnel is dug between two diametrically opposite points of the earth. If a particle is released in this tunnel, it will execute SHM. Calculate the time period of the motion. [ 5070 s ]
Hints: Extreme position of the particle is the surface of the earth where acceleration of the oscillating particle is equal to the acceleration due to gravity. Use the relation a = 2r. Where,  = angular velocity = 2/T a = acceleration due to gravity = 9.8 m/s2 r = radius of earth = 6.4 x106 m.

Rotational Dynamics

97. A disc of MI 10 kg m2 about its center rotates steadily about the center with an angular velocity 20 rad/s. Calculate (i) its rotational energy (ii) angular momentum about the center, (iii) the number of revolution per second of the disc. [2000J, 200Kgm2/s, 3.2/s] Hints: (i) Rotational energy E = ½ I2 , (ii) Angular momentum L = I ,(iii) and number of revolution per second or frequency is calculated by  = 2 f.
98. A constant torque of 200 Nm turns a wheel about its center. The moment of inertia about its axis is 100 kg m2. Find (i) angular velocity gained in 4 sec. (ii) the KE gained after 20 revolutions. [8 rad/s, 25133 J Hints: Determine angular acceleration by relation  = I  and use (i)  = 0 + t for angular velocity and (ii) E = ½ I2 for KE. In second case,  is the angular velocity gained after 20 revolution i.e. after the displacement 40. It is determined by using formula 2 = 02 + 2.
99. A flywheel has KE 200 J. Calculate the number of revolutions it makes before coming to rest if a constant opposing couple 5 Nm is applied to the wheel. If the MI of the flywheel about its center is 4 kg m2, how long does it take to come to rest? [6.4, 8] Hints: Use formula E = ½ I2 for KE and determine 0 , the initial angular velocity. Determine retardation and use formula  = 0 + t to calculate time.
100. A ballet dancer spins with 2.4 rev / s with her arms outstretched when MI about the axis of rotation is I. With her arm folded, the MI about the same axis becomes 0.6 I. Calculate the new rate of spin. [4rev/s] Hints: Apply the principle of conservation of angular momentum, i.e. I11 = I22 . Or, I1  2f1 = I2  2f2 . Where, I1 = I, I2 = 0.6 I f1 = 2.4 Hz, f2 = ?
101. A disc rolling along a horizontal plane has a MI 2.5 kg m2 about its center and a mass of 5 kg. The velocity along the plane is 2 m/s. If the radius of the disc is 1m, find (i) the angular velocity, (ii) the total energy of the disc. [2, 15J] Hints: (i) Angular velocity is determined by relation v =  r. (ii) Total energy of the disc = KE due to translation + KE due to rotation =
102. A horizontal disc rotating freely about a vertical axis makes 100 r.p.m. A small piece of wax of mass 10g falls vertically onto the disc and adheres to it at a distance of 9 cm from the axis. If the number of revolutions per minute is thereby reduced to 90, calculate the moment of inertia of the disc. [7.3 x 10-4 kg m2]
Hints: Apply the principle of conservation of angular momentum. The initial angular momentum of the disc = I11 = I1 x 2f = I1 x 2 x ………….(i) Final angular momentum of disc & wax = ( I1 + I2) 2 = (I1 + mr2) x2f2 = [I1 + ] x 2 x …………..(ii) Equating equation (i) and (ii), I1 can be determined.
103. If a ring, a solid sphere, a hollow sphere and disc of equal mass and radius roll down on a frictionless incline, which will reach the bottom first? Hints: Determine acceleration of all bodies one by one. The body which has higher acceleration reaches the bottom earlier. The acceleration of a rolling body is given by For ring, radius of gyration k = r, So, a = ½ g Sin For solid sphere, radius of gyration k = r, So, a = 5/7 g Sin For hollow sphere, radius of gyration k = r, So, a = 3/5 g Sin For disc, radius of gyration k = r, So, a = 2/3 g Sin Since acceleration of the solid sphere is maximum, it reaches the bottom earlier.

104. A roller whose diameter is 1.0 m weighs 360 N. What horizontal force is necessary to pull the roller over the brick 0.1 m high when the force is applied at the centre? Hints: CW moment of force = F x AC CCW moment of force = mg x BC Equating CW and CCW moments, F x AC = mg x BC Or, F = mg x
[ 270 N ]
105. A sphere rolls up a frictionless incline of inclination 300 . At the bottom of the incline the translation velocity of CM is 5 m/s. How far does the sphere move up the plane? Given , I of sphere = 2/5 mR2.
[ 3.5 m]
Hints: At the bottom of the inclined plane, the total energy of the sphere = KE due to translation + KE due to rotation = Finally, the entire energy converts into the PE. Hence, according to conservation of energy, mgh =
Or, mgl Sin =

Hydrostatics

106. An alloy of mass 588 gm and volume 100 cc is made of iron of density 8.0 g / cc and aluminum of density 2.7 g / cc. Calculate the proportion by (i) by volume, (ii) by mass of the constituents of the alloy.
[3:2, 4.44:1]

Equating equation (i) and (ii) both volumes and masses can be determined.

107. A string support a solid iron object of mass 180 g immersed in liquid of density 800 kg / m3 . Calculate the tension in the string if the density of iron is 8000 kg / m3. [ 1.8 N]
Hints: Total upward force = T + U Total downward force = mg So, T + U = mg,
Or, T = mg – U
Upthrust U = V g, where V is the volume of the displaced liquid which is equal to the volume of the body,  be the density of liquid and g is acceleration due to gravity.

