Class 11 - Physics
Chapter 4 -Motion In A Plane
Top Block 1
Question :1.
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.
Answer :
Scalar: Volume, mass, speed, density, number of moles, angular frequency.
A scalar quantity is specified by its magnitude only. It does not have any direction associated with it. For example:- Volume, mass, speed, density, number of moles, and angular frequency are some of the scalar physical quantities.
Vector: Acceleration, velocity, displacement, angular velocity.
A vector quantity is specified by its magnitude as well as the direction associated with it. For example: – Acceleration, velocity, displacement, and angular velocity.
Question : 2.
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.
Answer :
Both Work and current are scalar quantities.
Work done is given by the dot product of force and displacement.
W = (F.S)
Since the dot product of two quantities is always a scalar, therefore 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.
Question : 3.
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.
Answer :
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.
Question : 4.
State with reasons, whether the following algebraic operations with scalar and vector physical quantities are meaningful:
(a) adding any two scalars, (b) adding a scalar to a vector of the same dimensions ,
(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.
Answer :
- Adding any two scalars is meaningful because only the scalars of same dimensions or same unit can be added.
- Adding a scalar to a vector of the same dimensions is not meaningful because vector cannot be added to a scalar quantity.
- Multiplying any vector by any scalar is meaningful because when a scalar is multiplied to a vector quantity we get a scalar quantity. For example:-Force when multiplied with time gives impulse.
- Multiplying any vector by any scalar is meaningful can be multiplied with another scalar having the same or different dimensions.
- Multiplying any two scalars is not meaningful because only vectors having same dimensions and unit can be added.
- Adding a component of a vector to the same vector is meaningful as we can add component of vector to a vector.
Question :5.
Read each statement below 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.
Answer :
- True. The magnitude of a vector is a number. Hence, it is a scalar.
- False. Each component of a vector is also a vector.
- False. 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.
- True. Because the total path length is always greater than or equal to the magnitude of displacement of a particle.
- True. The resultant of two vectors will lie in a plane containing two vectors, as the third vector is not in the plane of the resultant of two vectors, they cannot give a null vector.
Question : 6.
Establish the following vector 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 do the equality sign above apply?
Answer :
- Let two vectors a ⃗ and b ⃗ be represented by the adjacent sides of a parallelogram OMNP, as shown in the figure.
| MN ⃗| = | OP ⃗| = | b ⃗| (ii)
| ON ⃗| = | a ⃗ + b ⃗ | (iii)
In a triangle, each side is smaller than the sum of the other two sides.
Therefore, in ΔOMN, we have:
ON < (OM + MN)
| a ⃗ + b ⃗ | < | a ⃗| + | b ⃗| (iv)
If the two vectors a⃗ and b⃗ act along a straight line in the same direction, then we can write:
| a ⃗ + b ⃗ | = | a ⃗| + | b ⃗| (v)
Combining equations (iv) and (v), we get:
| a ⃗ + b ⃗ | ≤ | a ⃗| + | b ⃗|
- Let two vectors a⃗ and b⃗ be represented by the adjacent sides of a parallelogram OMNP, as shown in the given figure.
Mddle block 1
| MN ⃗| = | OP ⃗| = | b ⃗| (ii)
| ON ⃗| = | a ⃗ + b ⃗ | (iii)
In a triangle, each side is smaller than the sum of the other two sides.
