At Saralstudy, we are providing you with the solution of Class 11 Physics Motion in a Plane according to the latest NCERT (CBSE) Book guidelines prepared by expert teachers. Here we are trying to give you a detailed answer to the questions of the entire topic of this chapter so that you can get more marks in your examinations by preparing the answers based on this lesson. We are trying our best to give you detailed answers to all the questions of all the topics of Class 11 Physics Motion in a Plane so that you can prepare for the exam according to your own pace and your speed.
Scalar: Volume, mass, speed, density, number of moles, angular frequency
Vector: Acceleration, velocity, displacement, angular velocity
A scalar quantity is specified by its magnitude only. It does not have any direction associated with it. Volume, mass, speed, density, number of moles, and angular frequency are some of the scalar physical quantities.
A vector quantity is specified by its magnitude as well as the direction associated with it. Acceleration, velocity, displacement, and angular velocity belong to this category.
Work and current are scalar quantities.
Work done is given by the dot product of force and displacement. Since the dot product of two quantities is always a scalar, work is a scalar physical quantity.
Current is described only by its magnitude. Its direction is not taken into account. Hence, it is a scalar quantity.
Impulse
Impulse is given by the product of force and time. Since force is a vector quantity, its product with time (a scalar quantity) gives a vector quantity.
(a) Meaningful
(b) Not Meaningful
(c) Meaningful
(d) Meaningful
(e) Meaningful
(f) Meaningful
Explanation:
(a)The addition of two scalar quantities is meaningful only if they both represent the same physical quantity.
(b)The addition of a vector quantity with a scalar quantity is not meaningful.
(c) A scalar can be multiplied with a vector. For example, force is multiplied with time to give impulse.
(d) A scalar, irrespective of the physical quantity it represents, can be multiplied with another scalar having the same or different dimensions.
(e) The addition of two vector quantities is meaningful only if they both represent the same physical quantity.
(f) A component of a vector can be added to the same vector as they both have the same dimensions.
(a) True
(b) False
(c) False
(d) True
(e) True
Explanation:
(a) The magnitude of a vector is a number. Hence, it is a scalar.
(b) Each component of a vector is also a vector.
(c) Total path length is a scalar quantity, whereas displacement is a vector quantity. Hence, the total path length is always greater than the magnitude of displacement. It becomes equal to the magnitude of displacement only when a particle is moving in a straight line.
(d) It is because of the fact that the total path length is always greater than or equal to the magnitude of displacement of a particle.
(e) Three vectors, which do not lie in a plane, cannot be represented by the sides of a triangle taken in the same order.
(a) Let two vectors 
and 
be represented by the adjacent sides of a parallelogram OMNP, as shown in the given figure.
Here, we can write:
(b) Let two vectors and be represented by the adjacent sides of a parallelogram OMNP, as shown in the given figure.
In a triangle, each side is smaller than the sum of the other two sides.
Therefore, in ΔOMN, we have:
(c) Let two vectors and be represented by the adjacent sides of a parallelogram PORS, as shown in the given figure.
In a triangle, each side is smaller than the sum of the other two sides. Therefore, in ΔOPS, we have:
OS < OP + PS
(d) Let two vectors and be represented by the adjacent sides of a parallelogram PORS, as shown in the given figure.
(a) Incorrect
In order to make a + b + c + d = 0, it is not necessary to have all the four given vectors to be null vectors. There are many other combinations which can give the sum zero.
(b) Correct
a + b + c + d = 0
a + c = – (b + d)
Taking modulus on both the sides, we get:
| a + c | = | –(b + d)| = | b + d |
Hence, the magnitude of (a + c) is the same as the magnitude of (b + d).
(c) Correct
a + b + c + d = 0
a = (b + c + d)
Taking modulus both sides, we get:
| a | = | b + c + d | … (i)
Equation (i) shows that the magnitude of a is equal to or less than the sum of the magnitudes of b, c, and d.
Hence, the magnitude of vector a can never be greater than the sum of the magnitudes of b, c, and d.
(d) Correct
For a + b + c + d = 0
a + (b + c) + d = 0
The resultant sum of the three vectors a, (b + c), and d can be zero only if (b + c) lie in a plane containing a and d, assuming that these three vectors are represented by the three sides of a triangle.
If a and d are collinear, then it implies that the vector (b + c) is in the line of a and d. This implication holds only then the vector sum of all the vectors will be zero.
Displacement is given by the minimum distance between the initial and final positions of a particle. In the given case, all the girls start from point P and reach point Q. The magnitudes of their displacements will be equal to the diameter of the ground.
Radius of the ground = 200 m
Diameter of the ground = 2 × 200 = 400 m
Hence, the magnitude of the displacement for each girl is 400 m. This is equal to the actual length of the path skated by girl B.
