NCERT Solutions Class 11 Physics Chapter 4

NCERT Solutions Class 11
Physics
Chapter 4 Motion in A Plane

Q.1 (a) Show that for a projectile the angle between the velocity and the x – axis as a function of time is given by θ (t) = tan −1 ( v 0y −gt v 0x ) 

(b) Show that the projection angle  θ 0  for a projectile launched from the origin is given by θ 0 = tan −1 ( 4 h m R ) Where the symbols have their usual meaning. 

Ans.

(a) Let initial components of velocity of the projectile along horizontal (OX) and vertical (OY) directions be  v 0x  and  v 0y  the respectively. Let the horizontal and vertical components of velocity at point P be  v x  and  v y  respectively. Let time taken by projectile to arrive at point P = t Using the first equation of motion along vertical and horizontal directions, we have: v y = v 0y −gt v x = v 0x ∴tanθ= v y v x = v 0y −gt v 0x ∴θ= tan −1 ( v 0y −gt v 0x ) In angular projection Maximum vertical height,  h m = u 2 sin 2 θ 0 2g →(i) Horizontal range is given as: R= u 2 sin2 θ 0 g →(ii) Dividing equation (i) by equation (ii), we obtain: h m R = sin 2 θ 0 2sin2 θ 0 = sin 2 θ 0 4sin θ 0 cos θ 0 = sin θ 0 4cos θ 0 ∴ h m R = tan θ 0 4   ∴ θ 0 = tan −1 ( 4 h m R )

Q.2 A cyclist is riding with a speed of 27 kmh-1. As he approaches a circular turn on the road of radius 80 m, he applies brakes and reduces his speed at the constant rate of 0.50 ms-1 every second. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn?

Ans.

Given, speed of the cyclist, v = 27  kmh -1 = 7.5  ms -1 Radius of circular turn, r = 80 m ∴Centripetal acceleration,  a c = v 2 r = 7.5 2 80 ms -2 = 0.7  ms -2 Suppose the cyclist applies brakes at point P of the circular turn then the tangential component of  acceleration  a T  will act opposite to velocity. Here, tangential component of acceleration,   a T = 0.5  ms -2 Angle between  a c  and  a T   = 90 o

∴Magnitude of resultant acceleration a is given as: a = a c 2 +a T 2 a = 0 .7ms -2 2 + 0 .5ms -2 2 = 0.86  ms -2 Let the resultant acceleration makes an angle θ with the direction of velocity ∴tanθ = a c a T tanθ = 0 .7 ms -2 0 .5 ms -2 = 1.4 ∴ θ = tan -1 1.4 = 54 .46 o

Q.3 A fighter plane flying horizontally at an altitude of 1.5 km with speed 720 kmh-1 passes directly overhead an anti-aircraft gun. At what angle from the vertical should the gun be fired for the shell with muzzle speed 600 ms-1 to hit the plane? At what minimum altitude should the pilot fly the plane to avoid being hit? (Take g = 10 ms-2)

Ans.

Given, altitude of plane = 1.5 km = 1500 m Speed of plane, v = 720  kmh -1 = 200  ms -1 Let the shell hits the plane when fired at an angle θ. The described situation is shown in the given figure. Here, muzzle velocity of gun, u = 600  ms -1 Let time taken by the shell to hit the fighter plane = t Horizontal distance covered by the  shell = u x t Distance covered by the plane = vt For the shell to hit the plane, these two distances must be equal. u x t = vt ∴usinθ = v ∴sinθ = v u = 200  ms -1 600  ms -1 = 0.33 θ = sin -1 0.33 = 19 .3 o In order to avoid being hit by the shell, the pilot must fly the plane at a minimum height  which is the maximum height attained by the shell. ∴H = u 2 sin 2 90 o -θ 2g = 600  ms -1 2 cos 2 θ 2g H = 600  ms -1 2 cos 2 19 .3 o 2g = 18000× 8 3 2  m H = 16000 m = 16 km

Q.4 A bullet fired at an angle of 30° with the horizontal hits the ground 3.0 km away. By adjusting its angle of projection, can one hope to hit a target 5.0 km away? Assume the muzzle speed to be fixed, and neglect air resistance.