Therefore, in ΔOMN, we have:
(ON +MN) > OM
(ON + OM) > MN
| ON ⃗| > | OM ⃗ – OP ⃗| (because OP=MN)
| a ⃗ + b ⃗ | > || a ⃗| – | b ⃗|| (iv)
If the two vectors a⃗ and b⃗ act along a straight line in the same direction, then we can write:
| a ⃗ + b ⃗ | = || a ⃗| – | b ⃗|| (v)
Combining equations (iv) and (v), we get:
| a ⃗ + b ⃗ | ≥|| a ⃗| – | b ⃗||
- c) Let two vectors a⃗ and b⃗ be represented by the adjacent sides of a parallelogram PORS, as shown in the given figure.
| OP ⃗| = | a ⃗| (ii)
In a triangle, each side is smaller than the sum of the other two sides. Therefore, in ΔOPS,
OS < OP + PS
| a ⃗ – b ⃗ | < | a ⃗| + | -b ⃗|
| a ⃗ – b ⃗ | < | a ⃗| + | b ⃗| (iii)
If the two vectors act in a straight line but in opposite directions, then we can write:
| a ⃗ – b ⃗ | = | a ⃗| + | b ⃗| (iv)
Combining equations (iii) and (iv), we get:
| a ⃗ – b ⃗ | ≤ | a ⃗| + | b ⃗|
- d) Let two vectors a⃗ and b⃗ be represented by the adjacent sides of a parallelogram PORS, as shown in the figure.
(OS +PS) > OP (i)
OS < (OP- PS) (ii)
| a ⃗ – b ⃗ | > | a ⃗| – | b ⃗| (iii)
The quantity on the LHS is always positive and that on the RHS can be positive or negative. To make both quantities positive, we take modulus on both sides as:
|| a ⃗ – b ⃗ || > || a ⃗| – | b ⃗||
| a ⃗ – b ⃗ | > || a ⃗| – | b ⃗|| (iv)
If the two vectors act in a straight line but in the opposite directions, then we can write:
| a ⃗ – b ⃗ | = || a ⃗| – | b ⃗|| (v)
Combining equations (iv) and (v), we get:
| a ⃗ – b ⃗ | ≥ || a ⃗| – | b ⃗||
Question :7.
Given a + b + c + d = 0, 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 magnitudes 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 ?
Answer :
- 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) i.e. their magnitude of (a + c) is equal to (c + d) but their directions are opposite. Therefore the statement is true.
- Correct.
Since (a + b + c + d = 0), therefore a = – (b + c + d)
This implies that the magnitude of a is equal to magnitude of vector
(b + c + d). The sum of the magnitudes of b, c, d may be greater than or equal to vector a. Hence, the statement that magnitude of a can never be
greater than sum of magnitudes of b, c and d is correct.
- Correct.
Since (a + b + c + d = 0),
(b + c) + (a + d) =0. The resultant sum of three vectors b + c, a + d can be 0 only if (b + c) is in the plane of a and d.
If (a) and (d) are collinear (b + c) must be line of a and d therefore the given statement is correct.
Question : 8.
Three girls skating on a circular ice ground of radius 200 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?
Answer :
Displacement is the minimum distance between the initial and final positions of a particle. In this 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.
Question : 9.
A cyclist starts from the centre O of a circular park of radius 1 km, reaches the edge P of the park, then cycles along the circumference,
and returns to the centre along QO as 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
Answer :
- 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.
- 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 +1) + 1)
= (2 + (1/2)) = 3.570km.
Time taken = 10 min = (10/60) = (1/6) h
Therefore, Average speed = (3.570)/ (1/2) = 21.42km/h
Question : 10.
On an open ground, a motorist follows a track that turns to his left by an angle of 600 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.
Answer :
The path followed by the motorist is a regular hexagon with side 500 m, as shown in the given figure:
The motorist takes the third turn at S. The displacement vector at S = PS
Therefore, 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.
Therefore displacement vector is null vector.
Therefore, Magnitude of displacement = 500 + 500 = 1000m
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.
Displacement vector = RS which is represented by the diagonal of the ||gm PQRV
Magnitude of displacement = PR
= √ ((PQ) 2 + (QR) 2+2 (PQ) (QR) cos600)
=√ (500)2 + (500)2 + (2 x 500 x 500 x cos 600)
=√ (250000) + (250000) + (500000 x (1/2))
=866.03m
β = tan-1(500 sin 600)/ (500 + 500 cos600)
β =300
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
Comparison of magnitude of displacement with total path length:-
- 3rd turn = (Displacement)/ (Path length) = (1000m)/ (1500m) =0.666.