(a) Displacement is given by the minimum distance between the initial and final positions of a body. In the given case, the cyclist comes to the starting point after cycling for 10 minutes. Hence, his net displacement is zero.
(b) Average velocity is given by the relation:
Average velocity = Net Displacement / Total Time
Since the net displacement of the cyclist is zero, his average velocity will also be zero.
(c) Average speed of the cyclist is given by the relation:
Average speed = Total Path Length / Total Time
Total path length = OP + PQ + QO
= 1 + 1/4 (2π x1) + 1
= 2 + 1/2 π = 3.570 km
Time taken = 10 min = 10/60 = 1/6 h
∴Average speed = 3.570 / 1/6 = 21.42 km/h
The path followed by the motorist is a regular hexagon with side 500 m, as shown in the given figure
Let the motorist start from point P.
The motorist takes the third turn at S.
∴Magnitude of displacement = PS = PV + VS = 500 + 500 = 1000 m
Total path length = PQ + QR + RS = 500 + 500 +500 = 1500 m
The motorist takes the sixth turn at point P, which is the starting point.
∴Magnitude of displacement = 0
Total path length = PQ + QR + RS + ST + TU + UP
= 500 + 500 + 500 + 500 + 500 + 500 = 3000 m
The motorist takes the eight turn at point R
∴Magnitude of displacement = PR
Therefore, the magnitude of displacement is 866.03 m at an angle of 30° with PR.
Total path length = Circumference of the hexagon + PQ + QR
= 6 × 500 + 500 + 500 = 4000 m
The magnitude of displacement and the total path length corresponding to the required turns is shown in the given table:
| Turn | Magnitude of displacement (m) | Total path length (m) |
| Third | 1000 | 1500 |
| Sixth | 0 | 3000 |
| Eighth | 866.03; 30° | 4000 |
(a) Total distance traveled = 23 km
Total time taken = 28 min = 28/60 h
∴Average speed of the taxi = Total Distance Travelled / Total Time Taken
= 23 / (28/60) = 49.29 km/h
(b) Distance between the hotel and the station = 10 km = Displacement of the car
∴Average velocity = 10 / (28/60) = 21.43 km/h
Therefore, these two : average speed and average velocity are not equal.
The described situation is shown in the given figure.
Here,
vc = Velocity of the cyclist
vr = Velocity of falling rain
In order to protect herself from the rain, the woman must hold her umbrella in the direction of the relative velocity (v) of the rain with respect to the woman.
v = vr + (-vc)
= 30 + (-10) = 20 m/s
tanø = vc / vr = 10/30
ø = tan-1 (1/3)
= tan-1 (0.3333) = 18°
Hence, the woman must hold the umbrella toward the south, at an angle of nearly 18° with the vertical.
Speed of the man, vm = 4 km/h
Width of the river = 1 km
Time taken to cross the river = width of the river / speed of the river
= 1/4 h = 15 minutes
Speed of the river, vr = 3 km/h
Distance covered with flow of the river = vr × t
= 3 x 1/4 = 3/4 km
= 3/4 x 1000 = 750 m
Velocity of the boat, vb = 51 km/h
Velocity of the wind, vw = 72 km/h
The flag is fluttering in the north-east direction. It shows that the wind is blowing toward the north-east direction. When the ship begins sailing toward the north, the flag will move along the direction of the relative velocity (vwb) of the wind with respect to the boat.
The angle between vw and (–vb) = 90° + 45°
tan β = 51sin (90+45) / 72 + 51cos(90+45)
∴ β = tan-1 (1.0038) = 45.11°
Angle with respect to the east direction = 45.11° – 45° = 0.11°
Hence, the flag will flutter almost due east.
Speed of the ball, u = 40 m/s
Maximum height, h = 25 m
In projectile motion, the maximum height reached by a body projected at an angle θ, is given by the relation:
h = u2 sin2Ø / 2g
25 = 402 sin2Ø / 2 x 9.8
sin2 θ = 0.30625
sin θ = 0.5534
∴θ = sin–1(0.5534) = 33.60°
Horizontal range, R = u2 sin2Ø / g
= 402 x sin2 x 33.60 / 9.8
= 1600 x sin 67.2 / 9.8
= 1600 x 0.922 / 9.8 = 150.53 m
Maximum horizontal distance, R = 100 m
The cricketer will only be able to throw the ball to the maximum horizontal distance when the angle of projection is 45°, i.e., θ = 45°.
The horizontal range for a projection velocity v, is given by the relation:
R = u2 sin2Ø / g
100 = u2 / g sin 90°
u2 / g = 100 ....(i)
The ball will achieve the maximum height when it is thrown vertically upward. For such motion, the final velocity v is zero at the maximum height H.