Ans.

Given, angle of projection of the bullet,  θ = 30 o

Range of the bullet, R = 3 km Acceleration due to gravity, g = 9.8  ms -2 Horizontal range for projection velocity  u 0 , R = u 0 2 sin2θ g ∴3 km = u 0 2 sin60 o g = u 0 2 g 3 2∴ u 0 2 g = 2 3

(i) The bullet achieves the maximum range  (R max ) when fired at an angle of  45 o  with the horizontal, R max = u 0 2 g

(ii) From equation (i) and (ii), we obtain R max = 3 3  km = 2×1.732 km = 3.46 km ∴The bullet cannot hit the target 5 km away.

Q.5 Can you associate vectors with (a) the length of a wire bent into a loop, (b) a plane area, (c) a sphere? Explain.

Ans.

(a) No, we cannot associate a vector with the length of a wire bent into a loop.

(b) Yes, we can associate a vector with a plane area. Such a vector is called an area vector and its direction is normal, inward or outward to the plane area.

(c) No, we cannot associate a vector with the volume of a sphere. But, we can associate an area vector with the area of a sphere.

Q. 6 A vector has both magnitude and direction. Does it mean that anything that has magnitude and direction is necessarily a vector? The rotation of a body can be specified by the direction of the axis of rotation, and the angle of rotation about the axis. Does that make any rotation a vector?

Ans.

A physical quantity having both magnitude and direction is not necessarily a vector. There are many physical quantities that have both magnitude and direction, but they are not vectors. The essential condition for a physical quantity to be a vector is that it should follow the laws of vector addition. For example, the rotation of a body about an axis is not a vector quantity, because it does not follow the laws of vector addition. However, the rotation of a body by a definite small angle follows the laws of vector addition and is therefore considered a vector.

Q.7 A vector has magnitude and direction. Does it have a location in space? Can it vary with time? Will two equal vectors a and b at different locations in space necessarily have identical physical effects? Give examples in support of your answer

Ans.

A vector has no fixed location in space because it remains invariant when displaced in such a way that its magnitude and direction remain the same.

A vector can change with time. For example, the displacement vector of an accelerated particle changes with time.

Two equal vectors at different locations in space do not necessarily produce the same physical effect. For example, two equal forces acting at two different points on a body can cause the rotation of the body, but their combination cannot produce the same turning effect.

Q.8 An aircraft is flying at a height of 3400 m above the ground. If the angle subtended at a ground observation point by the aircraft positions 10.0 s apart is 30°, what is the speed of the aircraft?

Ans.

The given situation is described in the figure. Given, height of the aircraft from ground, OS = 3400 m     Angle subtended between the positions of aircraft, ∠POS = 30°  Time taken by the aircraft =10 s In ΔPSO, tan 15° = PS OS ∴PS = OStan15° ∴PS = 3400 m×tan15° = 3400 m×0.2679 = 910.86 m PQ = PS + SQ PQ = 2PS = 2×910.86 m = 1821.72 m ∴Speed of aircraft, v = distance PQ time v = 1821.72 m 10 s = 182 .172 ms -1

Q.9 Read each statement below carefully and state, with reasons and examples, if it is true or false:

A scalar quantity is one that

(a) is conserved in a process

(b) can never take negative values

(c) must be dimensionless

(d) does not vary from one point to another in space

(e) has the same value for observers with different orientations of axes.

Ans.

(a) The given statement is false because, temperature being a scalar quantity is not conserved during inelastic collisions.

(b) The given statement is false because, temperature being a scalar quantity can take negative values.

(c) The given statement is false because, distance is a scalar quantity, yet it has the dimension of length.

(d) The given statement is false because, intensity of light being a scalar quantity change from one point to another in space.

(e) The given statement is true because, the value of a scalar does not change for observers with different orientations of axes.