108. A solid weighs 237.5 g in air and 12.5 g when immersed in liquid of density 0.9 g / cc . Calculate (a) density of the solid, (ii) the density of the liquid in which the solid would float with one- fifth of its volume exposed above the liquid surface.
[ 0.95 g/cc, 1.1875 g/cc]
109. A piece of wood of mass 124 g floats on water, What minimum mass of lead is to be attached to the wood so as to case it to sink. Given relative density of wood and lead are 0.5 and 11.3. [ 136 g ]
Hints: Let mass of the lead attached = x So, total weight W = ( x + 124 ) g total upthrust U = weight of displace liquid = ( Vw + Vl )  g =  1 g Since the body just sinks, weight of the body is equal to the upthrust. So, ( x + 124 ) g =  1 g
110. An iceberg has volume 200 m3. Calculate its volume that can be seen above the water surface. Given densities of sea water and ice are 1025 kg/m3 and 920 kg/m3 respectively. [ 20.49 cc] Hints: Let volume of ice above the surface be x. So, volume of ice below the surface is 200 – x. For floating body, Weight = upthrust mg = weight of displaced liquid. V    g = (V – x )    g Or, 200  920 = ( 200 – x )  1025
111. An alloy of mass 588 gm and volume 100 cc is made of iron of density 8.0 g/cc and aluminum of density 2.7 g/cc. Calculate the proportion by (i) volume, (ii) by mass of the constituents of the alloy. [(i) 6×10-5 m3, 4×10-5 m3 (ii) 0.48kg, 0.108 kg] Hints: Let mass of iron = x and mass of the aluminum = y. Then, according to question, x + y = 588 …………(i) Volume of iron = and volume of aluminum = So, + = 100 …………….(ii) Solving equation (i) and (ii), x and y can be determined.

Elasticity

112. A wire 2m long and cross-section area 10-6 m2 is stretched by 1mm by a force of 50 N in the elastic region. Calculate (i) strain, (ii) Young’s modulus, (iii) energy stored in the wire. [ 5 x 10-4, 1011 N/m2, 0.0125 J ] Hints: (i) Strain = (ii) Young’s modulus = (iii) Energy stored U = ½ F x e
113. What force must be applied to a steel wire 6m long and diameter 1.6mm to produce the extension of 1 mm. ( Y of steel = 1.1  1011 N / m2 ) [ 36.86 N ] Hints: Young’s modulus Y = . So, force F =
114. A spring is extended by 30mm when a force of 1.5 N is applied to it. Calculate the energy stored in the spring when hanging vertically supporting a mass of 0.2 kg if the spring was unstretched before applying the mass. Also calculate the loss of PE of the mass. [0.04 J, 0.08 J ] Hints: From Hook’s law, F1  e1 ……….(i) and F1  e1 ……….(ii) Dividing . The expression gives the value of e2. Now, energy stored U = ½ F x e2 and loss of PE = m x g x e2
115. What stress would cause a wire to increase in length by one-tenth of one percent if the young’s modulus of the wire is 12  1010 N / m2. What force produces this stress if the diameter of the wire is 0.56 mm? [ 12 x 107 N/m2, 31.7 N ] Hints: According to question, elongation e = one-tenth of one percent of L = = So, strain = and stress = Y x strain Now, stress = . Therefore, F = stress x A
116. Calculate the work done in stretching a wire 100 cm in length and a cross sectional area 0.03 cm2 when a load of 100 N is applied within the elastic limit. Given Young’s modulus of the wire is 1.1 x 1011 N/m2 ] [ 0.3 mm ] Hints: Young’s modulus Y = . So, force e =
117. How much force is required to punch a hole 1 cm in diameter in steel 3cm thick whose sheering strain is 2.76  108 N /m2.
118. A copper and a steel wire each of length 1.5m and diameter 2mm are joined at one end to form a composite wire 3m long. The wire is loaded until its length becomes 3.003m. Calculate the strain in the copper and steel wire and the force applied to the wires. ( Y of Cu = 1.2  1011 N / m2 and that of steel = 2  1011 N /m2 ) [ 1.25 x 10-3, 7.5 x 10-4] Hints: Let e1 and e2 are elongation produced on copper and steel wire. Then, e1 + e2 = 0.003 ………………(i) Also, e1 = and e2 = . Dividing, we get = 5/3 So, 3 e1  5 e2 = 0 ……………….(ii) Solving equation (i) and (ii), e1 and e2 can be determined.
119. A steel wire of density 7800 kg / m3 weighs 16g and is 250 cm long. It lengthens by 1.2 mm when a force of 80 N is applied. Calculate the (i) Y of steel (ii) energy stored in the wire. [ 2 x 1011 N/m2, 0.048J ] Hints: Density  = = . So, cross-section area of the wire A = = 8.2×10-7m2 Now, Young’s modulus Y = and energy stored U = ½ F e
120. The rubber chord of catapult has cross sectional area 1mm2and the total unstretched length 10 cm. It is stretched to 12 cm and then released to project a missile of mass 5g. Calculate the velocity of projection taking Y for the rubber as 5  108 N / m2. [ 20 m/s ] Hints: Energy stored U = ½ F x e = ½ . The potential energy stored on the wire converts into kinetic energy. So, ½ m v2 = U
121. Calculate the minimum tension with which platinum wire of diameter 0.1mm must be mounted between two points in a stout inner frame if the wire is to remain taut when the temperature rises 100 K. Platinum has linear expansivity 9 x 10-6 /K and Y = 17  1010 N / m2. [1.2N] Hints: Elongation due to increase in temperature e = . Now, strain Now, tension T = Y x A x strain. So, energy stored U = ½ F x e = ½
122. A railway track uses long steel rails which are prevented from expanding by the clamps. If the cross sectional area of each rail is 75 cm2 what is the elastic energy stored per km of track when its temperature rises by 100C.
(  of steel = 1.2  10-5 / K, Y = 2  1011 N / m2 ) [ 10.8 J ] Hints: Elongation due to increase in temperature e = . So, energy stored U = ½ F x e = ½ = ½
123. A uniform wire of unstretched length 2.49 m is attached to two points A and B which are 2 m apart. When 5 kg mass is attached to the midpoint of the wire C, the equilibrium position of the C is 0.75 m below the AB line. Neglecting the weight of the wire, and taking Y = 2  1011 N/m2, find (i) strain in the wire (ii) stress in the wire (ii) energy stored in the wire. [ 4.01 x 10-3, 8.03 x 108 N/m2, 0.083 J]
124. Two wire of equal cross-section but made of steel and the other copper are joined end. When the combination is kept under tension, the elongations in the wire are found to be equal. Given young’s modulii of steel and copper are 2.0×1011 Nm-2 and 1.1×1011 Nm-2. Find the ratio between the lengths of steel and copper wire. [20/11] Hints: Elongation produced on steel wire e1 = Elongation produced on copper wire e2 = According to question, e1 = e2
Surface tension & Viscosity