Acceleration, a = –g
Using the third equation of motion:
v2 - u2 = -2gH
H = ½ x u2 / g = ½ x 100 = 50m
Length of the string, l = 80 cm = 0.8 m
Number of revolutions = 14
Time taken = 25 s
Frequency, v = Number of revolutions / time taken = 14 / 25 Hz
Angular frequency, ω = 2πν
= 2 x 22/7 x 14/25 = 88 / 25 rad s-1
Centripetal acceleration, ac = ω2 r
= (88 / 25) 2 x 0.8
= 9.91 m/s2
The direction of centripetal acceleration is always directed along the string, toward the centre, at all points.
Radius of the loop, r = 1 km = 1000 m
Speed of the aircraft, v = 900 km/h = 900 x 5 /18 = 250 m/s
Centripetal acceleration, ac = v2 / r
= (250)2 / 1000 = 62.5 m/s2
Acceleration due to gravity, g = 9.8 m/s2
ac / g = 62.5 / 9.8 = 6.38
ac = 6.38g
(a) False
The net acceleration of a particle in circular motion is not always directed along the radius of the circle toward the centre. It happens only in the case of uniform circular motion.
(b) True
At a point on a circular path, a particle appears to move tangentially to the circular path. Hence, the velocity vector of the particle is always along the tangent at a point.
(c) True
In uniform circular motion (UCM), the direction of the acceleration vector points toward the centre of the circle. However, it constantly changes with time. The average of these vectors over one cycle is a null vector.
(b) 8.54 m/s, 69.45° below the x-axis
Direction, ø = tan-1 (vy / vx )
= tan-1 (-8 / 3 )
= - tan-1 (2.667)
= -69.45°
The negative sign indicates that the direction of velocity is below the x-axis.
(b) and (e)
(a) It is given that the motion of the particle is arbitrary. Therefore, the average velocity of the particle cannot be given by this equation.
(b) The arbitrary motion of the particle can be represented by this equation.
(c) The motion of the particle is arbitrary. The acceleration of the particle may also be non-uniform. Hence, this equation cannot represent the motion of the particle in space.
(d) The motion of the particle is arbitrary; acceleration of the particle may also be non-uniform. Hence, this equation cannot represent the motion of particle in space.
(e)The arbitrary motion of the particle can be represented by this equation.
(a) False
Despite being a scalar quantity, energy is not conserved in inelastic collisions.
(b) False
Despite being a scalar quantity, temperature can take negative values.
(c) False
Total path length is a scalar quantity. Yet it has the dimension of length.
(d) False
A scalar quantity such as gravitational potential can vary from one point to another in space.
(e) True
The value of a scalar does not vary for observers with different orientations of axes.
The positions of the observer and the aircraft are shown in the given figure.
Height of the aircraft from ground, OR = 3400 m
Angle subtended between the positions, ∠POQ = 30°
Time = 10 s
In ΔPRO:
tan 15° = PR /OR
PR = OR tan 15°
= 3400 x tan15°
ΔPRO is similar to ΔRQO.
∴PR = RQ
PQ = PR + RQ
= 2PR = 2 × 3400 tan 15°
= 6800 × 0.268 = 1822.4 m
∴Speed of the aircraft = 1822.4 / 10 = 182.24 m/s
Does it have a location in space? No;
Can it vary with time? Yes;
Will two equal vectors a and b at different locations in space necessarily have identical physical effects? No
Generally speaking, a vector has no definite locations in space. This is because a vector remains invariant when displaced in such a way that its magnitude and direction remain the same. However, a position vector has a definite location in space.
A vector can vary with time. For example, the displacement vector of a particle moving with a certain velocity varies with time.
Two equal vectors located at different locations in space need not produce the same physical effect. For example, two equal forces acting on an object at different points can cause the body to rotate, but their combination cannot produce an equal turning effect.
Does it mean that anything that has magnitude and direction is necessarily a vector? No
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? No
A physical quantity having both magnitude and direction need not be considered a vector. For example, despite having magnitude and direction, current is a scalar quantity. The essential requirement for a physical quantity to be considered a vector is that it should follow the law of vector addition.
Generally speaking, the rotation of a body about an axis is not a vector quantity as it does not follow the law of vector addition. However, a rotation by a certain small angle follows the law of vector addition and is therefore considered a vector.
No; Yes; No
(a) One cannot associate a vector with the length of a wire bent into a loop.
(b) One can associate an area vector with a plane area. The direction of this vector is normal, inward or outward to the plane area.
(c) One cannot associate a vector with the volume of a sphere. However, an area vector can be associated with the area of a sphere.