Q.10 For an arbitrary motion in space, which of the following relations are true : 

(a)  v → average = 1 2 v → t 1 + v → t 2                                     

(b)  v → average = r → t 2 + r → t 1 t 2 – t 1 

(c)  v → t = v → 0 + a → t  

(d)  r → t = r → 0 + v → 0 t + 1 2 a → t 2  

(e)  a → average = v → t 2 + v → t 1 t 2 – t 1 [The average stands for the average of the quantity over the time interval  t 1  and  t 2 ] 

Ans.

Out of the given relations, the relations (b) and (e) are true. The relations (a), (c) and (d) are false as they hold for uniform acceleration only.

Q.11 i ∧  and  j ∧  are unit vectors along x - and y – axis respectively. 

(a) What is the magnitude and direction of the vectors ( i ∧ + j ∧ ) and ( i ∧ − j ∧ ) ? 

(b) What are the components of a vector B → = 2 i ∧ +3 j ∧  along the directions of ( i ∧ + j ∧ ) and ( i ∧ − j ∧ ) ? 

Ans.

(a) Magnitude of  i ∧ + j ∧ = i ∧ + j ∧ = 1 2 + 1 2 = 2 Consider that the vector  i ∧ + j ∧  makes an angle θ with  i ∧ , ∴cosθ= i ∧ + j ∧ . i ∧ i ∧ + j ∧ i ∧ = 1 2 1 = 1 2 = cos45 o ∴ θ = 45 o Therefore, the vector  i ∧ + j ∧  makes an angle of  45 o  with the x-axis.

(b) Given, B → = 2 i ∧ +3 j ∧ Let  a ∧  be the unit vector directed along vector  i ∧ + j ∧ . ∴ a ∧ = i ∧ + j ∧ i ∧ + j ∧ = i ∧ + j ∧ 1 2 + 1 2 = i ∧ + j ∧ 2 Magnitude of the component vector of  B →  along vector  i ∧ + j ∧ = B →  . a ∧   = 2 i ∧ +3 j ∧ .  1 2 i ∧ + j ∧ =  1 2 2+3 = 5 2 ∴Component of  B →  along the direction of  i ∧ + j ∧ = B →  . a ∧ a ∧     = 5 2 i ∧ + j ∧ 2 = 5 2 i ∧ + j ∧ Let  b ∧  be the unit vector directed along  vector  i ∧ − j ∧ . ∴ b ∧ = i ∧ − j ∧ i ∧ − j ∧ = i ∧ − j ∧ 1 2 + −1 2 = 1 2 i ∧ − j ∧ Magnitude of the component vector of  B →  along the direction of  i ∧ − j ∧ = B → . b ∧ = 2 i ∧ +3 j ∧ . 1 2 i ∧ − j ∧ =− 1 2 ∴Component of  B →  along the direction of  i ∧ − j ∧ = B →  . b ∧ b ∧ = − 1 2 1 2 i ∧ − j ∧ =− 1 2 i ∧ − j ∧

Q.12 A particle starts from origin at t = 0 with a velocity of 10.0 j ∧ ms -1  and moves in the x – y plane with a  constant acceleration of ( 8.0 i ∧ +2.0 j ∧ )  ms -2 . At what time is the x – coordinate of the particle 16 m? 

(a) What is the y – coordinate of the particle at that time? 

(b) What is the speed of the particle at that time? 

Ans.

Given, initial velocity of particle,  u → =10.0 j ∧  m s −1 Acceleration of particle,  a → = d v → dt =( 8.0 i ∧ +2.0 j ∧ ) m s −2 ∴d v → =( 8.0 i ∧ +2.0 j ∧ )dt Integrating it within the limits as time changes from 0 to t and velocity changes from u to v, we obtain: v → ( t )=( 8.0t i ∧ +2.0t j ∧ + u → ) m s −1 As velocity,  v → = d r → dt