125. A rectangular plate of dimension 6 cm by 4 cm and thickness 2mm is placed with its largest face flat on the surface of the water. Calculate downward force on the plate due to surface tension assuming zero angle of contact. What is the downward force if the plate is placed vertical so that its longest side just touches the water? [ Surface tension of water is = 7 x 10-2 N/m ] [ 0.014 N, 0.0112 N ] Hints: Surface tension T = where l is the contact length. In first case, l = perimeter = 2 x (l+b) = 2 x ( 0.06 + 0.04) = 0.2 m. In second case, the contact length l = 2 x ( 0.06 + 0.002 ) = 0.124 m So, force due to surface tension F = T x l
126. A capillary tube of 0.4 mm diameter is placed vertically inside (i) water of surface tension 0.065 N/m and zero angle of contact, (ii) a liquid of density 800 kg/m3 and surface tension 0.05 N/m and angle of contact 300 .Calculate the height at which the liquid rises in the capillary tube in each case. [6.5cm, 5.4 cm] Hints: Height of liquid column in capillary tube h = .
127. What is the terminal velocity of the steel ball of radius 2mm falling through a tall jar containing glycerin? Densities of steel and glycerin are 8.5 g/cc and 1.32 g / cc respectively and the viscosity of the glycerin 0.85 poise. [ 0.75 m/s] Hints: Weight of the ball = Viscous force + up thrust V    g = 6rv + V    g . Solving, the terminal velocity is found to be v =
128. When a capillary tube of diameter 10-3m was dipped inside a clean liquid of density 1000 kg/m3, it was found that liquid rises to a height of 0.03m in the tube. Calculate the surface tension of the liquid. [ 0.075 N/m] Hints: Height of liquid column in capillary tube h = . So, surface tension T =
129. Castrol oil at 200C has a coefficient of viscosity 2.42 Nsm-2 and a density 940 kg/m3, calculate the terminal velocity of a steel ball of radius 2.0mm falling under gravity, taking density of steel 7800 kg/m3. [ 0.05 m/s] Hints: Weight of the ball = Viscous force + up thrust V    g = 6rv + V    g
130. What should be the pressure inside air bubble of 0.1mm radius situated just below the water surface? Surface tension of water 7.2×10-2 Nm-1 and atmospheric pressure =1.013.105Nm-2. [ 1.027x105Nm-2]
131. If wind blows at rate 30 m/s, over the house what is the total force on the roof if its area is 200 m2. Given density of air is 1.29 kg/m3. [ 1.16 x105 N]
Hints: Let P1 and P2 are pressure inside and outside the roof. Then applying Bernoulli’s principle, P1 +  v12 = P2 +  v22 P1 – P2 =  ( v12 – v22). It gives the pressure difference between inside and outside the roof. Now, net force on the roof F = A x (P1 – P2)

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MECHANICS

Unit and Dimension
1. The distance covered by a particle in time t is given by s = a + bt + ct2. Find the dimension of a, b and c.
Hints: Additive quantities must have same dimensional formula, known as principle of homogeneity. So, dimensional formula of s is same as that of a, bt and ct2. Therefore, [ a ] = [ s ] = [ L ] [ bt ] = [ s ] = [ L ], So, [ b ] = [ LT-1] [ ct2 ] = [ s ] = [ L ] So, [ c ] = [ LT-2 ] [ L], [ LT-1], [ LT-2 ]
2. The Vander Waal’s gas equation is What are the dimensions of ‘a’ and ‘b’ ? [ ML5T-2], [ L3]
Hints: Multiply the given expression and then apply principle of homogeneity.
3. Check the correctness of physical expression (a) Centripetal force , (b) Vertical distance covered by a body S = ut + ½ g t2. (c) Lorenz force F = B q v Sin Hints: Determine dimensional formula of LHS and RHS separately. If both are same, the expression will be correct otherwise not. In the third expression B is magnetic field intensity which is the force per unit pole strength and Sin is dimensionless. The unit of pole strength is Ampere-meter. So, [ B ] = = = [ ML0T-2A-1] [All correct]
4. The centripetal force experienced by revolving body of radius ‘r’ depends on mass of the body, velocity and radius of the circular path. Obtain the relation between them. [ x = 1, y = 2, z = -1; ]
Hints: dimensional formula of force is equal to the dimensional formula of mxvyrz. Equating power of M, L and T, value of x, y and z are obtained.
5. The time period of simple pendulum is given by t = 2 lx gy. Determine the value of x and y and write the equation. [ x = ½, y = -½ ; ]

6. If force, length and time are considered to be fundamental quantities, express dimension of (a) universal gravitational constant G and (b) density in terms of [FLT]. (a) [ F-1L4T-2] (b) [FL-4T2]
Hints: (a) The Newton’s universal law of gravitation is F = or, . Express mass in terms of force and acceleration according to Newton’s second law of motion i.e. F = m a. So, mass m = F/a. Replace the both masses and write dimension.

7. Convert 1 Joule into erg. [ 107 erg]
Hints: Write dimensional formula of energy and obtain conversion factor As, dimensional formula of energy is [ ML2T-2]
We have , 1 kg mass = 1000 gm 1 m length = 100 cm and 1 s time = 1 s So, conversion factor K =(conversion of mass) (conversion of length)2 (conversion of time)-2 Or, K = 1000  1002  1 = 107 So, 1 J is equal to 107 erg.
8. Density of mercury is 13600 kg/m3. Convert it into cgs unit. [ 13.6 g/cc]
Hints: Write dimensional formula of density and obtain conversion factor. Multiply the given density by the conversion factor.

9. Value of gravitational constant G is 6.67 x 10-11 N m2 /Kg2. Convert it into dyne cm2/g2. [6.67 x 10-08]
Hints: Write dimensional formula of G and obtain conversion factor. Multiply the given value by the conversion factor.
10. Assuming that the mass ‘m’ of the largest stone that can be moved by a flowing river depends on the velocity ‘v’ of water, the density ‘’ of water and ‘g’. Show that ‘m’ varies as the sixth power of the velocity.
Hints: The mass of the largest stone m  vx ……….(i)
m  y ……….(ii)
m  gz ……….(iii)
Combining all equations, we get m  vx y gz
Take dimensional formula both side and obtain x = 6.

