Complete Numerical Collections Of Class -11 Hseb with Hints | Neb Hseb Notes

Complete Numerical Collections Of Class -11 Hseb with Hints

Complete Numerical Collections Of Class -11 Hseb with Hints



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

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)

Revision for ambitious students

132. A body moving through air at high speed ‘v’ experiences a retarding force given by f = k A  vx, where A is the surface area of the body,  is density of air and k is dimensionless constant. Deduce the expression.
133. The position of a particle moving along on X – axis is given by x = A t2 + Bt + C. The numerical value of A, B and C are 1, -4, 2. Find (i) dimension of A, B and C, (ii) the velocity of the particle at t = 4 sec, (iii) acceleration of the particle when t = 4 sec, (iv) the average velocity during the interval t = 0 to t = 5 sec.
134. A stone is dropped from a balloon going up with a velocity 5 m / s. If the balloon was 50m high when a stone was dropped, find its height when the stone hits the ground. [ 68.5 m]
135. A football is kicked horizontally with uniform velocity towards a vertical wall. If the ball rebound after colliding at the wall with same velocity, draw (i) speed – time (ii) velocity – time curve.

136. Human body can survive a negative acceleration trauma incident if the magnitude of acceleration is less than 250 m/s2. If you are in automobile accident with a initial speed 105 km/hr and are stopped by an airbag that inflates from the dash-board, over what distance must the air bag stop you to survive the crash ?
[ 1.7 m ]
137. A canoe has velocity of 0.40 m/south-east relative to the earth. The canoe is on the river that is flowing 0.50 m/s east relative to the earth. Find the velocity of the canoe relative to the river. [ 0.354m/s, -52.50, south of west ]
138. A 8 kg ice block, released from the top of a 1.5 m long frictionless ramp, sliding downhill, reaching a speed of 2.5 m/s at the bottom. What is the angle between the ramp and the horizontal? If the coefficient of friction is 0.05, what is the new angle of inclination?

139. Thorium decays by the emission of alpha particle ( A = 4 ) to an isotopes of Radium ( A = 226 ). Calculate the ratio of the speed of the alpha particle to the radium and hence calculate recoil KE of the Radium if the ejected energy of the alpha particle is 4.6 MeV.
[113:2, 0.082 MeV]

140. A man of mass 70 kg is standing on a large sheet of frictionless ice and holding a large rock of mass 15 kg. In order to get off the ice, the man throws the rock at speed 12 m/s relative to earth at an angle 350 above the horizontal. What is his initial speed after he throw the rock ? [ 2.11 m/s]
141. Rain falls vertically onto a plane roof 1.5 m square, which is inclined to the horizontal at an angle of 300. The rain drops strike the roof with a vertical velocity of 3 m/s and a volume of 2.5 x 10-2 m3 of water is collected from the roof in one minute. Assuming that the conditions are steady and the velocity of raindrop after impact is zero, calculate (a) vertical force exerted on the roof by the impact of the rain and (b) pressure normal to the roof due to the impact of the of the rain. (c) If, instead, the roof were subject to a rain of hard spheres, which collided elastically, what would be the normal pressure on the roof then be? ( Density of water 1000 kg/m3) [1.25N, 0.48N/m2, 0.96Pa]
142. A fire engine pumps water at such a rate that the velocity of the water leaving the nozzle is 15 m/s. If the jet be directed perpendicularly on to a wall and rebound of the water be neglected, calculate the pressure on the wall. 1 m3 of water has mass 1000 kg. [ 2.25 x 105 N/m2]
143. In a nuclear collision, an alpha particle A of mass 4 unit is incident with velocity v on a stationary helium nucleus B of 4 mass unit. After collision, A moves in the direction BC with velocity v/2, where BC makes angle 600 with initial direction AB and the helium nucleus moves along BD. Calculate the velocity of rebound of the helium nucleus along BD and angle made with the direction AB. [ 0.87v, 300 ]
144. The diagram shows identical simple pendulums of length 0.8m . Bob A is raised with the string taut to the horizontal position A’ and released. Calculate (a) Velocity with which A strikes B. (b) velocities of A and B just after A makes a perfectly elastic collision with B. [ 4m/s, 4m/s]
145. Sand is deposited at rate 20 kg/s in a conveyor belt moving horizontally at 10m/min. Find (i) force required to maintain constant velocity, (ii) Power required to maintain constant velocity, (iii) Rate of change of KE of the sand. [10/3N, 5/9W, 5/18W]

146. Calculate the acceleration and tension in each case.

147. If coefficient of friction is 0.1, determine new acceleration in each case.

148. A bullet of mass 0.01 kg moving with velocity 500 m / s strikes a block of mass 2 kg which is suspended by a string of length 5 m. If the block rises the vertical height 0.1 m , calculate the emergent velocity of the bullet. [ 220 m/s]
149. According to ‘Chandrasekhar limit’ burnt-out star of size three times the solar mass undergoes into black hole. What is the radius of the event horizon?
150. Mass of sun is 330000 times greater than that of earth. For a person at the surface of earth, the average distance from the center of sun is 23500 times the distance to the center of the earth. What is the ratio of the sun’s gravitational force to that of earth’s?
151. Two people are carrying a uniform wooden board that is 3m long and weighs 160 N. If one person applies an upward force equal to 600 N at one end, at what point does the other person lift?
152. A mass X of 0.1 kg is attached to the free end of a vertical helical spring whose upper end is fixed and the spring is extended to 0.04m. X is now pulled down to 0.02 m and then released. Find its (i) Period, (ii) Maximum force during oscillation, (iii) Maximum KE. [0.4 sec, 0.5 N, 0.005 J]
153. A solid body floats with one-half of its volume outside the water and floats 3/8 of its volume outside in another liquid. What is the density of solid and liquid?
[ 0.5 and 0.8 g /cc ]
154. A disc of moment of inertia 0.1 kg m2 about its center and radius 0.2 m is released from rest on a plane inclined at 300 to the horizon. Calculate angular velocity after it has rolled 2m down the plane if its mass is 5 kg. [ 18.3 rad / s ]
155. What is the power output in horse power of an electric motor turning at 4800 rev/min and developing a torque of 4.30 Nm?
156. A braided nylon rope, 2.5 cm in diameter has a breaking strength of 1.24 x 105 N. Find the breaking strength of similar ropes 1.25 cm in diameter.
157. A compressed tank of rocket contains 0.25 m3 of kerosene, with mass 205 kg. The pressure at the top of the kerosene is 2.01 x 105 Pa. The kerosene exerts a force 16.4 N at the bottom of the tank, which has area 0.07 m2. Find the depth of the kerosene.
158. A soap bubble of in a vacuum has a radius 3 cm and another at vacuum has 6 cm. If the two bubbles coalesce under isothermal condition, calculate the radius of the formed bubble.
159. Water flows steadily along a uniform tube of cross-section 30 cm2. The static pressure is 1.2 x 105 Pa and the total pressure is 1.28 x 105 Pa. Calculate the flow velocity and the mass of the water per second flowing.
Revision
HSEB Exam Questions

160. A bullet of mass 20gm, moving with velocity 500 m/s passes through a wooden block of mass 100 kg, initially at rest. The bullet emerges out with a speed 100 m/s and the block slides 20 cm. Find coefficient of sliding friction. [ 0.16 ]
161. A ball of mass 4kg moving with a velocity 10 m/s collides with another body of mass 16 kg moving with 4 m/s in opposite direction. If both coalesce into a single body, determine loss of energy on impact. [ 313.6 J ]
162. A ball A of mass 0.1 kg moving with a velocity of 6 m/s collides with B of mass 0.2 kg at rest. Calculate their common velocity if both ball move off together. If A had rebounded with a velocity of 2 m/s, in the opposite direction, what would be the new velocity of B? [ 2m/s, 4m/s ]
163. A bullet of mass 20 g traveling horizontally at 100 m/s embeds in the wooden block of mass 1 kg which is suspended by vertical string of length 1m. Calculate the maximum inclination of the string. [ 36.10 ]
164. A car of mass 2000 kg moves at the speed 20 m/s along a horizontal road where the frictional force is 200 N. Calculate the power developed by the engine. If the car now travels in an inclined road of inclination 150, what will be the new power developed? [4 Kw,107 Kw]
165. A mass of gas emitted from the rear of toy rocket is initially 0.2 kg/s. If the speed of the gas relative to the rocket is 40 m/s, and the mass of the rocket is 4 kg, what is the initial acceleration of the rocket? [ 2 m/s2 ]

166. A disc rolling along a horizontal plane has a moment of inertia 4 kgm2 about its center and mass of 5 kg. The velocity along the plane is 2 m/s and radius is 2m, find (i) angular velocity (ii) total energy of the disc. [ 1rad/s, 12J ]
167. A roller whose diameter is 1m weights 360N. What horizontal force is necessary to pull the roller over the brick 0.1m high when the force is applied at the center? [ 270 N ]
168. A 200 kg satellite is lifted to an orbit of 2.2×104 km radius. If radius and mass of the earth are 6400km and 6 x 1024 kg respectively, how much additional PE is required to lift the satellite? [ 8.87 x 109 J ]
169. The density of ice is 971 kg/m3 and the density of sea water is 1025 kg/m3. What fraction of iceberg is beneath the water surface? [ 0.947 : 1 ]
170. An iceberg having volume 2060 cc floats in sea water of density 1030 kg/m3 with a portion of 224 cc above the surface. Calculate the density of the ice. [ 0.89 g/cc ]
171. A bullet of mass 10g is fired from a gun of mass 1 kg with a velocity of 100 m/s. Calculate the ratio of the KE of the bullet and the gun. [ 100:1]
172. A coin placed on a disc rotates with speed of 331/3 rev/min provided that the coin is not more than 10 cm from the axis. Calculate the coefficient of static friction.
173. Speed of a body spinning about an axis increases from rest to 100 rev/min in 5 sec if a constant of 20 NM is applied. The torque is removed and the body comes to rest 100 sec due to friction. Calculate the frictional torque.
174. A 25 cm thick block of ice floating on fresh water can support a 80 kg man standing on it. What is the smallest area of the ice block? ( Sp gravity of ice = 0.917)
175. A small mass rests on a horizontal platform which vibrates in a SHM with a period of 0.25 s. Find the maximum amplitude of the motion which will allow the mass to remain in contact with platform thought the motion.
176. Calculate the period of the revolution of the satellite revolving at a distance of 20 km from the earth surface. ( R = 6400km, g = 10 m/s2 ) [ 5050.13 s ]