∴d r → = v → dt =( 8.0t i ∧ +2.0t j ∧ + u → )dt Integrating it within the limits: as time changes from 0 to t,displacement changes from 0 to r, we have r → = u → t+ 1 2 ×8.0 t 2 i ∧ + 1 2 ×2.0 t 2 j ∧ = u → t+4.0 t 2 i ∧ + t 2 j ∧   =( 10.0 j ∧ )t+4.0 t 2 i ∧ + t 2 j ∧ or x i ∧ +y j ∧ =4.0 t 2 i ∧ +( 10t+ t 2 ) j ∧ Here, we have, x=4.0 t 2  and y=10t+ t 2 ∴t= ( x 4 ) 1 2 (a) when x = 16 m t= ( 16 4 ) 1 2 =2 s ∴y=10×2+ 2 2 = 24 m

(b) Velocity of the particle at time t is given as: v → ( t )=( 8.0t i ∧ +2.0t j ∧ + u → ) m s −1 at t = 2 s v → ( t )=8.0×2 i ∧ +2.0×2 j ∧ +10 j ∧ =( 16 i ∧ +14 j ∧ ) m s −1 ∴Speed of the particle = | v → | = ( 16 ) 2 + ( 14 ) 2 = 256+196 = 452 =21.26 m s −1

Q.13 The position of a particle is given by r  → = 3.0t  i ^  - 2 .0t 2 j ^  + 4.0 k ^  m, where t is in seconds and the coefficients have the proper units for  r →  to be in metres. (a) Find the  v →  and  a →  of the particle? (b) What is the magnitude and direction of velocity of the particle at t = 2s? 

Ans. 

Acceleration  a →  of the particle is given by the relation:   a →  =  d v → dt = d dt ( 3.0  i ^ −4.0t  j ^ ) ∴ a → =−4.0 j ^  m s −2 At time, t=2 s v → =3.0  i ^ −8.0t m s −1 The magnitude of velocity, v= ( 3.0 ) 2 + ( −8 ) 2 ∴ v= 73 =8.54 m s −1 If θ is the angle between  v →  and the x – axis, thentanθ= v y v x = −8 3 =−2.667 ∴θ= tan −1 ( v y v x ) = tan −1 ( −8 3 ) =− tan −1 ( 2.667 ) ∴θ=− 69.45 o The negative sign shows that the velocity is directed below the x – axis.

Q.14 Read each statement below carefully and state, with reasons, if it is true or false:

(a) The net acceleration of a particle in circular motion is always along the radius of the circle towards the centre.

(b) The velocity of a particle at a point is always along the tangent to the path of the particle at that point.

(c) The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null vector.

Ans.

(a) The given statement is false because the net acceleration of a particle is directed towards the centre only in case of uniform circular motion.

(b) The given statement is true because, when a particle leaves the circular path, it moves tangentially to the circular path. Therefore, the velocity vector of a particle at a point is always directed along the tangent at that point.

(c) The given statement is true because, the direction of the acceleration vector in a uniform circular motion is always directed towards the centre of the circle, but it constantly changes with time. The resultant of these vectors over one cycle is a null vector.

Q.15 An aircraft executes a horizontal loop of radius 1.00 km with a steady speed of 900 km/h, compare its centripetal acceleration with the acceleration due to gravity.

Ans.

Given, radius of horizontal loop, r = 1 km  or  1000 m Speed of aircraft, v = 900  kmh -1 = 250  ms -1 Centripetal acceleration is given by the  relation, a c = v 2 r a c = 250  ms -1 2 1000 m = 62.5  ms -2 Acceleration due to gravity, g = 9.8  ms -2 a c g = 62.5  ms -2 9.8  ms -2 = 6.38 ∴Centripetal acceleration,  a c = 6.38 g

Q.16 A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of acceleration of the stone?

Ans.

Given, length of the string, l = 80 cm = 0.8 m Number of revolutions = 14 Time taken = 25 s Frequency of revolution, ν = Number of revolutions Time taken ν = 14 25  Hz Angular frequency, ω = 2πν ω = 2× 22 7 × 14 25  Hz = 88 25   rads -1 Centripetal acceleration,  a c = ω 2 r a c = 88 25   rads -1 2 ×0.8 m = 9.91  ms -2 The direction of centripetal acceleration is always directed along the string,towards the centre of the circular path.

Q.17 The ceiling of a long hall is 25 m high. What is the maximum horizontal distance that a ball thrown with a speed of 40 ms-1 can go without hitting the ceiling of the hall?