- 6th turn = (0m)/(3000m) = 0
- 8th turn = (866.03)/(4000m) 0.216
Question : 11.
A passenger arriving in a new town wishes to go from the station to a hotel located 10 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?
Answer :
Given:-
- Total distance travelled = 23 km
Total time taken = 28 min = (28/60) h
Therefore, Average speed of the taxi = (distance travelled)/ (time taken)
= (23)/ (28/60) = 49.29km/hr.
- Distance between the hotel and the station = 10 km = Displacement of the car.
Therefore Average velocity = (10)/ (28/60) = 21.43km/hr
Therefore, the two physical quantities (average speed and average velocity) are not equal.
Question : 12.
Rain is falling vertically with a speed of 30 m s-1. A woman rides a bicycle with a speed of 10 m s-1 in the north to south direction.
What is the direction in which she should hold her umbrella?
Answer :
Velocity of the cyclist = vc
Velocity of falling rain = vr
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) = 20m/s
tan θ = (vc / vr) = (10/30)
θ = tan-1(1/3)
θ =tan-1(0.333) = 180
Hence, the woman must hold the umbrella toward the south, at an angle of nearly 18° with the vertical.
Question : 13.
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?
Answer :
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 = (1/4) x 60 = 15min
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 = 750m.
Question : 14.
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?
Answer :
Given:-
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.
Now, vwb = vw + (-vb)
Velocity of the boat,|- vb| = 51 km/h toward north
Velocity of the wind, vw = 72 km/h along N-E direction
The angle between vw and (–vb) = (90° + 45°) = 1350
tan β = 51 sin(900 + 450)/((72 + 51(-cos450))
= (51 sin450)/ (72 + 51(-cos450))
= (51 x (1/√2))/ (72 – (51 x (1/ √2))
= (51)/ (72√2 – 51)
= (51)/ (72 x 1.414 – 51)
= (51)/ (50.800)
Therefore β = tan-1(1.0039)
=45.10
Angle with respect to the east direction = (45.1° – 45°) = (0.1°).
Hence, the flag will flutter almost due east.
Question : 15.
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 m s-1 can go without hitting the ceiling of the hall?
Answer :
Given:-
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 sin2 θ)
25 = ((40)2 sin2 θ)/ (2 x 9.8)
sin2 θ = 0.30625
sin θ = 0.5534
θ = sin–1 (0.5534) = 33.60°
Horizontal range, R = (u2 sin2 θ)/ (g)
= ((40)2 x sin 2 x 33.60)/ (9.8)
= (1600 x sin 67.2)/ (9.8)
= (1600 x 0.922)/ (9.8)
=150.53m
Question : 16.
A cricketer can throw a ball to a maximum horizontal distance of 100 m. How much high above the ground can the cricketer throw the same ball?
Answer :
Given:-
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 sin2 θ)/ (g)
100 = (u2 /g) sin 900
(u2 /g) = 100 (1)
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 = (1/2) x (u2/g)
= (1/2) x 100
=50m
Question : 17.
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?
Answer :
Given:-
Radius r = 80 cm = 0.8m
Number of revolutions = 14
Time taken = 25 s
Frequency, ν = (Number of revolutions)/ (Time taken) = (14/25) Hz
Angular frequency, ω = 2πν = (2 x (22/7) x (14/25)) = (88/25) rad-1.
Centripetal acceleration, ac = ω2r = (250)2 x 1000
Again (ac/g) = ((250)2/1000) x (1/9.8) = 6.38
The direction of centripetal acceleration is always directed along the string, toward the centre, at all points.
Question : 18.
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.