177. 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, 656 kw]
178. A constant torque of 500 Nm turns a wheel of moment of inertia 20 kgm2 about its center. Find the angular velocity and KE gained in 2 second and KE gained. [ 50 rad/s, 25000J]
179. A simple pendulum 4m long swings with amplitude of 0.2m. Determine (i) velocity at its lowest point, (ii) acceleration at the end of the path.
[ 0.32 m/s,0.5m/s2]
180. A piece of gold aluminum alloy weighs 100g in air and 80g in water. What is the weight of the gold in the alloy if the relative density of gold is 19.3 and that of aluminum is 2.5
181. An object of mass 8 kg is whirled round in a vertical circle of radius 2m with a speed 6m/s. Calculate the maximum and minimum tension in the string.
[ 224N, 64N]
182. A body is projected horizontally from the top of a tower of height 100m with a velocity 9.8 m/s. Find the velocity with which it hits the ground. [ 45.8 m/s ]
183. A particle of mass 0.3 kg vibrates with a velocity 2sec. If its amplitude is 0.5m, what is its maximum KE? [ 0.37 J ]
184. A string supports a solid Iron of mass 200g totally immersed in liquid of density 800kg/m3. If the density of the iron is 8000 kg/m3, calculate the tension in the string. [ 1.8 N ]
185. An object is dropped from the top of a tower of height 156.8 m and at the same time, another object is thrown vertically upward with velocity 78.4 m/s from the foot of the tower. When and where will they meet? [ 2s, 20m from top]
186. A constant torque of 200 Nm turns a wheel which has a moment of inertia 100 kg m2 about its center. Find KE gained after 20 revolutions. [
187. The displacement y of a mass vibrating with SHM is given by y = 20 sin 10t. Where y is in millimeter and t is in second. What is (a) amplitude (b) period (c) velocity when t = 0. [ 0.002m, 0.2 s, 0.682 m/s]
188. 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]



HEAT

Temperature
189. At what temperature degree Fahrenheit scale shows twice the reading of degree centigrade? [ 160 0C] Hints: The relation between C and F scale of temperature is . Take C = , then F = 2.
190. At what temperature degree Fahrenheit scale shows half of the reading of degree centigrade? [ – 24.62 0C] Hints: Take C = , then F will be /2 and put the value in relation
191. At what temperature Fahrenheit and Kelvin scale give the same reading?
[ 574.25K]
Hints: Take the temperature F = , then K is also  and put the value in relation
192. The normal temperature of human body is 37 0C. What is its value in 0F and Kelvin scale? [ 98.60F, 310 K]
193. A faulty thermometer has its fixed points –100C and 130 0C. This thermometer reads the temperature of an object 60 0C. Find the correct temperature of the body in Celsius scale. [ 500 C ] Hints: The term where L is lower fix point and U is upper fix point, remains constant for all scale of thermometer. Use same expression for both correct thermometer and faulty thermometer.
194. A faulty Celsius thermometer reads – 4 0C when placed in melting ice and reads 98 0C when placed in the contact with steam at normal pressure. What the correct temperature in that scale when it reads room temperature 32 0C ? [ 35.3 0C ]

Thermal Expansion
195. A steel meter scale has length 100.00 cm at temperature 10 0C. If the temperature rises to 20 0C, what is the length of the scale? (  for steel 1.6 x 10-6 / k). If a road of length 1 km is measured at 20 0C, what will be the error in the measurement? Hints: The length of the scale at temperature 20 0C is given by l2 = l1 [ 1+ (2 – 1)]. Divide length of the road by the length of the scale at 20 0C to obtain its reading.
196. An aluminum rod when measured with a steel scale, both being at 25 0C appears to be 1 m long. If the scale is correct at 0 0C, (a) what is the true length of the scale at that temperature? (b) What will be the length of the rod at 0 0C? Linear expansivity of aluminum is 26 x 20-6 /K and of steel is 12 x 10-6/K. [1.0003m, 99.96m] Hints:
197. A glass vessel of volume 200 cm3 is just filled with mercury at 100C. If the temperature raised to 120 0C, how much mercury overflow? ( for glass 1.2 x 10-6 / k and  for mercury 1.6 x 10 –4 /k ) [ 3.44 cm3] Hints: The overflowing mercury is the difference between the volume of mercury and the volume of vessel at 120 0C. So, determine volume of mercury and volume of vessel at 120 0C separately and calculate the difference between them.
198. A thermal tap used in certain apparatus consists a silica rod which fits tightly inside an Aluminum tube whose internal diameter is 8 mm at 00 C. When temperature is raised, the fit is no longer exact. Calculate what change of temperature is necessary to provide a channel whose cross-section is equal to that of a tube of 1mm internal diameter. Linear expansivity of Si and Al are 8 x 10-6/K and 26 x 10-6 /K. [ 434 K ]
Hints: Let at temperature , the channel area will be equal to the cross section area of a tube of 1 mm internal diameter which is given by . The channel area is equal to the difference between the cross-section areas of the cylinder(Al) to piston(Si). So, (A)Al – (A)Si = .

199. A brass pendulum clock gives correct time at 15 0C. How many second does it lose or gain per day when the temperature changes to 20 0C? ( for brass 0.000019 / k ) [ 4 s ]
Hints: Time period of pendulum at 15 0C is and that of 20 0C is . Since correct time of pendulum is 2 second, put T1 = 2 second. The relation between l2 and l1 is . Put the value of l2 and divide T1 to T2. Determine difference between T1 and T2 and finally determine gain or lose in time per day.
200. A steel wire 8m long and 4 mm in diameter is fixed to two rigid support. Calculate the increase in tension when the temperature falls by 100 C? Given linear expansivity of steel 12 x 10-6 /K, Young modulus for steel 2 x 1011 N/m2 . [ 300 N ]
Hints: Young’s modulus Y = , where F is force or tension, l is initial length, A is cross section area and e is extension or contraction of the wire. Use relation or Or, e = where e = is the contraction. Finally use the relation of young’s modulus to calculate tension.
201. A steel cylinder has an Aluminum piston and at temperature of 200 C, internal diameter of cylinder is exactly 10 cm, there is an around clearance of 0.05 mm between the piston and the cylinder wall. At what temperature will the fit be perfect? Given linear expansivity of steel and Aluminum are 1.2 x 10-5 /K and 1.6 x 10-5 /K respectively. [ 2700 C ]
Hints: To fit perfectly, the diameter of steel piston should be exactly equal to the diameter of aluminum cylinder. Consider the temperature  at which their diameters are exactly equal. So, .

202. Aniline is a liquid which doesn’t mix with water and when a small quantity of it is poured into beaker of water at 200 C, it sinks to the bottom, the densities of two liquids at 200 C are 1021 and 998 Kg/m3 respectively. At what temperature must the beaker and its contents be heated so that the aniline will form a globule which just floats in the water? Given absolute expansivity of aniline and water are 0.00085 /K and 0.00045 /K respectively. [ 790 C ]
Hints: To satisfy the mentioned condition, density of aniline must be equal to the density of water. So, let at temperature , their densities are exactly equal i.e. .

203. A certain Fortin barometer has its pointer, body and scale made from brass. When it is at 00 C, it records pressure 760 mmHg. What will it read when its temperature is increased to 200C if pressure of the atmospheric remains unchanged? Given cubical expansivity of mercury 1.8 x 10-4 /K, Linear expansivity of brass 2 x 10-5 /K. [ 762.4 mm Hg ] Hints: Use the expression of correction of barometer.
204. Using the following data, determine the temperature at which wood will just sink in benzene? Density of benzene at 00C = 900 kg / m3. Density of wood at 00C = 880 kg / m3. Cubical expansivity of benzene = 0.0012 / k
Cubical expansivity of wood = 0.0015 / k
Hints: To satisfy the mentioned condition, density of wood should be exactly equal to the density of the benzene. So, let at temperature , their densities are exactly equal i.e.
Calorimetry / Change of state
205. A steel ball of mass 80 gram is taken from a furnace of temperature 2000C and dropped in a copper calorimeter of mass 50 gram containing 65 gm of water at temperature 20 0C. If the final temperature of the mixture rises to 22.5 0C, calculate the specific heat capacity of the steel. ( Specific heat capacity of copper is 0.1 cal / gm k. Hints: In the above phenomenon, steel ball loss heat but calorimeter and water gains heat. According to principle of calorimetry, heat loss is equal to the heat gain. So, heat loss by steel ball Q1 = m  S   Heat gain by caloriemeter and water Q2 = (m  S  ) for water + (m  S   ) for calorimeter. From principle of calorimetry, Q1 = Q2
206. A copper ball of mass 15 g is held in a flame until it has acquired the temperature of the flame. It is then quickly transferred to a copper calorimeter of mass 66 g containing 50 g of water at 20 0C. If the final steady temperature of the mixture is 27.5 0C, calculate the temperature of the flame. [323 0C]
207. Two identical calorimeters each of heat capacity 12 J/K, one contains 8 x 10-5 m3 of water and takes 150 second to cool from 325 K to 320 K, and the other contains an equal volume of an unknown liquid which takes 50 second to cool over the same range of temperature. If the density of the liquid is 800 kg/m3, what is the specific heat capacity of liquid over the temperature range? Given , density of water 1000 kg/m3 and heat capacity 4200 J/K. [ 1625 L/kgK]
208. A lead bullet moving with velocity of 400m/s strikes a target and is brought to rest. If half of the kinetic energy goes to raise the temperature of the bullet, by how many degrees will its temperature rise? Given specific heat capacity of the bullet is 0.13 J/g 0C . [ 307.20C] Hints: According to question, half of the KE changes into heat. So, ( ½ mv2 ) = m x s x 
209. A substance takes 3 min in cooling from 50 0C to 45 0C and takes 5 min in cooling from 45 0C to 40 0C. What is the temperature of its surrounding? How much time will it take to cool from 40 0C to 35 0C ? [ 35 0C, 15 min ]
210. How much energy is required to convert 500 g ice of temperature –5 0C to steam of temperature 100 0C ? ( Specific and latent heat of ice are 0.5 cal /gm 0C and 80 cal /gm and latent heat of steam is 340 cal / gm. )
211. 5 gm of ice at 0 0C is dropped into a beaker containing 20 g of water at 40 0C. What will be the final temperature? [ 16 0C ]
212. What is the result of mixture when 400g of water and 100 g of ice at 00C are in a copper calorimeter of mass 500g and 100 g of steam is passed into it? [
213. What is the result of mixing 20g of ice at -10 0C and 50g of water at 300C ? [ Specific heat of ice 2100 J/kg K & latent heat of ice 3.34 x 105 J/kg ]
214. 1 g of steam at 100 0C can melt how much ice at 0 0C ? [ 8 g ]
Ideal Gas Equation
215. A container of gas has volume 0.1 m3 at a pressure of 2 x 105 N/m2 and a temperature of 27 0C. (a) Find the new pressure if the gas is heated at constant volume to 87 0C, (b) The gas pressure is now reduced to 1.0 x 105 N/m2 at constant temperature. What is the new volume of the gas? (c) The gas cooled to – 73 0C at constant pressure. Find the new volume of the gas. [2.4 x 105 N/m2, 0.24 m3, 0.13 m3 ]
216. A cylinder of gas has mass of 10.0 kg and pressure of 8.0 atmosphere at 27 0C. When some gas is used in a cooled room at – 3 0C, the gas remaining in the cylinder at this temperature has a pressure 6.4 atm. Calculate the mass of the gas used. [ 1.1 kg ] Hints: From ideal gas equation, P1V = m1 r T1 ………..(i) When some gas is used in a cool room, the equation for the remaining gas will be P2V = m2 r T2 ………..(ii) Dividing equation (i) by (ii),
217. Two glass bulb of equal volume are joined by a narrow tube and are filled with a gas at stp. When one bulb is kept in melting ice and the other is placed in hot bath, the new pressure is 877.6 mm of Hg. Calculate the temperature of the bath. [ 100 0C ] Hints: When both bulbs are in STP, the ideal gas equation will be P1 (2V) = m r T1 So, m = When one bulb is placed in melting ice, P1 V = m1 r T1 So, m1 = When another bulb is placed in hot bath, mass of gas in the bulb is m2 = Since mass of gas remains constant, m1 + m2 = m
218. Two vessels of capacity 1.00 litre are connected by a tube of negligible volume. Together, they contain 3.42 x 10-4 kg of helium at a pressure of 800 mm of mercury at temperature 27 0C. Calculate (a) value for the constant ‘r’ for helium, (b) the pressure developed in the apparatus if one vessel is cooled to 0 0C and next is heated to 100 0C. [ 2080 J/kg K, 842 mm ]
Hints: Both bulbs are in same pressure, the ideal gas equation will be P (2V) = m r T where, pressure P = 800 mm of Hg = 800 x 13600×9.8 Nm2 volume V = 1 litre = 10-3 m3 mass m = 3.42 x 10-4 Kg temperature T = 270C = 300 K
219. What volume of liquid oxygen( density 1140 kg/m3) may be made by liquefying completely the contents of a cylinder of gaseous oxygen containing 100 litre at 120 atm pressure at 20 0C? Assume that oxygen behaves as an ideal gas in this latter region of pressure and temperature. Given 1 atmosphere = 1.01 x 105 N/m2, molar gas constant is 8.31 J /molK and relative molecular mass of oxygen is 32. [ 0.014 m3 ]
220. A sealed bottle full of water is placed in a strong container full of air at standard atmospheric pressure and at a temperature of 10 0C. The temperature in the container is raised to and maintained at 100 0C. Neglecting the expansion of the bottle and the container, what is the new pressure in the container? If the bottle breaks, what will be the pressure be? [ 1.3 x 105 N/m2, 2.3 x 105]
221. A container of gas has a volume of 0. 1 m3 at pressure of 2 x 105 N / m2 and temperature 27 0C. (i) Find the new pressure if the gas is heated at constant volume to 87 0C. (ii) The pressure is now reduced to 1 atmosphere at constant temperature.
Kinetic theory of gas
222. Calculate the root – mean- square speed at 0 0C of (i) Hydrogen molecules and (ii) Oxygen molecules, assuming 1 mole of the gas occupies a volume of 2 x 10-2 m3 at 0 0C and pressure 105 N/m2. Relative molecular masses of hydrogen and oxygen are 2 and 32 respectively. [1732 m/s, 433 m/s]
Hints: The pressure exerted by gas in terms of root mean square speed is given by . Replace density by mass per unit volume and finally determine C.
223. Assuming helium molecules have a root mean square speed of 900 m/s at 27 0C and 105 pressure, calculate the rms speed at (i) 127 0C and 105 N/m2, (ii) 27 0C and 2 x 105 N/m2 pressure. [ 1039 m/s, 900 m/s]
Hints:

224. Air may be taken to consist of 80 % nitrogen molecules and 20 %oxygen molecules of relative molecular masses 28 and 32 respectively. Calculate (a) ratio of the rms speed of nitrogen molecules to that of oxygen molecules, (b) ratio of the partial pressure of nitrogen and oxygen molecules, (c) ratio of the rms speed of nitrogen molecules at 100C to that at 2000C. [ 1.07:1, 4:1, 0.87:1]
225. Calculate the pressure in mm of Hg exerted by Hydrogen gas if the number of molecules per cm3 is 6.80 x 1015 and the root mean square speed of the molecules is 1.90 x 103 m/s. Given Avogadro constant is 6.02 x 1023 and relative molecular mass of hydrogen is 2.02. [ 0.21 mm of Hg ]
226. Assuming that the density of Nitrogen at stp to be 1.251 kg /m3 find the root mean square velocity of nitrogen molecules at 127 0C. [ 597 m/s ]
Hints: At first determine rms speed of nitrogen at stp and then determine its value at another temperature by using relation .
227. Air at 273 K and 1.01 x 105 N/m2 pressure contains 2.7 x 1025 molecules per cubic meter. How many molecules per cubic meter will there be at a place where temperature is 223 K and pressure is 1.33 x 104 N / m2

Hygrometry
228. If the temperature of air is 16.5 0C and dew point is 6.5 0C , find the relative humidity of air. ( SVP at 6, 7 , 16 and 17 0C are 7.05, 7.51, 13.62 and 14.42 mm of Hg respectively ) [ 51.9%]
229. What is the dew point on a day if the humidity is 40% on that day when the temperature is 300 C.
[12 0C ]
230. On a certain day the dew point is 8.5 0C and the room temperature is 18.4 0C. Find the RH if the maximum vapour pressure for 8, 9, 18 and 19 0C are 8.04, 8.61, 15.46 and 16.46 mm of Hg respectively. [ 52.5 %]

Transfer of heat
231. A man, the surface area of whose skin is 2 m2, is sitting in a room where the air temperature is 20 0C. If the temperature of skin is 28 0C , find the rate at which his body loses heat. The emissivity of his skin is 0.97 and  = 5.7 x 10-8 W/m2 K4. [ 92.2 W]
Hints: The rate of heat loss due to radiation according to Stephen’s law is given by , where is emissivity.
232. Calculate the rate of loss of heat from an unclothed person standing in air at 23 0C . Assuming that the skin temperature is 34 0C and the body surface area is 1.5 m2 , emissivity = 0.7 and  = 5.7 x 10-8 W/m2 K4. [ 46.85 W]
Hints: The rate of heat loss due to radiation according to Stephen’s law is given by .
233. A closed metal vessel contains water (i) at 30 0C (ii) at 75 0C . The vessel has surface area 0.5 m2 and a uniform thickness of 4 mm. If the outside temperature is 15 0C , calculate the heat loss per minute by conduction in each case. ( Thermal conductivity of metal = 400 W / mK ) [ 4.5×107, 18 x107J]
Hints: Use the relation of rate of flow of heat due to conduction is

234. Assuming that the thermal conductivity of woolen glove is equivalent to a layer of quiescent air 3 mm thick, determine the heat lost per minute from the man’s hand of surface area 200 cm2 0n a winter’s day when atmospheric temperature is 3 0C. The skin temperature is taken 34 0C and thermal conductivity of air is 24 x 10-3 W/mK. [ 354 J/min]
Hints: Use the relation of rate of flow of heat due to conduction is

235. A bar of 0.2 m in length and cross-sectional area 2.5 x 10-4 m2 is ideally lagged. One end is maintained at 373 k and other is kept in melting ice. Calculate the rate at which ice melts. [ 1.47 x 10-4 Kg / s ]
Hints: At first calculate and the energy is used to melt ice. So, mL = Q.

236. One face of a sheet of cork is 3 mm thick is placed contact with one face of the of a sheet of glass 5 mm thick, both sheet being 20 cm square. The outer faces of this square composite sheet are maintained at 100 0C and 20 0C , the glass being at the higher temperature. Find (a) the temperature of the glass – cork interface, (b) the rate at which heat is conducted across the sheet, neglecting edge effect. Thermal conductivity of cork and glass are 6.3 x 10-2 W/mK and 7.2 x 10-1 W/mK respectively. [ 90 0C, 57.6 W ]
Hints: The rate of flow of heat through glass due to conduction, ………..(i) Similarly, rate of flow of heat through cork
………..(ii)
Rate of flow of heat through glass is equal to that of cork. So, equating (i) and (ii),  can be determined.
237. Estimate the rate of heat loss from a room through glass window of area 2 m2 and thickness 3 mm when temperature of room is 20 0C and that of air outside is 5 0C. Thermal conductivity of glass is 0.72 W/mK respectively. [ 7200 W]
Hints: Use the relation of rate of flow of heat due to conduction

238. The silica cylinder of a radiant wall heater is 0.6 m long and has a radius of 5 mm. If it is rated at 1.5 kW, estimate the temperature when operated. State two assumptions you have made in making your statement. Given Stephen constant is 6 x 10-8 W/m2K4. [1070 K]
Hints: Use relation . The assumptions are (a) The heater is considered as a perfectly black body, (b) The temperature of surrounding is considered to absolute zero.

239. What is the ratio of the energy per second radiated by the filament of a lamp at 2500 K to that radiated at 2000 K, assuming the filament is a black body radiator. The filament of a lamp can be considered as a 90% black body radiator. Calculated the energy per second radiated when its temperature is 2000 K and surface area is 10-6 m2. [ 2.44, 0.82] Hints: Apply relation for both lamps and divide them to get result. In second case, take efficiency 90 % or 0.9.
240. The sun is a black body of surface temperature 6000 K. If the sun’s radius is 7 x 108 m, calculate the energy per second radiated from its surface. The earth is about 1.5 x 1011 m from the sun. Assuming all the radiation from the sun falls on a sphere of this radius, estimate the energy per second per meter square ( solar constant ) received by the earth. Stephen constant is 5.7 x 10-8 W/m2K4. [ 1600 W/m2 ] Hints: Apply relation to obtain energy radiated by sun per second. The energy radiated by sun spherically distributes all over and solar constant is the energy received by earth per unit area per unit time. So divide the power radiated by sun ‘P’ by surface are of the sphere made by the radius 1.5 x 1011m. i.e Solar constant S =
241. A sphere of radius 2 cm with a black surface is cooled and then suspended in a large evacuated enclosure the black walls of which are maintained at 27 0C. If the rate of change of thermal energy of the sphere is 1.85 J/s, when its temperature is -73 0C, calculate the value for the Stephen constant. [5.7 x 10-8 W/m2K4 ]
Hints: The rate of heat loss by black body is given by .