Ans.

Given, speed of ball, u = 40  ms -1Maximum height, h = 25 m Let θ be the angle of projection with the horizontal. For projectile motion, Maximum height, H = u 2 sin 2 θ 2g 25 m = 40 ms -1 2 sin 2 θ 2×9 .8 ms -2 ∴ sin 2 θ = 0.30625 sinθ = 0.5534 ∴ θ = sin -1 0.5534 = 33.60° Horizontal range, R = u 2 sin2θ g R = 40 ms -1 2 ×sin2×33.6° 9 .8 ms -2 = 150.5 m

Q.18 In a harbour, wind is blowing at the speed of 72 km/h and the flag on the mast of a boat anchored in the harbour flutters along the N-E direction. If the boat starts moving at a speed of 51 km/h to the north, what is the direction of the flag on the mast of the boat?

Ans.

Given, velocity of the boat, v → b = 51  kmh -1 Velocity of the wind,  v → w = 72  kmh -1 As, the flag is fluttering in north-east direction, it implies that the wind is blowing towards the north-east direction. When the boat starts moving towards the north direction, the flag will flutter  towards the direction of relative velocity of wind w.r.t. the boat. Let the relative velocity of the wind w.r.t. the boat be v → wb Let angle between  v → wb  and   v → w be β Angle between  v w  and  -v b = 90° + 45° ∴tanβ = 51 kmh -1 ×sin 90° + 45° 72 kmh -1   +51 kmh -1 ×cos 90° + 45° = 1.0038 ∴ β = tan -1 1.0038 = 45.1° Angle formed w.r.t. east direction = 45.1° – 45° = 0.1° ∴The flag will flutter almost towards east direction.

Q.19 A man can swim with a speed of 4.0 km/h in still water. How long does he take to cross a river 1.0 km wide if the river flows steadily at 3.0 km/h and he makes his strokes normal to the river current? How far down the river does he go when he reaches the other bank?

Ans.

Given, speed of man = 4  kmh -1 Width of river = 1 km Time taken to cross the river, t = Width of river Speed of man = 1 4 h ∴t = 1 4 ×60 = 15 min Given, speed of river,  v r =3  kmh −1 Distance travelled along with the flow of river in time t= v r ×t =3× 1 4  km = 3 4 ×1000 m =750 m

Q.20 Rain is falling vertically with a speed of 30 ms-1. A woman rides a bicycle with a speed of 10 ms-1 in the north to south direction. What is the direction in which she should hold her umbrella?

Ans.

Let velocity of  cyclist be v c Let velocity of  rain be v r The woman must hold her umbrella in the opposite direction of relative velocity (v) of the rain w.r.t. herself,  in order to protect herself from the rain, v = v r + -v c v = 30 ms -1 + -10 ms -1 = 20  ms -1 From the given figure, tanθ = v c v r tanθ = 10 30 ∴ θ = tan -1 1 3 = tan -1 0.333 ≈18° ∴The woman should hold the umbrella at an angle of approximately 18° with the vertical towards south direction.

Q.21 A passenger arriving in a new town wishes to go from the station to a hotel located10 km away on a straight road from the station. A dishonest cabman takes him along a circuitous path 23 km long and reaches the hotel in 28 min. What is (a) the average speed of the taxi, (b) the magnitude of average velocity? Are the two equal?

Ans.

Given, actual distance travelled = 23 km Total time taken = 28 min = 28 60 h Displacement of the car = Distance between the hotel and the station = 10 km (a) Average speed of the taxi = Actual distance travelled Total time taken = 23 km 28 60 h = 49.29  kmh -1 (b) Magnitude of average velocity = Displacement Total time taken = 10 km 28 60 h = 21.43  kmh -1 ∴Average speed of the taxi is not equal to the magnitudeof average velocity.

Q.22 A cricketer can throw a ball to a maximum horizontal distance of 100 m. With the same speed how high above the ground can the cricketer throw the same ball?

Ans.