Answer :
Radius of the loop, r = 1 km = 1000 m
Speed of the aircraft, v = 900 km/h = 900 x (5/18) = 250m/s
Centripetal acceleration, ac = (v2/r)
= (250)2/ (1000) = 62.5m/s2
Acceleration due to gravity, g = 9.8 m/s2
(ac /g)= ((62.5)/ (9.8) = 6.38
ac =6.38g
Question : 19.
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 vector 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.
Answer :
- False. The net acceleration of a particle in circular motion is along the radius of the circle towards the centre only in uniform motion.
- Because while leaving the circular path, the particle moves tangentially to the circular path.
- The direction of acceleration vector in uniform circular motion is directed towards the centre of circular path.
- It is constantly changing with time. The average of all these vectors over one cycle is a null vector.
Question : 20.
The position of a particle is given by r = 3.0 î – 2.0t2ĵ + 4.0 k̂ 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 = 2.0 s?
Answer :
- Velocity, v = (dr/dt)
=d/dt (3.0 î – 2.0t2ĵ + 4.0 k̂) =3.0 î – 4.0t ĵ
Acceleration, a = (dv/dt)
a = d/dt (3.0 î -4.0 ĵ) = – 4.0 ĵ
8.54m/s, 69.50 below the axis
- At time t =2s,
v = (3.0 î – 2.0t2ĵ + 4.0 k̂)
v =√ [(3)2 + (-8)2] ms-1
=√ (73) = 8.54ms-1
If θ is the angle which v makes with x-axis:
tan θ =(vy/vx)= (-8/3)
=-2.667
= tan-1(2.667)
=- 69.50
The negative sign indicates that the direction of velocity is below the x-axis.
Question : 21.
A particle starts from the origin at t = 0 s with a velocity of 10.0 ĵ m/s and moves in the x-y plane with a constant acceleration of (8.0 î + 2.0 ĵ) m s-2.
(a) At what time is the x- coordinates of the particle 16 m? What is the y- coordinate of the particle at that time?
(b) What is the speed of the particle at the time?
Answer :
Given:-
Velocity of the particle v = 10.0m/s î
Acceleration of the particle a = (8.0 î + 2.0 ĵ) m s-2.
a = d (v)/dt = (8.0 î + 2.0 ĵ) m s-2
Also,
But, a = (dv/dt) = (8.0 î + 2.0 ĵ)
dv = (8.0 î + 2.0 ĵ)
Integrating both sides:
v(t) = (8.0 î + 2.0 ĵ + u)
Where,
Velocity vector of the particle at t = 0 = u
Velocity vector of the particle at time t = v
But, v = (dr/dt)
dr= v dt = (8.0 î + 2.0 ĵ + u)dt
Integrating the equations with the conditions: at t = 0; r = 0 and at t = t; r = r
But,
r = ut + (1/2)8.0t2î + (1/2) x 2.0t2ĵ
= ut + 4.0 t2î + t2ĵ
= (10.0 ĵ) t + 4.0 t2î + t2ĵ
(x î + y ĵ) = (4.0 t2î + (10t+ t2) ĵ
Since the motion of the particle is confined to the x-y plane, on equating the coefficients of î and ĵ, we get:
x = 4t2
t =(x/4)1/2
And y = 10t + t2
When x = 16 m:
t= (16/4)1/2 =2s
Therefore, y = 10 × 2 + (2)2 = 24 m
- b) Velocity of the particle is given by:
v (t) = (8.0 î + 2.0t ĵ + u)
At t=2s
v (t) = 8.0 x2 î + 2.0 x2 ĵ + 10 ĵ
= (16 î + 14 ĵ)
Therefore, speed of the particle;
|v| = √ (16)2 + (14)2
=√ (256 + 196) = √ (452)
=21.26m/s
Question : 22.
î and ĵ are unit vectors along x- and y- axis respectively. What is the magnitude and direction of the vectors î + ĵ and î – ĵ?