11. Diameter of a spherical ball measured by micrometer screw gauge is found to be 5.470 cm. Determine (a) LC of the micrometer screw gauge. (b) Permissible percentage error in the measurement of diameter. (c) Area up to proper significant figure. (d) Volume up to proper significant figure. (e) Permissible percentage error in the measurement of area. (f) Permissible percentage error in the measurement of volume.
[ 0.001 cm, 0.0183%, 94.00 cm2, 85.70 cm3, 0.0366%, 0.549 % ]
Hints: Permissible error in the measurement of diameter is determined by using formula where d is diameter and d is the LC of the instrument. The error in area and volume are two and three times that of diameter.
The given radius is upto four significant figures. So, both area and volume should be upto four significant figures.
12. Length, breadth and height of a glass slab are given to be 4.025 cm, 2.63 cm and 1.47 cm respectively. Determine its volume and permissible error (PE) in the measurement of volume. [ 15.6 cm3, 1.085%]
Hints: From the given data, the breadth is upto three decimal places, which is the least significant figures among the data. So the result should be in three significant figures. The PE in volume is the sum of PE in length, breadth and height. So, PE in volume =
13. Percentage error in the measurement of mass, length and time are respectively 1%, 1.5% and 2%. Find the error in the measurement of (i) force and (ii) power. [6.5 % , 10% ] Hints: Dimensional formula of force is [ MLT-2]. So, error in force = error in mass + error in length + 2 X error in time = 1% + 1.5 % + 2 X 2% = 6.5 %
14. Determine the angle between two equal vectors if their resultant is equal to the either of the vector. [ 1200 ] Hints: Use the expression of parallelogram law of vector addition r = and take p = q = r.
15. One of the rectangular components of vector 25 unit is 15 unit. Determine another rectangular component. [ 20 unit ]
Hints: Two rectangular components are v Cos and v Sin. Consider the parallel component of vector v Cos = 15. Determine another component v Sin.

16. The physical quantity which has only unit but not dimension is (a) momentum (b) angle (c) quantity of matter (d)specific heat capacity
17. The dimensional formula of calorie is (a) [ ML2T-2 ] (b) [ MLT-2 ] (c) [ ML2T-3 ] (d) [ M0LT-1 ]
18. The SI unit of magnetic moment is (a) Am (b) Tesla (c) Gauss (d) Am2
19. The dimensional formula of is same as that of (a) energy (b) momentum (c) speed (d) work
20. The dimensional formula of RC, where R is resistance and C is capacitance is (a) [ ML2T-2 ] (b) [ M0L0T1 ] (c) [ M0L0T-1 ] (d) [ M0LT-1 ]
21. A particle displaces by under the effect of force . The work done by the force is (a) 23 J (b) 17 J (c) 29 J (d) 11 J
22. The value of is always (a) 0 (b) 1 (c) (d) k3
23. The value of is always (a) 0 (b) 1 (c) (d)
24. The condition for is (a) A = B (b) (c) (d) is a unit vector
25. Which pair of displacement cannot give the resultant 5m ? (a) 12m and 5m (b) 8m and 4m (c) 9m and 6m (d) 4m and 2m
26. The incorrect relation is (a) (b) (c) (d)
27. If . What is the angle between two vectors?
(a) 0 (b) 450 (c) 900 (d) 1200

Linear Motion

28. A car moving with velocity 10 m/s accelerates uniformly at 1 m/s2 until it reaches a velocity of 15 m / s. Calculate (i) the time taken (ii) the distance traveled (iii) the velocity reached 100 m from the place where the accelerations begun.
Also draw v-t curve and find the above terms. [ 5s, 62.5m, 17m/s] Hints: Apply equation of motion v = u + at, s = ut + ½ at2 and v2 = u2 + 2as. In graphical representation, slope of the v-t curve gives acceleration and area enclosed by the curve gives displacement.
29. A ball is thrown vertically upward with initial velocity 20 m / s. Draw the velocity – time, speed – time and displacement – time graph. Also calculate (i) time to return to the thrower, (ii) the maximum height reached.
[ 4s, 20m] Hints: Initial velocity and acceleration are opposite in direction. So, take either of them negative. Solve the question by (i) considering upward half motion, (ii) considering whole motion, (iii) considering the projectile motion of angle of projection 900.
30. A ball is dropped from a height of 20 m and rebounds with velocity 3/4 of the velocity with which it hit the ground. What is the time interval between first and second bounces? Also draw its (a) Velocity – time and (ii) Displacement – time graphs.
[ 3 s ]
Hints: Determine velocity of the ball before the ball just touches the ground by applying equation of motion, v2 = u2 + 2as. The rebounding upward velocity will be ¾ of that velocity. The rebounding velocity is in opposite to the acceleration due to gravity. So, take it negative.

