MECHANICS

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

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

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

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

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

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

Linear Motion

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

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

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

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

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

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

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

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