242. The element of an electric fire, with an output of 1.0 kW, is a cylinder 25 cm long and 1.5 cm in diameter. Calculate the temperature when in use of it behaves as a black body. [ 1105 K ]
Hints: Use relation , where A is the curved surface area of cylindrical electric element, equal to 2 r h
243. A solid copper sphere of diameter 10 mm is cooled to a temperature of 150 K and is then placed in an enclosure maintained at 290 K. Assuming that all interchange of heat is by radiation, calculate the initial rate of rise of temperature of the sphere. The sphere may be treated as a black body. Density of copper 893 kg /m3, specific heat capacity of copper 3.7 x 102 J / kgK and Stephen constant 5.7 x 10-8 W/m2K4. [ 0.068 K/s]
Hints: Use relation . This energy is used to increase the temperature. So, . Equate the relations to obtain .
244. The element of 1 kW electric fire has a surface area of 0.006 m2. Estimate the working temperature. Stephen constant is 5.7 x 10-8 W/m2K4 [ 1300 K]
Hints: Use relation .
245. A roof measures 20m x 50m and is blackened. If temperature of the sun’s surface is 6000 K and its radius is 7.8 x 108 m, and the distance from the earth is 1.5 x 1011 m calculate how much solar energy is incident on the roof per minute. Assume that the half of the energy is lost in passing through the earth’s atmosphere and the roof is normal to the sun’s rays. [ 5.6 x 107 J]
Hints: Power radiated by sun and solar constant S = . The additional term ½ is taken according to question that half of the energy is lost in passing through the earth’s atmosphere. Finally multiply area of the roof with solar constant.
246. Calculate the apparent temperature of the sun from the following data. Sun’s radius = 7.04 x 105 Km. Distance from earth = 14.72 x 107 Km Solar constant = 1400 w m-2 Stephen constant = 5.7 x 10-8 W m-2 k-4 [ 5450 0C ]
Hints: Total power released by sun is P = S × 4 r2, where ‘r’ is the distance of sun from earth. Finally use the Stephen’s law , where A be the surface area of sun, given by A = 4 R2.

Thermodynamics

247. For hydrogen, the molar heat capacity at constant volume and pressure are respectively 20.5 J/molK and 28.8 J/molK. (a) Which heat capacity is related to internal energy ? (b) Calculate the molar gas constant. (c) Calculate the heat needed to rise the temperature of 8g hydrogen from
100 C to 150 C at constant pressure? (d) Increase in the internal energy of the gas. (e) External work done.
[ 8.3 J/molK, 576 J, 410 J, 166 J] Hints: (a) Specific heat capacity at constant volume Cv is related to the internal energy. (b) Molar gas constant R = Cp – Cv (c) dQ = N  Cp  dT, where N is the number of mole = 4/2 = 4. (d) Increase in internal energy dU = N  Cv  dT (e) External work done dW = dQ – dU
248. A gas has a volume of 0.02 m3 at pressure of 2 x 105 Pa and temperature of 27 0C. It is heated at constant pressure until its volume be 0.03 m3. Calculate the (a) External work done, (b) The new temperature of the gas, (c) Increase in internal energy of the gas. Mass of the gas is 16 g, molar heat capacity at constant volume is 0.8 J/molK and its molar mass is 32 g. [ 2000J, 450K, 60J] Hints: (a) External work done dW = P  V (b) Ideal gas equation (c) Increase in internal energy dU = N  Cv  dT
249. If ratio of the principal specific heat capacities of a gas is 1.4 and its density at stp is 0.090 kg/m3. Calculate the specific heat capacities at constant pressure and at constant volume. [ 144 x104, 1.03 x 104 J/kgK] Hints: Pressure exerted by gas P =  r T. So, gas constant r = Now, cp – cv = r …………(i) and cp = 1.4 cv ………..(ii) Solving equation (i) and (ii), cp and cv can be determined.
250. Given that the volume of a gas at stp is 2.24 x 10-2 m3 /mol , calculate the molar gas constant R and use it to find the difference between the quantities of heat required to raise the temperature of 0.01 kg of oxygen from 0 0C to 10 0C when (a) pressure (b) volume of the gas is kept constant. Given relative molecular mass of the oxygen = 32. [ 8.3 J/molK, 25.9 J] Hints: Molar gas equation PV = NRT gives the molar gas constant ‘R’. Energy required to heat at constant pressure dQ = N Cp  T …………..(i) Energy required to heat at constant volume dU = N Cv  T ……………..(ii) Subtracting, the obtained result is dQ – dU = N  T  ( Cp – Cv) = N  T  R.
251. An ideal gas at a temperature of 290 K and a pressure of 1.0 x 105 N/m2, occupies the volume of 1.0 x 10-3 m3. Its density under these conditions is 0.30 kg/m3. It expands at a constant pressure to a volume of 1.5 x 10-3 m3.Calculate the energy added. It is now compressed isothermally to its original volume. Calculate its final pressure and temperature and the difference between its initial and final internal energy. Given Cv = 7.1 x 102 J / Kg K . [ 81J, 1.5 x 105 N/m2, 435K, 30.9 J] Hints: Ideal gas equation gives the final temperature T2. Pressure exerted by gas P =  r T. So, gas constant r = . Also, specific heat capacity at constant pressure cp = cv + r and Heat energy added dQ = m  cp  T Again, Ideal gas equation P2V2 = P3V3 gives the final pressure P3. Difference between initial and final internal energy is dU = m x cv x dT
252. The density of ideal gas is 1.6 kg/m3 at 27 0C and 1.00 x 105 N/m2 pressure and specific heat capacity at constant volume is 312 J/ kgK. Find the ratio of specific heat capacities at constant pressure and constant volume. [1.67] Hints: Pressure exerted by gas P =  r T. So, gas constant r = . Now, specific heat capacity at constant pressure cp = cv + r. And, ratio of specific heat capacities  = cp / cv
253. A gas in a cylinder of temperature 17 0C and a pressure of 1.01 x 105 N/m2 is to be compressed to one-eighth of its volume. What would be the difference between the final pressures if the compression is done (a) isothermally, (b) adiabatically ? (c) What would be the final pressure in the later case? Given  = 1.40 [10 .48 x 105 N/m2, 666 K] Hints: When gas is compressed isothermally, P1 V1 = P2 V2 When gas is compressed adiabatically, P1 V1 = P2 V2
254. A liter of air initially at 20 0C and 760 mm of Hg is heated at constant pressure until its volume is doubled. Find (a) its final temperature, (b) external work done, (c) the quantity of heat supplied. Given density of air at stp is 1.293 kg/m3 and Cv = 714 J/kgK. [ 586 K, 101.2 J, 355J] Hints: (a) Ideal gas equation gives final temperature (b) External work done, W = P V, where V = 2 lit – 1 lit = 1 lit = 10-3 m3 (c) Volume of gas at 00C is given by Or, Or, V1 = 9.32 x 10-4 m3 Therefore , mass of the gas m =  x V = 1.293 x 9.32 x 10-4 = 1.205 x 10-3 kg Quantity of heat supplied dQ = dU + dW = m x cv x dT + P V
255. A mass of air occupying initially a volume 2 x 10-3 m3 at a pressure 760 mm of Hg and temperature 20 0C is expanded adiabatically reversibly to twice its volume, and then compressed isothermally and reversibly to a volume of 3 x 10-3 m3. Find the final temperature and pressure. Given  = 1.40
[ 222K, 384 mm of Hg] Hints: Expression for adiabatic expansion P1 V1 = P2 V2 Expression for isothermal compression P2 V2 = P3 V3. Where, P3 is the final pressure. Similarly, the expression for adiabatic expansion T1V1-1 = T2V2-1. In isothermal process, temperature remains constant. So, T2 is the final temperature after isothermal compression.
256. A petrol engines consumes 5 kg petrol per hour. If the power of the engine is 20 KW and the calorific value of petrol is 11 x 103 K Cal per kg, calculate the efficiency of the engine. [ 31.15%] Hints: Efficiency of an engine Where, output = 20 KW = 20000 W Input = 5 Kg petrol per hour = 5 / 60 x 60 kg per sec. = Watt
257. An ideal heat engine operates in a Carnot Cycle between 227 0C and 127 0C. It absorbs 6 x 104 J at the higher temperature . How much work per cycle is this engine capable of performing? [ 5 x 104 J ] Hints: Efficiency of an engine = So, = Where, T2 = temperature of sink in absolute scale T1 = temperature of source in absolute scale Input = 6 x 104 J
258. A Carnot engine operates between a hot reservoir at 320 K and a cold reservoir 260 K. If it absorbs 500 J of heat from hot reservoir, how much work does it deliver? If the same engine working in reverses as a refrigerator, how much work must be supplied to remove 1000 J of heat from the cold reservoir? [ 4333 J] Hints: Efficiency of a heat engine is , = where, T2 = temperature of sink, 260 K T1 = temperature of source, 320 K Input = 500 J Output = ? If the same engine works as a refrigerator, its efficiency Where,  = efficiency obtained from above , 18.75 % Q2 = heat taken from cold body, 1000 J W = work supplied, to determine.
259. A motor in a refrigerator has a power of output 200 W. If the freezing compartment is at 270 K and room temperature is 300 K, assuming ideal efficiency, what is the maximum amount of heat that can be extracted from the freezing compartment in 10 min? [ 1.1 x 106 ]

Revision
HSEB Exam Questions

260. From what height a block of ice is dropped in order that it may completely melt. It is assumed that 20% of energy of fall is retained by ice. [168000m] Hints: Actually the ice doesn’t melt completely when it is dropped from any height. But just in the view of examination, take 20% PE = mL and determine height.
261. Find the rms speed of Nitrogen at NTP. Density of nitrogen is 1.29 kg/m3 at NTP. [ 484.65 m/s ]
262. How much heat is needed to change 10 g ice at – 10 0C to steam at 100 0C. Given, specific heat capacity of ice is 0.5 cal/gK, Latent heat of ice 80 cal/gK and that of steam 540 cal/gK. [30450 J]
263. Helium gas occupies a volume of 0.04 m3 at a pressure of 2×105 N/m2 and temperature 300K. Calculate the mass of the helium and rms speed of its molecule. Relative molecular mass of helium = 4 , molar gas constant = 8.3 J/ mol K. [ 1366.57 m/s]
264. What is the result of mixing 100g of ice at 0 0C and 100 g water at 100 0C ? Latent heat of ice 3.36 x 105 J/K. [ 10 0C ]
265. Estimate the rate of heat loss through a glass window of area 2 m2 and thickness 4 mm when the temperature of the room is 300 K and that of outside is 5 0C? [ 11000K watt ]
266. A cylinder of gas has mass of 10.0 kg and pressure of 8.0 atmosphere at 27 0C. When some gas is used in a cooled room at – 3 0C, the gas remaining in the cylinder at this temperature has a pressure 6.4 atm. Calculate the mass of the gas used. [ 1.1 kg ]
267. Assuming that the thermal conductivity of woolen glove is equivalent to a layer of quiescent air 3 mm thick, determine the heat lost per minute from the man’s hand of surface area 200 cm2 0n a winter’s day when atmospheric temperature is 3 0C. The skin temperature is taken 34 0C and thermal conductivity of air is 24 x 10-3 W/mK. [ 354 J/min]
268. Density of a gas is 1.775 kg/m3 at 27 0C and 1 atm pressure. If the specific heat capacity at constant pressure is 846 J/kgK, find the ratio of the specific heat capacity at constant pressure to that at constant volume. [ 1.29]
269. In a gas at standard condition, what is the length of the side of a cube that contains a number of molecules equal to the population of the earth, 6 x 109 ?
OPTICS