31. A bus moves with uniform velocity 36 km/hr for 4 s. The bus then slows down and stops in 10 second. Draw its v-t curve and determine (a) Retardation and (b) total distance covered during the motion from the graph. [ 5/3 ms-2, 70m] Hints: The slope of the line gives retardation and the area enclosed by the curve gives the total distance covered.
32. A stone dropped from a balloon ascending with the velocity 10 m / s. If the stone reached the ground after 17 second, calculate the height of the balloon when the stone was dropped.
[ 1275m]
Hints: Consider downward velocity. The initial upward velocity of the balloon will be negative. Then apply equation of motion S = ut + ½ at2 to obtain S.
33. A bus goes from A to B with uniform velocity 40 km /hr and returns from B to A with uniform velocity 60 km /hr. Calculate the average speed of the bus on the trip.
[ 48 km/hr]
Hints: Suppose the distance between A and B be x and determine the time interval required to reach from A to B and from B to A separately and the average velocity is the ratio of total distance covered ‘2x’ to the total time taken ‘x/24’.
i.e. Average velocity V = km/hr.
34. A bus was moving with velocity 50 m/s. Simultaneously, another bus starts from rest from the same point with acceleration 4 m/s2. When and where were they met? [25 sec, 1250 m]
Hints: Since both buses are moving simultaneously from the same point, distance covered ‘s’ and time interval ‘t’ are same for both motions. For first bus, distance covered S = ut ……..(i) For second bus, distance covered S = at2 ……(ii) , since u is zero. Equating two equations, ‘t’ and ‘s’ can be determined.
35. A stone is dropped from the top of a building and one second later a second stone is thrown vertically down with velocity 20 m /s. When and where will they meet? [ 1.5 s, 11.25m]
Hints: Both bodies are released from the same point. So, distance ‘s’ is equal for both. Second stone is thrown one second later. So, if ‘t’ be the time of fall for first, that of second will be one second less i.e. ‘t-1’. For first body, S = ½ gt2 …………(i) For first second body, S = u (t – 1 ) + ½ g(t -1) 2 …………(ii) Solving equations, time and distance can be determined.
36. A stone is dropped from the top of a building of height 45m. Simultaneously a second stone is thrown vertically upward from the bottom of the building with velocity 30 m/s. When and where will they meet? [ 1.5sec, 33.75 from bottom]
Since both stones are thrown simultaneously, time of flight for both are equal i.e. ‘t’. If ‘x’ be the distance covered by first body, that of second will be 45 – x. For first body, x = ½ gt2 …………(i) For first second body, 45 – x = ut – ½ gt2 …………(ii) Equating equations, time and distance are determined.
37. A body moving with uniform acceleration covers 5 m in third second and 9 m in fifth second of its motion. Find the distance covered in 6th second. [a = 2, u = 0, S6 = 11m]
Hints: Use the formula of distance traveled in nth second Snth = u + and obtain two equations. Solve them to calculate ‘u’ and ‘a’. Again apply same formula to calculate distance covered in 6th second.
38. A ‘moving sidewalk’ in an airport terminal building moves at 1.5 m/s and is 35 m long. If a man steps on at one end and walks at 1.0m/s, how much time does he takes to cross the side walk (a) in the same direction, (b) in the opposite direction Hints: (i) When the man walks along the same direction of sidewalk, the velocity of the man with respect to earth V = 1.5 + 1 = 2.5 m/s. So, the time taken t = (ii) When the man walks in opposite direction, the resultant velocity will be the differenced of the two velocities.
39. A man is walking horizontally with velocity 3m/s and rain is falling vertically with velocity 4 m/s. What is the velocity of the rain with respect to the man? Also calculate the inclination of his umbrella to the vertical that he should hold to protect from the rain. [ 5 m/s]
Hints: Make man at rest taking equal but opposite velocity of the man and then apply parallelogram or triangle law of vector addition.

40. A stone attached to a string is whirled round in a horizontal circle with a uniform speed 10 m/s. Calculate the different in the velocity when the stone is (i) at opposite ends of a diameter, (ii) in two positions A and B where angle AOB is 900 and O is the center of the circle. [20 m/s, 14.1 m/s] Hints: (i) At opposite ends of diameter, the direction of two velocities will be in opposite direction. (ii) Apply triangle law.
41. Two ships A and B are 4 km apart. A is due west of B. If A moves with velocity 8 km/hr due east and B moves with 6 km/hr due south, calculate (i) velocity of A relative to B, (ii) closest distance between A and B.
[10km/hr, 2.4km]
Hints: Make B stationary taking equal and opposite velocity. Apply triangle law to determine velocity of A relative to B i.e. AC in the diagram. The perpendicular distance from B to AC gives the shortest distance between them.

42. A small object slides down from rest down a smooth inclined plane inclined at 300 to the horizon. What is (i) acceleration down the plane? (ii) the time to reach the bottom if the plane is 5 m long. The object is now thrown up the plane with an initial velocity of 15 m / s, (iii) how long does the object take to come to rest? (iv) how far up the plane has the object then traveled?
[ 5m/s2,1.4s, 3s, 22.5m ]
Hints: The acceleration down the plane is gsin and in second case it will be negative.

Projectile
43. The distance covered by a uniform accelerating body starting from rest in time ‘t’ is proportional to (a) (b) t (c) (d) t2
44. A ball is thrown horizontally from the top of a tower with a velocity 10 m/s. the height of the tower is 45 m . Calculate (i) time to reach ground, (ii) horizontal distance covered by the body (iii) the direction of the ball when it just hits the ground. [3s, 30m, 71.50 with horizontal]
Hints: (i) Consider vertical downward motion and determine time. Time along vertical downward path will be same as that of parabolic path. (ii) Consider horizontal motion and determine horizontal range.(iii) Determine vertical velocity Vy and horizontal velocity Vx separately and angle with the horizontal line is given by .
45. A projectile is fired with a velocity 320 m / s at an angle 300 to the horizon. Find (i) the time taken to reach the greatest height, (ii) its horizontal range. What is the maximum possible range with the same velocity? [ 16s, 8868m, 10240m ]
Hints: Total time of flight is given by and time to reach the maximum height is the half of the time of flight. Also, horizontal range . For maximum horizontal range, the angle of projection should be 450.
46. A projectile is fired from ground with velocity 500 m / s, at 300 to the horizontal. Find the horizontal range, the greatest vertical height, and time to reach the greatest height. What is the least speed with which it would be projected in order to achieve the same range?
[ 21675m, 3125m, 25s,465 m/s ] Hints: Horizontal range . The time to reach the maximum height is half of the total time of flight. So, and for the calculation of least speed required to achieve the same range, take angle of projection 450.
47. A projectile is fired making certain angle to the horizon. If it attains the same height at times t1 and t2, show that the sum of these times is equal to the total time of flight.
Hints: Consider vertical upward motion and determine vertical displacement or height using formula s = ut + ½ at2 . The height will be same for two different times t1 and t2. Equating these two expressions, obtain t1 + t2 = i.e. total time of flight.