Maximum horizontal distance,R = 100 m The ball will cover maximum horizontal distance when, angle of projection = 45° As, horizontal range is given by the relation, R = u 2 sin2θ g 100 m = u 2 g sin90° ∴ u 2 g = 100 m The ball will gain maximum height when height when it is thrown vertically upward. For such motion, the final velocity is zero at maximum height (H). Acceleration, a = -g From third equation of motion, v 2 – u 2 = -2gH 0 – u 2 = -2gH ∴H = 1 2 × u 2 g = 1 2 ×100  m = 50 m

Q.23 On an open ground, a motorist follows a track that turns to his left by an angle of 60° after every 500 m. Starting from a given turn, specify the displacement of the motorist at the third, sixth and eighth turn. Compare the magnitude of the displacement with the total path length covered by the motorist in each case.

Ans.

In this problem, the path followed by the motorist is a regular hexagon of length of each side of the regular hexagon = 500 m Let A be the starting point of the motorist. At D, the motorist will take the third turn .  ∴Magnitude of displacement at D = AG + GD = 500 m + 500 m = 1000 m Total path length from A to D = AB + BC + CD = 500 m + 500 m + 500 m = 1500 m At A, the motorist takes sixth turn, which is the starting point ∴Magnitude of displacement = 0 Total path length from A to A = AB + BC + CD + DE + EF + FA = 6×500 m = 3000 m At C, the motorist takes eighth turn. ∴Magnitude of displacement = AC AC = AB 2 +BC 2 +2 AB . BC ∴AC = 500 m 2 + 500 m 2 +2× 500 m × 500 m ×cos60 o = 866.03 m tanβ = 500 m× sin60 o 500 m +500 m× cos60 o = 1 3 = tan30 o ∴ β = 30 o Therefore, the magnitude of displacement (AC) is 866.03 m at an angle of 30° with AC. Total path length = AB + BC + CD + DE + EF + FA + AB + BC = 8×500 m = 4000 m

Q.24 A cyclist starts from the centre O of a circular park of radius 1 km, reaches the edge Pof the park, then cycles along the circumference, and returns to the centre along QOas shown in Fig. 4.21. If the round trip takes 10 min, what is the (a) net displacement,(b) average velocity, and (c) average speed of the cyclist?

Ans.

(a) In the given case the cyclist comes back to the starting point after cycling for 10 minutes. Time taken = 10 min = 1 6  h ∴Total displacement of the cyclist = 0 (b) Average velocity is given as: Average velocity = Total displacement Total time taken = 0 10 60 h =0 (c) Average speed is given as: Average speed = Total path length Total time taken Total path length = OP + PQ + QO Given, OP = QO = 1 km PQ = 1 4 2πr = 1 4 2π×1 km = 3.570 km ∴Average speed = 3.570 km 1 6 h = 21.42  kmh -1

Q.25 Three girls skating on a circular ice ground of radius200 m start from a point P on the edge of the ground and reach a point Q diametrically opposite to P following different paths as shown in Fig. 4.20. What is the magnitude of the displacement vector for each? For which girl is this equal to the actual length of path skate?

Ans.

( a ) The given statement is incorrect, because in order to make a → + b → + c → + d → = 0, it is not necessary to have all the four vectors to be null vectors. There are many other combinations whose sum is equal to zero. ( b ) The given statement is correct. As, a → + b → + c → + d → =0 ∴ a → + c → =−( b → + d → ) Taking modulus on both sides, | a → + c → |=| −( b → + d → ) | =| ( b → + d → ) | Therefore, magnitude of ( a → + c → ) is equal to the magnitude of ( b → + d → ). Given  a → + b → + c → + d → = 0,

Q.26 Which of the following statements are correct? 

(a)  a → ,  b → ,  c →  and  d →  must each be a null vector. 

(b) The magnitude of ( a → + c → ) equals the magnitude of ( b → + d → ). 

(c) The magnitude of a can never be greater than the sum of the magnitude of  b → ,  c →  and  d → 

(d) b + c must lie in the plane of  a → and d → , if  a →  and  d →  are not collinear and in the line of  a →  and  d → , if they are collinear. 

Ans.