Reflection of Light
270. A ray of light falls normally on a plane mirror. What are (i) angle of incidence, (ii) angle of reflection, (iii) glancing angle, (iv) angle of deviation? [0, 0, 900, 1800] Hints: (i) Angle of incidence is the angle made by incident ray with normal. (ii) Angle of incidence and angle of reflection are equal. (iii) Glancing angle is the angle made by incident ray with the plane of surface. (iv) Angle of deviation is the angle between incident and angle of reflected ray.
271. A man is walking towards a plane mirror with velocity 5 m/s (a) normally, (b) making angle 300 to the normal. What is the relative velocity between the man and his image? [ 10 m/s, 8.7 m/s] Hints: (a) The image also moves with same velocity as object moves but the motion of image is in opposite direction. So, relative velocity between them is 2v. (b) In second case, component of velocity along the normal is v Cos. So, relative velocity is 2v Cos.
272. An object is placed 4 cm from a concave mirror of radius of curvature 12 cm. Calculate the image position and respective magnification. [ 12 cm, 3 ] Hints: Focal length is half of the radius of the curvature. Use mirror formula to obtain image position.
273. An object is placed 16 cm from a convex mirror of focal length 10 cm. Calculate the image distance and magnification. [ 6 cm, 2/5 ] Hints: Take focal length of convex mirror negative and apply mirror formula.
274. A 10 cm long stick is placed along an axis of a convex length of focal length 15 cm. If the nearer end of the stick is 5 cm away from the mirror, calculate the length of the image.
[ 3.75 cm]
Hints: Take two point object distances u1 and u2 5 cm and 15 cm and determine respective image position v1 and v2. The difference in image position v gives the length of image.
275. A mirror forms an erect image 30 cm from the object and twice its height. Where the object must be situated? What is the radius of curvature? Assuming the object is real, determine whether the mirror is concave or convex? [ 10 cm, 40 cm ] Hints: Draw ray diagram of the formation of magnified and virtual image and use mirror formula. The mirror must be concave because convex mirror always forms diminished image. Object distance = u, Image distance = v and given that u + v = 30 cm Magnification m = v / u So, v = 2u Use mirror formula to calculate focal length. Since image is virtual, image distance is negative.
276. A coin 2.54 cm in diameter held 254 cm from the eye just covers the full moon. What is the diameter of image of the moon formed by the concave mirror of radius of curvature 1.27 m? [ 6.35 x 10-3 m ]
277. A driving mirror consists of a cylindrical mirror of radius 10 cm and length over the curved surface is 10 cm. If the eye of the driver be assumed at a great distance from a mirror, find the angle of view. [ 1140 ]
278. A small convex mirror is placed 60 cm from the pole and on the axis of a large concave mirror, radius of curvature 200 cm. The position of the convex mirror is such that a real image of distance object is formed in the plane of the hole drilled through the concave mirror at its pole. Calculate (i) R of the convex mirror, (ii) the height of the real image if the distance object subtends an angle of 0.5 0 at the pole of the concave mirror. [ 240 cm, 1.3 cm ]
279. A concave mirror of focal length 40 cm and a plane mirror are placed at a distance of 60 cm. A point object is placed midway between the two mirrors. Find the final position of image if the ray first meets the plane mirror. [ 72 cm ] Hints : Object distance (u) = 60 + 30 = 90 cm and focal length (f) = 40 cm. Obtain image distance from the mirror formula.
280. An object is situated 40 cm from a convex mirror. When a plane mirror is inserted between object and convex mirror at a distance 32 cm from the object, the image in the two images coincide. What is the focal length of the convex mirror?
[ – 60 cm ]
Hints: Object distance (u) = 40 cm and image distance (v) = – (32-8) = – 24 cm. Then apply mirror formula.

281. A small object placed in front of a convex mirror gives a real image four times the size of the object. When the object is moved 10 cm towards the mirror, a virtual image of the same magnification is formed. Find the focal length of the mirror. [20 cm] Hints: From first condition, u = x and m = 4. So from relation m = v/u, v = 4x . Use mirror formula and obtain ……….(i) From second condition, u = x – 10 , m = 5. So, v = – 4 (x – 10 ). Use mirror formula and obtain ……………(ii). Equate both equations to obtain focal length.
282. It is desired to cast the image of a candle magnified 4 times upon a wall 40 cm distance from the lamp. What type of spherical mirror is required and what is its position? [ 10.67 cm ]
Hints: Concave mirror is required to form magnified image. Let, u = x then v = x + 40. Obtain x from relation m = v/u and use mirror formula to obtain the object and image distance.

Refraction of Light

283. A ray of light is incident at 600 in air on an air – glass plane surface. Find the angle of refraction in the glass (  for glass= 1.5 ) Hints: Use Snell’s law to obtain angle of refraction.
284. A ray of light is incident in water at an angle of 300 on water air plane surface. Find the angle of refraction in the air (  for water = 4 / 3 ) Hints: Use Snell’s law , where is the refractive index of air with respect of water which is the reciprocal of refractive index of water with respect to air, .
285. A ray of light is incident in glass at an angle of (i) 300 , (i) 700 on a glass – water plane surface. Determine the angle of refraction in the water in each case ( g = 1.5, w = 1.33) Hints: Determine refractive index of glass with respect of water and apply snell’s law. In second case determine critical angle for water-glass interface and check whether there is total internal reflection or not. If TIR takes place, the ray reflects on the same denser medium.
286. What is the apparent position of an object below a rectangular block of glass 6cm thick if a layer of water 4 cm thick is on the top of the glass? (Refractive index of glass and water are 3/2 and 4/3 respectively)? Hints: Determine separately the apparent position of the object due to glass and water by using relation and finally add the apparent depths.
287. A tank whose bottom is a mirror is filled with water to a depth of 20 cm. A small object hangs motionless 8 cm under the surface of water. What is the apparent depth of the image when viewed at normal incidence? ( refractive index of water = 4/3) [ 24 cm ] Hints: Total real depth of the image = 20 + 12 = 32 cm. Using relation , apparent depth of the image can be determined.
288. Calculate the critical angel for (i) an air – glass surface, (ii) an air water surface, (iii) a water – glass surface. Draw diagrams in each case illustrating the total internal reflection of a ray incident on the surface (g = 1.5, w = 1.33) Hints: For critical angle, the angle of refraction in rarer medium will be 900. So, the snell’s law will be .So,
289. A microscope is focused on a scratch on the bottom of a beaker. Turpentine is poured in to the beaker to a depth of 4 cm, and it is found necessary to raise the microscope through a vertical distance of 1.28 cm to bring the scratch again into focus. Find the refractive index of the turpentine. Hints: Real depth = 4 cm. Apparent depth = 4 – 1.18 = 2.82 Now , refractive index
290. A concave mirror of small aperture and focal length 8 cm lies on a bench and a pin is moved vertically above it. At what point will image and object coincide if the mirror is filled with water of refractive index 4/3? [ 12 cm ]
291. Sky subtends angle 1800 to a man’s eye. What is the angle subtending by the sky with the fish’s eye? [ 840 ] Hints: Light coming from horizon HH’ refract into the water making angle of refraction equal to the critical angle. So, angle subtend by sky is twice of the critical angle.
292. Light from a luminous point on the lower face of a rectangular glass slab, 2.0 cm thick, strikes the upper face and the totally reflected rays outline a circle of 3.2 cm radius on the lower face. What is the refractive index of the glass?

293. ABCD is the plane of a glass cube. A horizontal beam of light enters the face AB at grazing incidence. Show that the angel with any ray emerging from BC would make with the normal to BC is given by Sin  = Cot c , where c is the critical angel. What the greatest value that the refractive index of the glass may have if any of the light is to emerge from BC?
Hints: In first face AB, applying Snell’s law, ………….(i) In second face ,angle of incidence = 900 – c and angle of refraction =  . Applying Snell’s law , .
Or, ………………(ii)
Equating equation (i) and (ii), the required expression can be obtained.

Prism
294. A ray of light is refracted through a prism of angle 700. If the angle of refraction in the glass at the first face is 280, what is the angle of incidence in the glass at the second face?
[ 420 ] Hints: The relation between angle of prism and angles of refraction is A = r1 + r2
295. (i) The angle of a glass prism is 600, and the minimum deviation of light through the prism is 390. Calculate (i) the refractive index of the glass. (ii) The refractive index of a glass prism is 1.66, and the angle of the prism is 600. Find the minimum deviation. [ 1.52, 52.20]
Hints: The relation between refractive index and minimum deviation is
296. A narrow beam of light is incident normally on one face of an equilateral prism (refractive index 1.45) and finally emerges from the prism. The prism is new surrounded by water (refractive index 1.33). What is the angle between the directions of the emergent beam in the two cases?
[ 49.2] Hints: In first case, angle of incidence on the second face will be 600 ( can be obtained geometrically) and the light undergoes into total internal reflection. When the prism is surrounded by water, the Snell’s law in the second face of prism will be , where is the refractive index of water with respect to glass. It is obtained by relation It gives angle of emergence on water r = 70.70 From geometry, angle between emergent rays in two conditions  E1OE2 = 300 + (900 – 70.760) = 49.20
297. A certain prism is found to produce a minimum deviation of 510. While is produces a deviation of 620 48’ for two values of the angle of incidence, namely 400 6’ and 820 42’ respectively. Determine the refracting angle of the prism, the angle of incidence at minimum deviation and the refractive index of the material of the prism.
[600, 300, 1.65] Hints: The prism formula is A +  = i1 + i2 , here, angle of emergence e is replaced by angle of incidence. It is due to the principle of reversibility of light. Also, in the case of minimum deviation r1 = r2. So, r = A/2.
Refractive index of the material of prism
298. PQR represents a right – angle in a prism of glass of refractive index 1.50. A ray of light enters into the prism through the hypotenuse QR at an angle of incidence I, and is reflected a the critical angle from PQ to PR calculate and draw a diagram showing the path of the ray through the prism . (Only rays in the plane of PQR need be considered.)

299. A is the vertex of a triangular glass prism, the angle at a being 300. A ray of light OP is incident at P on one of the face enclosing the angle A, in a direction such that the angle OPA = 400. Show that, if the refractive index of the glass is 1.50 the ray cannot emerge from the second face. [ 65.20]
Hints: For face AC, angle of incidence (I1 ) = 600 . Snell’s law gives the angle of refraction (r1) = 35.20. From geometry, angle of incidence in AB face (i2) = 65.20 The angle is more than the critical angle. So the light cannot escape from the second face.
300. The refracting angle of a prism is 62.00 and the refractive index of the glass for yellow light is 1.65. What is the smallest possible angle of incidence of a ray of this yellow light which is transmitted without total internal reflection? Explain what happens if white light is used instead and the angle of incidence is varied? [ 39.540 ] Hints: In second face, since grazing emergence is taking place, the angle of incidence on the face should be ‘c’. The relation between angle of prism and angles of incidence is A = r1 + r2 , which gives r1 = 22.70. Applying Snell’s law in first face, angle of incidence can be determined.