Laws of Motion
48. A car of mass 1000 kg is accelerating at 2 m/s2 what resultant force act on the car? If the resistance of the motion is 1000 N, what is the force due to engine? [ 3000 N ]
Hints: Resistance of motion or friction opposes the motion and to overcome the opposing force, engine has to generate more force. So, add opposing force with the force given by Newton’s second law of motion.
49. A box of mass 50 kg is pulled up from the hold of a ship with an acceleration of 1 m/s2 by a vertical rope attached to it. Find the tension in the rope. What is the new tension when the box moves up with a uniform velocity 1 m/s? [550N, 500N]
Hints: Force required to lift a body of mass ‘m’ is mg and force required accelerate a body is ma. Hence total force required to accelerate a body in vertical upward direction is equal to F = mg + ma = m(g+a). In second case, the box is moving with uniform velocity i.e. zero acceleration.
50. A lift moves (i) up and (ii) down with an acceleration 2 m / s2 . In each case, calculate the reaction of the floor on a man of mass 50 kg standing in the lift. [ 600 N, 400 N ] Hints: Force required to lift a body of mass ‘m’ is mg and force required accelerate a body is ma. Hence total force required to accelerate a body in vertical upward direction is equal to F = mg + ma = m(g+a). According to third law of motion, this force is equal to the reaction of the floor. Therefore, when lift moves upward, reaction of the floor R = m ( g + a) and when lift moves downward, the reaction R = m ( g – a ).
51. A ball of mass 0.2 kg falls from a height of 45 m. On striking the ground, it rebounds in 0.1 sec with 2/3 of striking velocity. Calculate the momentum change and (ii) the force on ball due to impact. [10 Ns, 100 N ] Hints: At first determine the velocity of the ball before it just touches the ground. Rebounding velocity is in vertically upward direction which is opposite to the falling downward velocity. So, take it negative. The change in momentum per unit time gives force.
52. A ball A of mass 0.1 kg moving with a velocity 6 m /s directly collides with ball B of mass 0.2 kg at rest. Calculate their common velocity if both ball move off together. If A had rebounded with velocity of 2 m/s in opposite direction after collision, what would be the new velocity of B? [ 2 m/s, 4m/s] Hints: Use the concept of conservation of linear momentum i.e. m1u1 + m2u2 = (m1 + m2)V . In second case, the expression of conservation of linear momentum will be m1u1 + m2u2 = m1v1 + m2v2. Since A rebounded with velocity 2 m/s, it would be negative.
53. An object of mass 10 kg is moving with velocity 6 m/s. If a constant opposing force 20 N is applied to the body, find time it takes to come to rest and the distance through which it moves. [ 3s, 9m]
Hints: Apply equation of motion. Acceleration of the object is obtained by force per unit mass.

54. A bullet of mass 20 g is fired horizontally into a stationary wooden block of mass 380 g with velocity 200 m /s. What is the common velocity of the bullet and block if it embedded the block? (b) If the block and the bullet experience a constant opposing force of 2N, find the time taken by them to come to rest. (c) If the wooden block is suspended in a weightless thread of length 10m, what is the angle made by the string with the vertical? [ 10 m/s, 2s, 300 ] Hints: Using principle of conservation of linear momentum, i.e. m1u1 + m2u2 = (m1 + m2)V, determine common velocity. (b) Take opposing force negative and then determine retardation a = F/m. Time to come into rest can be determined by using equation of motion, i.e. v = u + at (c) Vertical height gained by the body is determined by v2 = u2 + 2a h and angle with vertical can be determined geometrically.
55. A hose directs a horizontal jet of water moving with a velocity of 20 m/s on to a vertical wall. The cross section area of the jet is 5 X 10-4 m2 . If the density of the water is 1000 kg / m3, calculate the force on the wall assuming the water is brought to rest there. [ 200 N ] Hints: Force on the wall . But m = volume × density. So, and volume of water per second can be determined by area × velocity. Therefore, force

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Women’s Business: A Successful Nepali Businesswoman

Sulo Shrestha-Shah

Interview with Sulo Shrestha-Shah, Lotus Holdings

A successful businesswoman boosts business exports in Nepal by starting an investment company built on the principle of corporate social responsibility.

Lotus Holdings is an investment company and business incubator that has already helped to establish five manufacturing companies and six other companies in information technology (IT) and other service areas.

How it all began

Sulo Shrestha-Shah, president and founder of Lotus Holdings, began trading in carpets and textiles in 1991. As she had a German designer as business partner, she exported to Germany from the outset. She set up her own manufacturing company, Formation Carpets, when it became difficult to find high-quality goods.

Ms Shrestha-Shah’s experiences as a businesswoman in a man’s world, and the obstacles that she identified as hampering development in her native Nepal, shaped her vision in setting up Lotus Holdings. “It was the realization that I needed to look beyond myself which led to investing in other companies and sharing the market,” she explains.

Ms Shrestha-Shah says that although there are some women entrepreneurs currently exporting from Nepal, most women are unable to engage in business activities because their families prevent them from working.

Apart from the cultural resistance to women working outside the home, she sees the laws governing property rights as the main problem for would-be women entrepreneurs, since only males can inherit property in Nepal. If this were not the case, she is convinced that more Nepalese women would become entrepreneurs.

Blocks to export competitiveness

Although some barriers are specific to women, other obstacles to competitiveness are gender neutral. Ms Shrestha-Shah identifies skill shortages as a major difficulty for Nepalese companies, as 50% of the country’s population is illiterate.

Another problem is obtaining finance, since banks are unwilling to issue loans against companies, as is the practice in many countries.

Investments for success

Lotus Holdings and its affiliates believe that research and development activities are the only path to success in exports. After it was set up in 1998, Lotus Holdings began to research markets, such as France, Italy and the United States of America, with help from the Nepalese Government and the US Agency for International Development (USAID).

It plans to accelerate its current development rate by investing in companies in areas identified as productive and in need of assistance. It helps its members to export efficiently by providing advice and information on shipments and legal requirements.

To address the skills shortage problem, Lotus Holdings and its affiliates have introduced educational programmes for staff and their children. They have also invested in technology that they feel has the potential to increase exports.

Lotus Holdings has a strong philosophy of corporate social responsibility, and will only invest in companies that believe in ethical business, treat their employees fairly and invest in education. It has founded a non-governmental organization, Hoste Hainse, to focus on social issues. All companies within the group operate an equal rights policy, always employing the best person for the job. Ms Shrestha-Shah feels that the business community as a whole would have a better image if it focused more on corporate social responsibility.

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Company: Lotus Holdings

Sector: Manufacturing, trading, investment management, services

Location: Nepal

Employees: Over 400, of whom 20 are in the head office at Kathmandu

Yearly turnover: US$ 427,000

Export sales as % of total turnover: 50%

Current export markets: France, Germany, Italy, United Kingdom, United States of America

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What is “fetal alcohol syndrome”? What is a “buzz word”?

Fetal alcohol syndrome refers to a condition in new-born babies caused by excessive intake of alcohol by the mother during pregnancy: characterized by various defects including mental retardation.

Buzz word refers to a word or phrase, often sounding authoritative or technical, that is a vogue term in a particular profession, field of study, popular culture, etc. The buzz word in the adoption scene in the 1970s was matching.

What is “matching” and how is it done? Provide examples of matching in the text.