( a ) The given statement is incorrect, because in order to make a → + b → + c → + d → = 0, it is not necessary to have all the four vectors to be null vectors. There are many other combinations whose sum is equal to zero.

( b ) The given statement is correct. As, a → + b → + c → + d → =0 ∴ a → + c → =−( b → + d → ) Taking modulus on both sides, | a → + c → |=| −( b → + d → ) | =| ( b → + d → ) | Therefore, magnitude of ( a → + c → ) is equal to the magnitude of ( b → + d → ).

(c) The given statement is correct. As  a →  +  b →  +  c →  +  d →  =0 ∴ a →  = -( b →  +  c →  +  d → ) Taking modulus on both sides  | a → |=| -( b →  +  c →  +  d → ) | =| b →  +  c →  +  d → | ∴| a → |≤| b → |+| c → |+| d → |→(i) Equation (i) indicates that the magnitude of  a →  can be less than or equal to the sum of the magnitudes of b → ,  c →  and  d → . ∴The magnitude of  a →  can never be greater than the sum of the magnitudes of  b → ,  c →  and  d → .

(d) The given statement is correct, As,  a →  +  b →  +  c →  +  d →  =0  ∴ a → +( b →  +  c → )+ d → =0 The resultant sum of three vectors  a → , ( b →  +  c → ) and  d →  can be zero only if ( b →  +  c → ) lie in the plane of  a →  and  d →  and these three vectors are represented by the three sides of a triangle taken in the same order.

If  a →  and  d →  are collinear, then ( b →  +  c → ) should be in line of  a →  and  d → , only then the vector sum of all the vectors will be zero.

Q.27 Establish the following inequalities geometrically or otherwise: (a) | a → + b → |≤| a → |+| b → | (b) | a → + b → |≥| | a → |−| b → | | (c) | a → − b → |≤| a → |+| b → | (d) | a → − b → |≥| | a → |−| b → | | When does the equality sign above apply?

Ans.

(a) Consider two vectors,  a →  and  b →  represented by the sides OR →  and  OT →  of a parallelogram OTSR as shown in the  given figure (a). Here, we have  | OT → |=|   a → |→(i) | TS →   |=| OR → | =|   b → |→(ii) | OS → |=|   a → +  b → |→(iii) In case of a triangle, each side is smaller than the sum of  other two sides. ∴From ΔOTS, we obtain OS<( OT+TS ) |   a → +  b → |<|   a → |+|   b → |→(iv)

If two vectors   a →  and  b →  are acting along a straight line in the same direction, then we have: |   a → +  b → |=|   a → |+|   b → |→(v) From equation (iv) and (v), we obtain |   a → +  b → |≤|   a → |+|   b → | (b) Consider two vectors,  a →  and  b →  represented by  the sides  OR →  and  OT →  of a parallelogram OTSR as  shown in the given figure (a). Here,   | OT → |=|   a → |→(i) | TS →   |=| OR → | =|   b → |→(ii) | OS → |=|   a → +  b → |→(iii) In case of a triangle, each side is smaller than the sum  of other two sides. ∴From ΔOTS, we obtain OS+TS>OT OS+OT>TS | OS → |>| OT → − OR → |( ∵OR=TS ) |   a → +  b → |<|   a → |+|   b → |→(iv)

If two vectors   a →  and  b →  are acting along a straight line in  the same direction, then we have: |   a → +  b → |=|   a → |+|   b → |→(v) From equation (iv) and (v), we obtain |   a → +  b → |≤|   a → |+|   b →

(c) Consider two vectors,  a →  and  b →  represented by the sides  OQ →  and  OP →  of a parallelogram OPQN as  shown in the given figure (b). Here,  | OP → |=|   a → |→(i) | PN →   |=| OQ → | =|   b → |→(ii)

