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.
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.
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.
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 .
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 – mr2 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 = 2f.
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 = 2f and then determine number of revolution per second, i.e. frequency.
Gravity
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 – mr2 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 = 2f.
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 = 2f 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.
[ 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.
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
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. m1r12 = m2r22 and obtain r1 = 2/3 x106m and r2 = 1/3 x 106 m.
Then equate centrifugal force with gravitational force, and determine angular velocity.
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. m1r12 = m2r22 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,
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.
. 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 =
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 10t, 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.
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 10t, 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 I2 , (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 I2 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 I2 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. I11 = I22 . Or, I1 2f1 = I2 2f2 . 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 = I11 = I1 x 2f = I1 x 2 x ………….(i) Final angular momentum of disc & wax = ( I1 + I2) 2 = (I1 + mr2) x2f2 = [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.
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 I2 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 I2 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. I11 = I22 . Or, I1 2f1 = I2 2f2 . 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 = I11 = I1 x 2f = I1 x 2 x ………….(i) Final angular momentum of disc & wax = ( I1 + I2) 2 = (I1 + mr2) x2f2 = [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 =
[ 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]
[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.
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.
[ 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
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 = 6rv + 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 = 6rv + 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)
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 = 6rv + 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 = 6rv + 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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