Matching  here refers to bringing two objects, ideas, or people together. Matching was a buzz word used in the American adoption practices in the 1970s. The social workers who worked at finding suitable adoptive parents for the homeless and parentless children living in foster homes first assessed the child’s characteristics, which involved getting information about the child’s personality, cultural background, existing relationship with biological or foster family, and emotional state. Based on these factors, the worker would draw up a profile of an appropriate family.

The first example that Traugot cites is of a 15-year old boy who had a bad history of

disrupted placements, did badly in academics and fought a lot. The appropriate family for this boy is a single male family, who could give him the latitude to be free but also circumscribe him when need be. Similarly, an 11-year old boy with Down’s Syndrome, a weak heart and hearing disability could be accepted by a deeply religious, working class family having older children. The child welfare specialists were really very optimistic and hopeful of the children finding permanent homes for these kinds of almost unadoptable children.

           

Student Assignment

Write an essay on ” the rising cases of divorces”. Spot the trend in divorce and trace the causes of the change.

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Comment on the influence of the Black Civil Rights Movement and the Women’s Movement in helping reform the adoption practices in America.

The Black Civil Rights Movement and the Women’s Movement had far-reaching and transformative effect on the adoption practices in America. The first movement helped in making America a more integrated and discrimination-free society. It helped to change the formerly vindictive attitude of the Whites against the Afro-Americans. In the changed American society, the liberal whites gathered the black and mixed-race infants and toddlers into their families. As well as that, the Blacks started to enjoy the riches of justice and decency.

The second movement, i.e. the Women’s Movement, gave women the reproductive rights. There was the easy availability of birth control  methods to them. Also, abortion was legalized. There was also a changed attitude toward sexual behaviour and marriage. Women didn’t have to get married to have sex. Even unwed mothers faced less societal stigma: they reared their child and were supported by their family members. Women’s rights advocates pointed out that a mature single woman could care for a child as well as two-parents could. All these changes reduced the birth of unwanted babies, and thereby children who could find a passage to foster homes. Thus, the two movements had a positive impact on adoption.

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Describe Tammy and her problems.

Tammy is a five and half-year old child who has the smile of Mona Lisa and the cuteness of a kitten. She is petite (very small) with brown eyes, and has dark, curly hair. Her complexion is light brown. However, behind this outward veneer is a girl who is suffering from fetal alcohol syndrome, and whose intellectual growth could stop at any time. Also, she is a lot older for adoption and not a white child – she is black. She has passed through the procedures set by the adoption agency and the regional or state exchange but has found no compassionate family to take her. That is why her profile has been advertised in the “Sunday Child” section of the Boston Globe for the potential family, who could adopt  and give her warmth, love and support.

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The Children Who Wait -Marsha Traugot September 13, 2013

The Children Who Wait by Marsha Traugot

The Children Who Wait is an essay by Marsh Traugot. Marsha is a social reformer and she describes the condition of the foster house in the USA. She also suggests reasons for a new trend in adoption in America. Now a wider verity of families can open their house to children who in the past had been labeled unadoptable.

The essay begins with an example of five and a half years old, black, handicapped girl, Tammy. She is suffering from fetal alcohol syndrome. Twenty years ago or until about 1960 the process of adoption was strict. A baby like Tammy was unadoptable and she was treated as waste material. But since the 1970s, she has also a market. She could be adopted. That is why a great departure in the field of adoption was seen in the 70s of the last century.

The field of adoption became very easy because of new technique, various civil rights movement, birth control, changing social values, legalized abortion and changes in views on sexual behaviour and marriage. Black civil right movement encouraged inter-racial adoption. The unwed mothers increased the number of adoptive infants. All these factors were responsible for the drastic change in the field of adoption. Due to lack of baby food, some interested families could not adopt children. As well as that, the black marketing of doctors also created scarcity of adoptable children. The attention inevitably shifted towards the children at foster houses.

Child welfare specialists were very worried about the growing children in foster houses. Earlier research had shown that children once sent to foster care for more than 18 months remain there until they grew up. This long stay led to dreadful results: children used to be the victims of mental and physical perversion. Of course, they could spoil their childhood, and their adult lives could be disturbed as well.

Funding was needed to improve the system but funding for children services had always been scarce because they did not cast their votes. The cost of keeping a child in foster house could run very high.

The conception of ideal adoptive family changed because the traditional family type has almost disappeared. A social worker makes the list of characteristics and looks for a good match. But for this, the social worker must change his attitude. He should believe firmly that even a disturbed or multiple handicapped child is adoptable. The worker should be flexible in his attitude towards a family of different socio-economic group. The specialists have started values clarification workshops for placements. Workers and their supervisors must be trained.

Adoption agencies find a potential family step-by-step. The process of matching is taken very seriously. The adoption agencies keep the list of families living in their periphery. Efforts are made to find a home from among the listed families, but if it fails then the adoption agencies refer the child to the State Adoption Exchange which gives information about the concerned child to other agencies. Monthly meetings and informal meetings are held for matching families. If they cannot find a matching family through any of the means they apply, the child welfare organisation and adoption exchange advertise through media, TV and the newspaper.

For example – Tammy is on search for an adoptive family. Because of the changes in attitudes in different aspects as well as in the field of adoption many children have got the supportive families and writer also hopes that Tammy will also get a warm supportive family life in the near future.

According to Traugot, what changes are transforming the American adoption scene? What factors are responsible for the changes?

Traugot wrote this essay in the 1970s. This decade saw transformation in the adoption of children, especially the possibility of integrating those from black, minority and mixed racial background into a wide variety of homes that became possible because of the disappearance of the traditional  middle-class, home-owning, two-parent, one-career families in America.

Until the 70s, only the upper class white childless couples adopted healthy white infants, but they did not adopt handicapped, black or mixed/minority, and older children. However, the late 70s marked a heartening change in the children adoption scene. The factors that contributed primarily for this transformation were the various civil rights movements, birth control, changing social values, harsh economic reality and research in social science.

Who are the children who wait and why do they wait?

The children who wait are children waiting in foster homes to find ideal prospective parents. They are mostly homeless children who come predominantly from black, mixed or minority background. Some of them could be orphans and some others could be living in foster homes because of familial problems, away from their biological parent(s). Whatever their background, these children are waiting for adoptive parents who could give them lot of love, affection, security and support. Traugot’s introduction of Tammy in the first paragraph draws the thesis of the essay, and readers instantly realize the possibility of Tammy’s adoption into a permanent  home should a family with a kind heart appear to take her  their home. She has been legally freed for adoption, so her profile has been advertised in a newspaper. Tammy is a representative of homeless and parentless children who come from diverse cultural background, have handicaps, have varying temperament and personalities.