301. Determine minimum deviation of an equatorial glass prism of refractive index 1.5 when it is immersed in a liquid of refractive index 1.33.
Hints: The refractive index , where is the refractive index of glass with respect to water. It is calculated by relation = x

Lens
302. An object is placed (i) 12 cm (ii) 4 cm from a converging lens of focal length 6 cm. Calculate image position and magnification in each case. [ 12cm, 1 ; 12cm, 3 ] Hints: (a) Use lens formula (b) In second case, take focal length of concave lens negative.
303. The image of a real object in a diverging lens of focal length 10 cm is formed 4 cm from the lens. Find the object distance and the magnification. [ 6.67 cm, 0.6 ]
Hints: Use lens formula , take focal length of concave lens negative.

304. The image obtained with the converging lens is erected and three times magnified. The focal length of the lens is 20cm. Calculate object distance and image distance. [ 13.33, 40 cm ]
Hints: Magnification m = . Since m = 3, v = 3 u. Image is erected, so it must be virtual. Therefore image distance is negative. Now the lens formula will be , . Substituting value of f, object distance and hence image distance can be determined.
305. A beam of light converges to a point 9 cm behind (i) a converging lens of focal length 12 cm (ii) diverging lens of focal length 15 cm. Find the image position in each case. [ 5.1 cm, 22.5 cm]
Hints: In the absence of lens, the rays converge 9 cm behind it and image forms there. But when lens is introduced, the point where image forms, behaves as virtual object and image is formed at new position. So, take virtual object distance, u = – 9 cm.
306. If the moon subtends the angle 9.1 x 10-3 radian at the center of the lens of focal length 20 cm. Calculate the diameter of this image. With the screen removed, a second converging lens of focal length 5 cm is placed coaxial with the first and 24 cm from it opposite to the moon. Find position, nature and size of final image. [ 1.82mm, 20cm, 9.1 mm ] Hints: Since moon is very far, image forms at principal focus. If ‘d’ be the diameter of the image, then So, d = 9.1 x 10-3 x 20 = 1.82mm For Concave lens, Object distance (u) = 4 cm Focal length (f) = 5 cm Image distance (v) = ? Lens formula gives the virtual image distance 20 cm. Magnification m = v/u = 5 So, size of final image = 1.82mm x 5 = 9.1 mm
307. A converging lens of focal length 20 cm is arranged coaxially with the diverging lens of focal length 8 cm. A point object lies on the side of first lens and very far from it. What is the smallest possible distance between the lenses if the combination is to form a real image of the object? [ 12 cm ]
Hints: For converging lens, Object distance (u) =  Focal length ( f ) = 20 cm Lens formula gives the image distance (v) = 20 cm. Let distance between two lenses be x. For diverging lens, Object distance (u) = – (20 – x ), negative sign for virtual image. Image distance (v) =  , final rays are parallel to each other. Focal length (f) = – 12 cm
Lens formula gives the value of x
308. A converging lens of 6 cm focal length is mounted at 10 cm from screen placed right angles to the axis. A diverging lens of 12 cm focal length is then placed coaxially between the lens and the screen so that an image of an object 24 cm from the converging lens is focused on the screen. What is the distance between the two lenses? [ 4 cm ] Hints: For converging lens, Object distance (u) = 24 cm Focal length ( f ) = 6 cm Lens formula gives the image distance (v) = 8 cm. Let distance between two lenses be x. For diverging lens, Object distance (u) = – (8 – x ), negative sign for virtual image. Image distance (v) = 10 – x Focal length (f) = – 12 cm
Lens formula gives the value of x
309. Light from an object passes through a thin converging lens of focal length 20 cm, placed 24 cm from the object and then thin diverging lens of focal length 50 cm forming a real image 62.5 cm from the diverging lens. Find (a) position of the image due to first lens, (b) distance between the lenses, (c) magnification of the final image. [ 120 cm, 92.2 cm, 3.4 ] Hints: For converging lens, Object distance (u) = 24 cm Focal length ( f ) = 20 cm Lens formula gives the image distance (v) = 120 cm. Let distance between two lenses be x. For diverging lens, Object distance (u) = – (120 – x ), negative sign for virtual image. Image distance (v) = 62.5 cm Focal length (f) = – 50 cm
Lens formula gives the value of x Magnification of the final image m = m1 x m2, where m1 and m2 are magnification of converging and diverging lenses respectively.
310. An object is paced right angles to the axis of a converging lens of focal length 15 cm, and 22.5 cm from it. On the other side of the lens, a diverging lens of focal length 30 cm is placed coaxially. Find the position of the final image when (a) lenses are 15 cm apart and plane mirror is placed perpendicular to the axis 40 cm beyond the diverging lens, (b) when the mirror is removed and the lenses are 35 cm apart. [beside O,72.5 cm from O] Hints: For converging lens, Object distance (u) = 22.5 cm Focal length ( f ) = 15 cm Lens formula gives the image distance (v) = 45 cm. Given, distance between two lenses = 15 cm For diverging lens, Object distance (u) = – (45 – 15 ) = – 30 cm, negative sign for virtual image. Focal length (f) = – 30 cm Image distance (v) = ?
Lens formula gives the value of image distance = . It means final rays are parallel. When they fall on a plane mirror perpendicularly, they return along the same path. It is due to the reversibility of light. The returning light follows the same path and final image coincides with the object.
311. A thin equiconvex lens of refractive index 1.5 whose surface have radius of curvature 24 cm is placed on a horizontal plane mirror. When the space between the lenses and the mirror is filled with a liquid, a pin held 40 cm vertically above the lens is found to coincide with its own image. Calculate the refractive index of the liquid. [ 1.4 ] Hints: From Lens maker formula the focal length of convex lens = 1/24 cm Similarly, focal length of liquid plano-concave lens = , since 1/R2 = 0 Now, the combined focal length F = 40 cm, since object and image coincide at the same point. So,
312. A converging equiconvex lens of glass of refractive index 1.5 is laid on a horizontal plane mirror. A pin coincide with it’s inverted image when it is 1.0m above the lens. When some liquid is placed between the lens and the mirror the, pin has to be raised by 0.55m for the coincidence to occur again. What is the refractive index of the liquid? [ 1.35 ] Hints: Focal length of equiconvex lens = 1 m = 100 cm Focal length of combined lens = 0.55 m = 55 cm
313. The curved face of a planoconvex lens (µ=1.5) is placed in contact with a plane mirror. An object at 20 cm distance coincides with the image produce by the reflection by the mirror. If the gap between the lens and the mirror is filled by liquid, the coincident object and image will be at distance 100 cm. What is the index of the liquid? [ 1.4 ] Hints: From Lens maker formula, the focal length of glass plano-convex lens …………………………..(i)
Similarly, focal length of liquid plano-concave lens = ………………………… (ii) Now, dividing (i) by (ii) .
, which gives the value of l

314. An achromatic lens having a focal length 20 cm is to be constructed by combining a crown lens and a flint lens. What must be the focal length of the component lenses if the dispersive power of the crown and flint glasses are 0.0158 and 0.0324 respectively? [ 10.2 cm, -21 cm] Hints: Apply the condition of achromatism and obtain f1 = – 0.488 f2. Also, . Substituting value of f1 and F = 20, f1 and f2 can be determined.
Defect of vision

315. A person can focus objects only when they lie between 50 cm and 300 cm from his eye. What spectacles should he use (i) to see up to infinity, (ii) to reduce his least distance of distinct vision to 25 cm. Find the range of vision using each pair. Hints: (i) To extend far point from 300 cm to infinity, concave lens should be used. For concave lens, object distance (u) =  image distance (v) = – 300 cm , virtual. Lens formula gives the focal length of the lens, equal to – 300 cm. (i) To reduce near point from 50 cm to 25 cm, convex mirror should be used. For concave lens, object distance (u) = 25 image distance (v) = – 50 cm , virtual. Lens formula gives the focal length of the lens, equal to 50 cm.
316. A certain person can see clearly objects lying between 20 cm and 200 cm from his eye. What spectacle is required to enable him to see distance object clearly? Also calculate the least distance of distinct vision after wearing them?
[-0.5D, 22.2cm]
Hints: (i) To extend far point from 200 cm to infinity, concave lens should be used. For concave lens, object distance (u) =  image distance (v) = – 200 cm , virtual. Lens formula gives the focal length of the lens, equal to – 300 cm and the power of the lens P = 1/f = – 0.5 D

317. In order to correct his near point to 25 cm , a man given spectacles with converging lens of 50 cm focal length an to correct his far point to infinity he is given diverging lens of focal length 200 cm. Determine his near and far point not wearing the glass. [50 cm, -200 cm]
318. A person can not see objects in nearer than 500 cm from the eye. Determine the focal length and the power of the glass which enable him to read a book 25 cm from his eye. [ 3.8 D ]
319. An elderly person can not see clearly without the use of spectacles, objects nearer than 200 cm. What spectacles will he need to reduce this distance to 25 cm? If his eyes can focus rays which are converging to points not less than 150 cm behind them, calculate his range of distinct vision when using spectacles. [ 28.6 cm, 25-35.3cm ]
Optical instruments

320. A simple astronomical telescope in normal adjustment has an objective of focal length 100 cm and eyepiece of focal length 5 cm. (a) where is the final image formed ? (b) calculate the angular magnification. (c) how would you increase the resolving power of the telescope? [ infinity, 20] Hints: (i) In normal adjustment, final image forms at infinity. (ii) Angular magnification in normal adjustment m =
321. Calculate the distance of the eye ring from the eye piece of a simple astronomical telescope in normal adjustment whose objective and the eyepiece have focal lengths of 80 cm and 10 cm respectively. [ 11.25 cm ] Hints: Eye ring is that point where eye lens forms the image of objective lens. So, for eye lens, object distance (u) = 80 + 10 = 90 cm image distance (v) = ? focal length (f) = 10 cm
322. A refracting telescope has its objective of focal length 1.0 m and eye piece of focal length 2 cm. A real image of sun, 10 cm in diameter is formed in the screen 24 cm from the eye piece. What angle does the sun subtend the objective? [0.0091 rad] Hints: For eyepiece Object distance (u) = ? Image distance (v) = 24 cm Focal length (f) = 2 cm Lens formula gives u = 2.18 cm Now So, object size O=0.91cm Angle made by objective with lens Tan   = 0.91/100 = 0.0091 radian.
323. A telescope is made of an object glass of focal length 20 cm and eye piece of 5 cm, both conversing lenses. Find the magnifying power in the following cases (a) when the eye is focused to receive parallel rays. (b) when the eye sees the image situated at the nearest distance of distinct vision which may be taken 25 cm. [4, 4.8 ] Hints: (a) magnification of telescope when image is formed at  is m = (b) magnification when image is formed at near point is m =
324. A telescope consist two thin converging lenses of focal lengths 0.3 m and 0.03 m separated by 0.33 m. It is focused on the moon which subtends an angle 0.5 0 at the objective. Find the angle subtends to the observer’s eye by the image of the moon formed by the instrument. [ 50 ] Hints: Magnification of telescope when image is formed at  is m = So,
325. A telescope objective has focal length 96 cm and diameter 12 cm. Calculate the focal length and minimum diameter of a eyepiece lens for use with the telescope, if the magnifying power required is x24, and all the light transmitted by the objective from a distance point on the telescope axis is to fall on the eyepiece.
326. An astronomical telescope has objective of focal length 60 cm and an eyepiece of focal length 3 cm is focused on the moon so that the final image is formed at the minimum distance of distinct image from the eyepiece. Assuming that the diameter of the moon subtends an angle of 0.5 0 at the objective, calculate the (a) angular magnification, (b) actual size of the moon on seen, (c) How, with the same lenses, could an image of the moon 10 cm in diameter be formed on a photographic plate? [ 22.4, 4.95 cm] Hints: (a) Magnification when image is formed at near point is m = = 22.4 (b) Magnification m = . So,  = m x  = 11.20 If ‘d’ be the actual size of the image seen, then Tan 11.2 = h / D