In case of a triangle, each side is smaller than the sum of  other two sides. ∴From ΔOPN, we obtain ON<OP+PN |   a →  − b → |<|   a → |+| −  b → | |   a →  − b → |<|   a → |+|   b → |→(iii) If two vectors are acting along a straight line but  in opposite directions, then  |   a →  − b → |=|   a → |+|   b → |→(iv) From equation (iii) and (iv), we obtain |   a →  − b → |≤|   a → |+|   b → | (d) Consider two vectors,  a →  and  b →  represented by  the sides  OQ →  and  OP →  of a parallelogram OPQN as  shown in the given figure (b). In case of a triangle, each side is smaller than the sum  of other two sides. ∴From ΔOPN, we obtain ON+PN>OP→(i) ON>OP−PN→(ii) |   a →  − b → |>|   a → |−|   b → |→(iii)

The LHS of the above equation is positive but the RHS     can be positive or negative. In order to make the  quantities on both sides of the above equation positive,  we take modulus on both sides as: | |   a →  − b → | |>| |   a → |−|   b → | | |   a →  − b → |>| |   a → |−|   b → | |→(iv) If the two vectors are acting in a straight line but  in opposite directions, then  |   a →  − b → |=| |   a → |−|   b → | |→(v) Combining equation (iv) and (v), we get |   a →  − b → |≥| |   a → |−|   b → | |

Q.28 Read each statement carefully and state with reasons, if it is true or false:

(a) The magnitude of a vector is always a scalar.

(b) Each component of a vector is always a scalar.

(c) The total path length is always equal to the magnitude of the displacement vector of a particle

(d) The average speed of a particle (defined as total path length divided by the time taken to cover the path) is either greater or equal to the magnitude of average velocity of the particle over the same interval of time.

(e) Three vectors not lying in a plane can never add up to give a null vector.

Ans.

(a) It is true because the magnitude of a vector is a pure number.

(b) It is false because each component of a vector is also a vector.

(c) It is false because the total path length is either greater than or equal to the magnitude of displacement. It is true only if the particle is moving in a straight line.

(d) It is true because the total path length is always greater than or equal to the magnitude of displacement of a particle.

(e) It is true because three vectors that do not lie in a plane cannot be represented by the three sides of a triangle taken in the same order. Thus, it can never follow the triangle method of vector addition to give a null vector.

Q.29 State with reasons whether the following algebraic operations with scalars and vectors are meaningful:

(a) adding any two scalars

(b) adding a scalar to a vector of the same dimension

(c) multiplying any vector by any scalar

(d) multiplying any two scalars

(e) adding any two vectors

(f) adding a component of a vector to the same vector

Ans.

(a) No, the addition of any two scalars is meaningful only if they represent the same physical quantity.

(b) No, the addition of a vector physical quantity with a scalar physical quantity is not meaningful.

(c) Yes, a scalar quantity can be multiplied by a vector quantity. Velocity multiplied by time to give displacement is one of the examples of it.

(d) Yes, multiplication of two scalars is meaningful. For example, power is multiplied by time to give work, which is a useful operation.

(e) No, addition of two vectors is meaningful only if both of them represent the same physical quantity.

(f) Yes, addition of a component of a vector to the same vector is meaningful as both of them have the same dimensions.

Q.30 Pick out the only vector quantity in the following list :

Temperature, pressure, impulse, time, power, total path length, energy, gravitational potential, coefficient of friction, charge.

Ans.

Impulse = change in momentum = force × time

As force is a vector quantity, its product with time will also be a vector quantity. Thus, the impulse is a vector quantity.

Q.31 Pick out the two scalar quantities in the following list :

force, angular momentum, work, current, linear momentum, electric field, average velocity, magnetic moment, relative velocity.

Ans.

Work and current are the scalar quantities in the given list.

Work is defined as the dot product of force and displacement. As the dot product of two physical quantities is always a scalar physical quantity, so work is a scalar physical quantity.

Current is explained by its magnitude only. So current is a scalar physical quantity.

Q.32 State, for each of the following physical quantities, if it is a scalar or a vector :

volume, mass, speed, acceleration, density, number of moles, velocity, angular frequency, displacement, angular velocity.

Ans.

Out of the given physical quantities,

Scalar physical quantities: Volume, mass, speed, density, number of moles and angular frequency

Vector physical quantities: Acceleration, velocity, displacement and angular velocity