Describe the procedure of finding prospective adoptive parents/Why is Tammy advertised in the Boston Globe?

The Children Who Wait by Marsha Traugot discusses the adoption system and problems in child adoption in America in the twentieth century. In this essay there are various stakeholders: the children waiting for adoption, the families to adopt them, the agencies to look after and help find children suitable adoptive family.

The first step is at the adoption agency. An adoption agency in one particular location has a list of families who wish to adopt children. If any one of the families wants to have a child than they could have it, but if it doesn’t materialize, the next step is at the regional or state adoption exchange, where the unadopted child is registered. The exchange distributes the photo and description of the child to all other agencies. Some of these exchanges hold monthly meetings where placement workers discuss children or families, and they also sponsor parties where children, workers and prospective parents meet informally. If the exchange also doesn’t succeed in finding a permanent home for a child, then it along with other child welfare organizations go for aggressive media advertisement where the profiles of the waiting child is either printed in newspapers like the Boston Globe or a video shown on TV. It is hoped that the waiting child can eventually get a caring and loving family.

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A Child is Born

This essay provides information about differences in matters of pregnancy and child birth, parent-child relationship, child rearing and the use of modern technology in traditional, agricultural societies and in affluent Western societies. Written in 1984, this essay begins by talking about pregnancy.

 A traditional woman has accepted certain practices that have come from the past. These practices are different but are accepted by all, and they also free the pregnant women from t he mental burden of finding out new ways. Practices like taboos and prohibitions, which are often superstitious, help woman to lower her anxiety although there is a possible danger in child birth. She knows of the various child-birth related problems that have happened in her society, but she goes about her life observing the taboos and prohibitions, which will help her with the unseen problems of the future. Some of these practices are magical but some are unusual and insensible, but they play a role in a woman’s pregnancy. She also has her husband, family and members of the community who provide vital support as and when required. She is emboldened to get through with childbirth. A western woman, on the other hand, may also have her own set of communal practices but their uses are less. Their uses could come in controversy with modern practices as it happened with one of the students of the writer who had taken training for unmedicated childbirth which resulted in the hospital’s nurses non-cooperation with breastfeeding, and this ultimately resulted in the woman’s leaving the hospital with her daughter two days after the delivery. Also, the birth of her delivery was unattended, which is just the opposite in non-technocratic societies, where births are always attended except during remarkable accidents.

Greer agrees that child mortality as well as mother mortality rate are higher in agricultural communities but she thinks the use of obstetric care in the West doesn’t allow woman true freedom which are often regarded as being more important than their own survival.

The role of a child is crucial in traditional society. A married woman would not be truly accepted by her husband’s family until she begets a child. This child will provide her the same intimacy as was hers with her mother. This is considered backward, wrong and cruel in the Western society. Also it is felt that traditional mother-in-laws are unjust and disposed to seek revenge (vindictive) with their daughter-in-laws and husbands too are exploitative and perfunctory (as a formality only).

Similarly, women in traditional society have no identity because when she is married she is called the daughter of someone. After marriage she becomes the wife of someone and when she gives birth to a child she is the mother of someone. Moreover, the woman is given her husband’s surname. Everywhere she would lose her name and identity. This is considered outrageous in the West.

Commenting on the parent-child relationship, Greer claims that in the traditional society mother-child relationship is more important than the relationship between husband and wife. However, child’s relationship with the family is considered equal or even more important than either. In traditional families child’s relationship with the rest of t he kin-group even can come at the expense of the biological parents as is the case in Africa and India. The child’s aunts and uncles are permitted more physical intimacy with the child in public than its parents. A child is born not because of the wants of the married couple but in response to a broader pressure from the whole group. A Rajput bridegroom suckling at his mother’s breast is suggestive of the child’s relationship with his kins group.

A woman giving birth to a child is praised and congratulated in many ways in non-technocratic societies. Greer provides a first-person account of the practices in Sylheti Community of West Bengal. A Sylheti girl can go her mother’s house for the last few months of her pregnancy, and she is very well looked after. When she gives birth to a child she is given gifts like sari while her child gets new dresses. There is a wild celebration at home. Garlands of garlic and turmeric are worn to ward off evil spirits. During the naming ceremony, unmarried girls and elderly women sit together and eat pan and sing. There are jokes and much laughter. They look funny eating pan and singing, and their songs are almost always oral and celebrate their day-to-day lives. This kind of fun-filled activities is absent in Western society. It is a dry occasion. The women there don’t get opportunity to go to their mothers’ houses to visit their mothers and their family members.

Child rearing in traditional societies is a whole group activity. Because they live in joint and/or extended joint-family system bringing up baby is the business of the entire family. In Bangladesh, for example, one daughter-in-law may bathe the children in the pond, another could feed them, and still another perhaps cooks the meal. And then, in the afternoon, the children are told fairy tales (Rupthoka) perhaps by their favorite aunt. However, when it gets dark or they feel sleepy they return to their mothers and sleep in their embraces.

Greer also voices her opinion on the intervention of Western medicine in the traditional societies in an effort to make them modern. Drugs are expensive; medical technological tools are not easily operational. The establishments of hospitals in these societies have created horrific conditions as Greer mentions Sheila Kitzinger account in one of her visits in an enormous modern hospital for “Bantu patients” in South Africa. The hospital was a representation of the meeting place of the old Africa and the new technology of the West. Majority of women were laboring and there was a pool of blood there but nurses busied here and there carrying sophisticated machines but ignored the cries of the laboring women as far as possible because these nurses wanted these women to witness “the blessings of the earth”. Greer is against the idea of making personal experience of giving birth to a child a personal disaster as it happens in the West where more and more women are subjected to several cases of brutalities. There are no one at home to give the mother and the child an enthusiastic welcome, is not praised for their  courage to give birth to a baby, and is not given a hand in raising a child. All of these happen in the traditional societies. She thinks that people in the traditional societies are more levelheaded than the people in western societies. They maintain their lives irrespective of medicine and doctors.

Greer considers that the practices of the traditional societies have more advantages than those of the Western attitudes.

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