327. A telescope constructed from two converging lenses, one of focal length 250 cm and other of 2 cm, is used to observe a planet which subtends an angle of 5 x 10-5 radian. Explain how these two lenses would be placed for normal adjustment and calculate the angle subtended at the eye of the observer by the image. Hints: Magnification of telescope when image is formed at  is m = So,
In normal adjustment, final image forms at infinity. So, the distance between two lenses will be f0 + fe
328. A converging lens of focal length 5 cm is used as a magnifying glass. If the near point of the observer is 25 cm from the eye and the lens is held close to the eye, calculate (a) distance of the object from the lens, (b) angular magnification.
329. State and explain how you would arrange simple converging lenses, of focal lengths 2 cm and 5 cm, to act as a compound microscope with magnifying power x 42, the final image being 25 cm from the eye lens.
330. A point object is placed on the axis of, and 3.6 cm from, a thin converging lens of focal length 3 cm. A second thin converging lens of focal length 16.0 cm is placed coaxial with the first and 26.0 cm from it in the side of objective. Find the final image produced by the two lenses.
331. If the final image formed coincides with the object, and is at the least distance of distinct vision when the object is 4 cm from the objective, calculate the focal length of the objective and the eye lenses, assuming magnifying power of the microscope 14.
332. A compound microscope has lenses of focal length 1 cm and 3 cm. An objective is placed 1.2 cm from the object lens. If virtual image is formed 25 cm from the eye, calculate the separation of the lenses.

Photometry
333. A lamp whose luminous intensity is 100 candela is placed one meter above the horizontal round table of 1 m radius. Calculate the illuminance at a point directly under the table and also the edge of the table. [ 100, 35.36 lm/m2]
334. Two lamps 25 and 50 candela are 100 cm apart. At which point between them will a screen be equally illuminated on both side ? [ 41.4 cm]
335. Two identical lamps are placed at 60 cm apart. Where should a screen be placed between them so that the intensity on one side of the screen is four times that of the other side? [ 20 cm]
336.
Revision
HSEB Exam Questions
337. How long will the light take in traveling a distance of 500 m in water? Given refractive index of water is 1.33. [ 2.2 x 10-6 sec]
338. A glass prism of angle 720 and index of refraction 1.66 is immersed in a liquid
339. Calculate the focal length of a concave mirror when an object placed at a distance of 40 cm makes image equal to the size of the object. [ 20 cm ]
340. If two identical lens are 1 m apart, where should be the screen be placed between them so that the intensity on one side of the serene be four times the intensity of the other side. [ 0.29 m ]
341. An object 10 cm high is placed in front of a convex mirror of focal length 20 cm and the object is 30 cm from the mirror. Find the height of the image. [ 0.04 m]
342. What is the apparent position of an object below a rectangular block of glass 6 cm thick if a layer of water 4 cm thick is on the top of the glass ? Refractive index of glass 3/2 and that of water is 4/3. [ 0.07 m from top]
343. A microscope is focused on a scratch on the bottom of a beaker. Turpentine is poured into the beaker to a depth of 4 cm and it is found necessary to raise the microscope through a vertical distance of 1.25 cm to bring the scratch again into focus. Find the refractive index of the turpentine. [ 1.45 ]
344. An erect image, three times the size of the object is obtained with a concave mirror of radius of curvature 36 cm. What is the position of the object? [ 12 cm]
345. The radii of curvature of the faces of a thin converging meniscus lens of glass of refractive index 1.5 are 15 cm and 30 cm. What is the focal length of the lens when it is completely immersed in water of refractive index 4/3 ? [ 2.35 m] Hints: Apply Lens Makers formula where µ be the refractive index of glass with respect to water.
346. A glass prism of angle 720 and index of refraction 1.66 is immersed in a liquid of refractive index 1.33. Find the angle of minimum deviation for a parallel beam of light passing through the prism. [ 22.40 ]
347. Light from a luminous point on the lower face of a rectangular glass slab, 2 cm thick, strikes the upper face and the totally refractive rays outline a circle of 3.2 cm radius on the lower face. What is the refractive index of the glass? [ 1.6 ]
348. A narrow beam of light incident normally on one face of an equilateral prism of refractive index 1.45, being surrounded by water of refractive index 1.33. At what angle the ray of light emerges out? [ 710 ]

Electrostatics

349. A body is charged to 6 C. Calculate the number of electrons that are taken out from the body.
[ 3.75 x 1013] Hints: Use the mathematical expression of quantization of charge is q = ne, where ‘n’ be the number of basic charge and ‘e’ is the charge of an electron equal to 1.6 x 10-19 C.
350. A body of mass 5mg is charged to 1200C. Does its mass change after charging? If so, calculate the magnitude of the mass after charging.
[ 4.99 mg] Hints: The decrease in mass due to charging is m = n × me , where n is number of electrons that are taken out from the body and me is the mass of an electron. From quantization of charge, n = q/e. So, the decrease in mass be m = q/e x me
351. Calculate the total flux generated from a charge of magnitude 88.5 nC. If the charge is surrounded by a sphere, calculate the flux passing through a section of the sphere which subtends an angle 2/5 steradian to the charge.
[10000Wb,1000Wb]
Hints: Total flux generated from a charged body is given by Gauss’s law i.e.  = . The emitted flux scattered around 4 steradian angle. Applying unitary method flux through the given section can be determined.
352. If 250 weber flux passes through a cross-section of a cylinder having radius 20 cm, calculate the electric flux density at that point
[ 1990 wb/m2]
Hints: Flux density is the flux per unit area. So, E = /A
353. Two charges 6C and 9C are placed 10 cm apart in air. Calculate the electrostatic repulsive force between them. What will be the new force when they placed in water in same distance away? ( of water is 80 )
[48.6N, 0.6N]
Hints: Electric force between two charges, according to Coulomb’s law is F = When charges are placed in dielectric, the electric force decreases by dielectric constant times. So, the new force F’ = F/
354. Two electrons are placed 1m away in air. Compare the electric and gravitational force between them.
[ – 4.17 x 1042 :1 ]
Hints: Electric force between two electrons Fe = ……..(i), where e = 1.6 x 10-19 C Gravitational force between two electrons Fg = ………..(ii), where m = 9.1×10-31Kg Divide (i) by (ii) to obtain the result. Negative sign in ratio is required to show that electric force in this case is repulsive while gravitational force is attractive i.e. they are opposite in nature.
355. Two point charges 6 x 10-8 C and -9 x 10-8 C are 50 cm apart in vacuum. At what point on the line joining them are the (a) electric field intensity and (b) potential zero?
[222 cm, 20cm from first]

Hints: Since two charges are opposite in nature, potential will be zero at any point between them (at point N) and intensity will be zero outside them towards the smaller charge ( at point M). (a) Let at point M, ‘x’ distance from q1, resultant field intensity is zero. Intensity at M due to charge q1, E1 = =
Intensity at M due to charge q2, E2 = = According to question,
(b) Let at point N, ‘x’ distance from q1, resultant potential is zero. Potential at N due to charge q1, V1 = =
Potential at N due to charge q2,V2 = = According to question,

356. Two charged metal balls, each of mass 0.05 g are suspended from the same point by means of two silk threads each 5 cm long. Calculate the charge on each if they repel each other to a distance of 6 cm. [ 1.22×10-8 C] Hints: In equilibrium condition, T Cos = mg …………..(i) and T Sin = F ………….(ii). Dividing, the electrostatic force F = mg Tan. Or, = mg Tan, where Tan = p/b = 3/4 .
357. Two charges of magnitude 9C and -9C are placed two vertices of an equilateral triangle of side 1m. What is the electric field intensity at third vertex? If 6C charge is placed at the third vertex, calculate the force experienced by the charge. [ 81000N/C, 0.48N] Hints: Determine electric field intensity due to first and second charge separately. The angle between them is 1200. So, apply parallelogram law of vector addition to get resultant intensity. Intensity at C due to charge A, E1 = = = 81000 N/C Intensity at C due to charge B, E2 = = = 81000 N/C The angle between E1 and E2 is 1200. So, resultant intensity E = If 6µC charge is placed in third vertex, force experienced by the charge F = Eq.
358. Find the potential and field due to a charged hollow sphere of radius 20 cm at the following distances (a) 10 cm from the centre, (b) 25 cm from the sphere if charge on the sphere is 6C.
[2.7 x 105V, 2.16 x 105V]
Hints: (a) the region inside the hollow sphere is the equi-potential region and the potential is equal to that of the surface. So, take distance equal to the radius of the sphere. Since there is no charge inside the hollow sphere, due to skin effect, intensity is zero. (b) Outside the sphere, potential V = and field intensity E = , where r = 25 cm = 0.25m.
359. A charged oil drop remains stationary when situated between two parallel metal plates 25 mm apart and a pd of 1000V is applied to the plates. Find the charge on the oil drop if it has a mass of 5 x 10-15 kg . Also calculate the number of basic charge present.
[ 1.25×10-18C, 8 ]
Hints: Downward gravitational force = mg. Upward electrostatic force = Eq. Since the body is stationary, downward gravitational force ‘mg’ must be equal to the upward electric force ‘Eq’ So, Eq = mg Or, x q = mg Or, q = The number of basic charge, according to quantization of charge is, n = q/e.
360. A charged oil drop of radius 1.3 x 10-6 m is prevent from falling under gravity by the vertical field between two parallel plates charged to a pd of 8340 V. If the plate separation is 16 mm and density of oil is 920 kg/m3, calculate the magnitude of charge present. [ e ] Hints: Determine mass of the oil drop by mass = density x volume and follow the above process. i.e. m =  x v =  x 4/3  r3
361. An electron starts from a point of a conductor and reaches a second conductor with a velocity 107 m/s , calculate the pd in volts. [281V] Hints: Use principle of conservation of energy i.e. KE ½ mv2 = PE Ve So, pd V =
362. A high voltage terminal of a generator consist of a spherical conducting shell of radius 0.5 m. Estimate the maximum potential to which it can be raised in air for which electric breakdown occurs when the electric intensity exceeds 3 x 106 V/m? [1.5 x 106 V] Hints: Maximum electric field intensity ( dielectric strength of air) E = …….(i) Maximum PD due to the charge V = ……………..(ii) Dividing (i) by (ii), we get V = E r
363. Two positive point charges 10 and 14 micro coulomb are 20 cm apart. Calculate the work done if the distance between them was reduced to half the original distance? [ 6.3 J ] Hints: Make any one of the charges stationary and bring another charge to half of the distance. Use formula, Work done W = V q Where V is potential difference between two points due to stationary charge and q is the moving charge.
364. What is the potential gradient between two parallel plane conductors when their separation is 20 mm and pd 400 V is applied to them? Calculate the force on an oil drop between the plates if the drop carries charge 8 x 10-19 C?
[ 1.2 x 104 V/m, 1.6 x 10-14 N]
Hints: Potential gradient E = V / d, where V is the potential difference and d is the plate separation. Force experienced by charge particle in an electric field is F = E q