What are the components of a vector A= 2 î + 3 ĵ along the directions of (î + ĵ) and (î – ĵ)? [You may use graphical method]
Answer :
- Magnitude of (î+ ĵ) = | î + ĵ|
=√ (1)2 + (1)2 = √2
Let the vector (î + ĵ) make an angle θ with the direction of î, then
cos θ = ( (î + ĵ). Î)/ (| î + ĵ|| î|
= (1/√2) = cos 450 or θ = √2
Magnitude of (î – ĵ) = | î – ĵ|
=√ (1)2 + (-1)2 = √2
Similarly, if vector (î – ĵ) makes an angle θ with the direction of î, then
cos θ = ((î – ĵ). Î)/ (| î – ĵ|. Ĵ)
= (1/√2) (1)
= (1/√2)
cos 450 or θ =450
Here θ = -450 with x-axis.
- Given :- A = 2 î + 3 ĵ
Ax î + Ay ĵ = 2 î + 3 ĵ
On Comparing: – Ax = 2 and Ay = 3
| A| =√ (2)2 + (3)2
=√13
Let Axmake an angle θ with the x-axis, as shown in the following figure.
θ = tan-1(3/2)
=tan-1(1.5) = 56.310
Angle between the vectors (2 î + 3 ĵ) and (î + ĵ),
θ = (56.310 – 450) = 11.310
Component of vector A, along the direction of P, making and angle θ
= (A cos θ) P
= (A cos11.31) (î + ĵ)/ (√2)
= √13 x (0.9806) (î + ĵ) / (√2)
=2.5 (î + ĵ)
= (25/10) x (√2)
= (5/ (√2))
Let θ be the angle between the vector 🙁 2 î + 3 ĵ) and (î – ĵ),
θ = (450 +56.310) = 101.310
Component of vector A, along the direction of Q, making and angle θ.
= (A cos θ) Q
= (A cos θ) (î – ĵ)/ (√2)
= (√13) cos (901.310) (î – ĵ)/ (√2)
= (√13)/ (√2) sin (11.300) (î – ĵ)
=-2.550 x 0.196(î – ĵ)
=-0.5(î – ĵ)
=-(5/10) x (√2)
=-(1/ (√2))
Question : 23.
For any arbitrary motion in space, which of the following relations are true:-
(a) vaverage = (1/2) (v (t1) + v (t 2))
(b) vaverage = [r(t2) – r(t1) ] /(t2 – t1)
(c) v(t) = v (0) + a t
(d) r(t) = r (0) + v (0) t + (1/2) a t2
(e) aaverage =[ v (t2) – v (t1 )] /( t2 – t1)
(The ‘average’ stands for average of the quantity over the time interval t1 to t2)
Answer :
- False. It is given that the motion of the particle is arbitrary. Therefore, the average velocity of the particle cannot be given by this equation.
- True. The arbitrary motion of the particle can be represented by this equation.
- False. The motion of the particle is arbitrary. The acceleration of the particle may also be non-uniform. Therefore this equation represent the motion of the particle is space.
- False. The motion of the particle is arbitrary; acceleration of the particle may also be non-uniform. Therefore this equation cannot represent the motion of particle in space.
- True. The arbitrary motion of the particle can be represented by this equation.
Question : 24.
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.
Answer :
- False. Kinetic energy is a scalar but does not remain conserved in an inelastic collision.
- False. Temperature is a scalar quantity but still it can take negative values.
- False. Total path length is a scalar quantity but still it has dimension of length.
- True. Scalar quantity has the same value for observers with different orientations of axes.
Question : 25.
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?
Answer :
As shown in the figure: – O is the observation point at the ground.
Height of the aircraft OR=3400m
Time taken =10s.
In ΔPRO:
tan 150 = (PR/OR)
PR = OR tan150
=3400 x tan150
ΔPRO is similar to ΔRQO,
Therefore, PQ = PR + RQ
=2PR = 2 x 3400 tan150
=6800 x 0.268 = 1822.4m.
Therefore speed of the aircraft = (1822.4)/ (10) = 182.24m/s
Bottom Block 3
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