Capacitor
365. A capacitor has a charge of 50 C when a p.d. of 50 V is applied to it. Calculate the capacitance of the capacitor.
[ 1 F ] Hints: Charge stored by capacitor is given by q = CV. So, C = q / V
366. Three capacitors of capacitance 2F, 4F and 6F are connected in (a) series and (b) parallel. Calculate resultant capacitance. If the combination is connected to a source of 12 V, calculate total energy stored in each case.
[ 12/11 and 12F,0.00086J] Hints: At first, determine equivalent capacitance using relation for series and C = C1 +C2 + C3 for parallel combination. The energy stored by the capacitor is given by U = ½ CV2, where C be the equivalent capacitance in each case.
367. Three capacitors of capacitance 3F, 10F and 15F are connected in series with a 15 V battery. What is (a) the charge in 3 F capacitor? (b) pd across 10F capacitor? (c) energy stored in 15 F capacitor?
[3V, 30C, 3×10-5 J] Hints: (a) At first, determine equivalent capacitance by relation and apply q = CV to determine charge stored by the capacitor. In series combination, each capacitor stores equal amount of charge. (b) The relation q = C2 V2 gives the pd across the second capacitor and (c) U3 = q2/2C3 gives the energy stored by the third capacitor.
368. 8 drops of water are equally charged. They combine together to form a bigger drop. Compare the capacity and the potential of the bigger drop with that of smaller drop.
[ 2:1, 4:1 ]

Hints: Let, v be the volume of a smaller drop. Then the volume of the larger drop V = 8 x v
Or,
Or, R = 2r, i.e. radius of the larger drop is twice of the smaller drop.

Capacitance of smaller spherical drop C1 = 40r, where ‘r’ is radius of smaller drop. Capacitance of larger spherical drop C2 = 40 R, where ‘R’ is radius of larger drop. Dividing, Potential of a smaller drop ……….(i)
Potential of larger ……….(ii)
Divide to obtain result.
369. 2F and 5F capacitors are charged to 500V and 100 V battery respectively. The capacitors are then joined positive to positive and negative to negative plate. Calculate common potential and loss of energy.
[ 0.114 J] Hints: Common pd across the combination is and the loss of energy due to combination is
370. Two capacitors of 4F and 12F are connected in series and the combination is joined across a 200V battery. They are now isolated and connected in parallel. Calculate the common potential and energy loss. [ 75 V, 0.015 J ] Hints: At first determine equivalent capacitance using relation . In series combination, each capacitor stores equal magnitude of charge. So, charge stored by each capacitor q = CV. Now, Common pd across the combination is . Initial energy stored by capacitor U1 = , where C is the equivalent capacitance. Final energy stored by capacitor U2 = , where Vc is common PD. Loss of energy is the difference between initial and final energy, i.e. U =U1 – U2 Alternative method: Energy loss U = , where V1 and V2 are pd across C1 and C2. Since V1 and V2 are not known, they are replaced by charge as U =
371. A 21V battery is connected across 3F and 4F capacitors in series. Calculate the pd across each capacitor and their total energy. The capacitors are disconnected and then joined plates of like charge together. Calculate the charge on 3F and energy stored in 4F capacitor. [ 30.8C,2.1 x 10-4 J] Hints: Determine equivalent capacitance and the charge stored by each capacitor. Using relation q = CV, determine pd across each capacitor. Determine common pd across the combination, . Relation q = CV gives charge and U = ½ CV2 gives energy stored by capacitor, where V is the common pd.
372. A capacitor of capacitance 60F is connected to 100 V dc supply. (i) Calculate energy stored by the capacitor. (ii) If the capacitor is disconnected to source and connected to a chargeless capacitor of capacitance 30F, calculate the new energy on each capacitor and loss of energy. [ 0.3J, 0.13J, 0.067J, 0.0103J ] Hints: (ii) Use formula to calculate common pd. Since second capacitor is chargeless, pd across it is zero. Q = CV gives the new charge stored by each capacitor.
373. A sheet of paper 40 mm wide and 1.5 x 10-2 mm thick is used as a dielectric of capacitor of capacitance 2 F which has foil of same wide. If dielectric constant of paper is 2.5, calculate the length of paper required. [ 34 m ] Hints: Capacitance of the parallel plate capacitor C = and the A = l × b gives the length of the sheet.
374. Two plane parallel conducting 15 mm apart are held horizontal one above the other. The upper plate is maintained +1500 V while the lower is earthen. Calculate the number of electrons which must be attached to a small oil drop of mass 4.9 x 10-15 Kg, if it remains stationary in the air between the plates. If the potential of the upper plate is suddenly changed to -1500V, what is the initial acceleration of the oil drop? [ 2g ] Hints: Since the oil drop is stationary, gravitational downward force must be equal to the upward electric force. That is mg = Eq. In second case, both forces are along the downward direction, total force is equal to F = ( mg + Eq ) and the total force divided by mass gives acceleration.
375. Calculate the equivalent capacitance of the shown figures, each capacitor has capacitance C.

376. In figure below PD is 100V, capacitors are 30 F and 60 F. If S2 is left open and S1 is closed, calculate the quantity of the charge on each capacitor in the given figure. If S1 is now opened and S2 is closed how many charge will flow through the 10 ohm resistor. [ 0.002C ] Hints: Two capacitors are is series, so determine equivalent capacitance. q = CV gives the charge stored by each capacitor. The equal amount of charge stored by one capacitor flows through the resistor. In series combination, charge or current doesn’t increase which increase in parallel combination.
377. A capacitor of capacitance 9F is charged from a source of emf 200V. It is now disconnected from the source and connected in parallel with 3F uncharged capacitor. The second capacitor is now removed and discharged. What charge remains in the first capacitor? How many times would the process have to be performed to reduce the charge on 9F capacitor below 50% of its initial value? [ 1012 C ] Hints: Determine common pd after the combination of second capacitor. q =CV gives the charge remaining in the first capacitor. Repeat the process.

Revision for ambitious students
378. Two charges 10 C and 20 C are placed at a separation of 2 m. Find the electric potential due to the pair at the middle of the line joining the two charges. [2.7 x105 V ]

379. A charge Q is to be divided on two parts. What should be the value of charge so that the force between them maximum. [ Q / 2 each ]

380. A charged particle of mass 1.0 g is suspended through a silk thread of length 40 cm in a horizontal electric field of 4 x 104 N/C. If the particle stays at a distance 24 cm from the wall in equilibrium, find the charge on the particle. [ 1.8 x 10-7 ]

381. Charges 2 x 10-6 and 1 x 10-6 are placed at A and B points of a square of side 5 cm. How much work will be done against the electric field in a moving of a charge 10-6 from C to D? [ 0.053 J ]

382. Two charged particles are placed at a distance 1.0 cm away from each other . What is minimum possible value of force between them? [ 2.3 x 10-24 N ]

383. A particle having the charge 2 C and the mass 100 g is placed at the bottom of the smooth inclined plane of inclination 300 . Where should the another particle having same charge and mass be placed on the incline so that it may remain in equilibrium? [ 26.8 cm ]

384. Two particles A and B, each having the charge Q are placed d cm apart. Where a particle of charge q should be placed on the perpendicular bisector of AB so that it experience maximum force? Also calculate the force. [ d/22, 3.08F ]

385. A charge of magnitude 1.0 C is placed at the corner of a cube. How much flux passes through the cube? [ 1.4 x 103 wb]

386. An inclined plane is making an angle of 300 with the horizontal electric field E of 100 V / m. A particle of mass 1kg and charge 0.01C allowed to slide down from a height of 1m . If the coefficient of friction is 0.2, find the time it will take the particle to reach the bottom. [2.13m/s2 and 1.3 s]

387. A parallel plate capacitor has plates of area 200 cm2 and separation between the plates 1.00 mm. What PD will be developed if a charge of 1.00 nC is given to the capacitor? If the plate separation is now increased to 2.00 mm, what will be the new PD ? [ 5.6V, 11.3V ]

388. A 50 V battery is connected to a 2 F capacitor. Calculate energy in the capacitor. If the terminals of the battery are removed and mica sheet of dielectric constant 5 is placed between the plates, find the loss of energy. [ 2.5 x 10-3, 2 x 10-3 J]

389. Find the capacitance between A and B. Find the charges on the three capacitors. Taking the potential of B is zero, find the potential of D. [ 6 F, 48 & 96 C, 16 V ]

390. Calculate the equivalent capacitance of the given circuit if each has capacitance ‘C’. [ Ans : 2C ]

391. A pendulum bob of mass 50g is suspended by a thread of length 1m in a region having vertical upward electric field 100 N / C. If the bob has charge 6 micro coulomb, what is the time period of oscillation ? What is the new time period if the field is horizontal ?

392. In adjacent figure, if value of each capacitor is C, calculate the equivalent capacitance.

393. Two capacitor of capacitance 4 and 12 micro coulomb are connected in series and the combination is further connected to a source of PD 200V. The charged capacitors are now isolated and connected in parallel. Calculate the common potential. [HSEB –2002]

Answer of Objective questions

Q.N Ans Q.N Ans Q.N Ans Q.N Ans Q.N Ans
15 b 20 b 25 a
16 a 21 d 26 c
17 d 22 b 27 c
18 c 23 a 28
19 b 24 b 29

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Neb Hseb Notes : Complete Numerical Collections Of Class -11 Hseb with Hints
Complete Numerical Collections Of Class -11 Hseb with Hints
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