Ans: 1
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.
Ans: 2
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.
Ans: 3
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.
Ans: 4
Answer:
(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.
Ans: 5
Answer:
(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.
Ans: 6
(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:
In a triangle, each side is smaller than the sum of the other two sides.
Therefore, in ΔOMN, we have:
ON < (OM + MN)
If the two vectors and act along a straight line in the same direction, then we can write:
Combining equations (iv) and (v), we get:
(b) Let two vectors and be represented by the adjacent sides of a parallelogram OMNP, as shown in the given figure.
Here, we have:
In a triangle, each side is smaller than the sum of the other two sides.
Therefore, in ΔOMN, we have:
… (iv)
If the two vectors and act along a straight line in the same direction, then we can write:
… (v)
Combining equations (iv) and (v), we get:
(c) Let two vectors and be represented by the adjacent sides of a parallelogram PORS, as shown in the given figure.
Here we have:
In a triangle, each side is smaller than the sum of the other two sides. Therefore, in ΔOPS, we have:
If the two vectors act in a straight line but in opposite directions, then we can write:
… (iv)
Combining equations (iii) and (iv), we get:
(d) Let two vectors and be represented by the adjacent sides of a parallelogram PORS, as shown in the given figure.
The following relations can be written for the given parallelogram.
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:
If the two vectors act in a straight line but in the opposite directions, then we can write:
Combining equations (iv) and (v), we get:
Ans: 7
Answer:(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.
Answer:(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).
Answer:(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.
Answer:(d) Correct
For a + b + c + d = 0
a + (b + c) + d = 0
The resultant sum of the three vectors a, (b + c), and dcan be zero only if (b + c) lie in a plane containing aand d, assuming that these three vectors are represented by the three sides of a triangle.
If aand 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.
Ans: 8
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.
Ans: 9
(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
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 = OP + PQ + QO
Time taken = 10 min
∴Average speed
Ans: 10
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: 11
a) Total distance travelled = 23 km
Total time taken = 28 min
∴Average speed of the taxi
(b) Distance between the hotel and the station = 10 km = Displacement of the car
∴Average velocity
Therefore, the two physical quantities (averge speed and average velocity) are not equal.
Ans: 12
Thedescribed 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.
Hence, the woman must hold the umbrella toward the south, at an angle of nearly 18° with the vertical.
Ans: 13
Speed of the man, vm= 4 km/h
Width of the river = 1 km
Time taken to cross the river
Speed of the river, vr= 3 km/h
Distancecovered with flow of the river = vr× t
Ans: 14
Velocity of the boat, vb= 51 km/h
Velocity of the wind, vw= 72 km/h
Theflag 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 vwand (–vb) = 90° + 45°
Anglewith respect to the east direction = 45.11°– 45° = 0.11°
Hence, the flag will flutter almost due east.
Ans: 15
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:
sin2 θ = 0.30625
sin θ = 0.5534
∴θ = sin–1(0.5534) = 33.60°
Horizontal range, R
Ans: 16
Maximum horizontal distance,R = 100 m
Thecricketer 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:
The ball will achieve the maximum height when it is thrown vertically upward. For such motion, the final velocity vis zero at the maximum height H.
Acceleration,a = –g
Usingthe third equation of motion:
Ans: 17
Length of the string, l = 80 cm = 0.8 m
Number of revolutions = 14
Time taken = 25 s
Frequency,
Angular frequency, ω =2πν
Centripetal acceleration,
The direction of centripetal acceleration is always directed along the string, toward the centre, at all points.
Ans: 18
Radius of the loop, r= 1 km = 1000 m
Speed of the aircraft, v= 900 km/h
Centripetal acceleration,
Accelerationdue to gravity, g = 9.8 m/s2
Ans: 19
(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
Inuniform 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.
Ans: 20
(a)
The position of the particle is given by:
Velocity
, of the particle is given as:
Acceleration
, of the particle is given as:
(b) 8.54 m/s, 69.45°below the x-axis
The magnitude of velocity is given by:
The negative sign indicates thatthe direction of velocity is below the x-axis.
Ans: 21
Velocity of the particle,
Acceleration of the particle
Also,
But,
Integrating both sides:
Where,
= Velocity vector of the particle at t = 0
= Velocity vector of the particle at time t
Integrating the equations with the conditions: at t = 0; r = 0 andat t = t; r = r
Since the motion of the particle is confined to the x-y plane, on equating the coefficients of
, we get:
(a) Whenx = 16 m:
∴y = 10 × 2 + (2)2 = 24 m
(b) Velocity of the particle is given by:
Ans: 22
Consider a vector
, given as:
On comparing the components on both sides, we get:
Hence,the magnitude of the vector
is
.
Let
be the angle made by the vector
, with the x-axis, as shown in the following figure.
Hence,the magnitude of the vector
is.
Letbe the angle made by the vector
, with the x- axis, as shown in the following figure.
It is given that:
On comparing the coefficients of
, we have:
Let
make an angle with the x-axis, as shown in the following figure.
Angle between the vectors
Component of vector
, along the direction of , making an angle 
Let
be the angle between the vectors
.
Component of vector, along the direction of, making an angle 
Ans: 23
Answer:(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.
Ans: 24
(a) False
Despitebeing a scalar quantity, energy is not conserved in inelastic collisions.
(b) False
Despitebeing 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.
Ans: 25
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:
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
Ans: 26
Answer:No; Yes; 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.
Ans: 28
Answer: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.
Ans: 29
Answer: No
Range,R = 3 km
Angle of projection, θ= 30°
Acceleration due to gravity, g = 9.8 m/s2
Horizontal range for the projection velocity u0, is given by the relation:
The maximum range (Rmax) is achieved by the bullet when it is fired at an angle of 45°with the horizontal, that is,
On comparing equations (i) and (ii), we get:
Hence, the bullet will not hit a target 5 km away.
Ans: 30
Height of the fighter plane = 1.5 km = 1500 m
Speed of the fighter plane, v = 720 km/h = 200 m/s
Let θ be the angle with the vertical so that the shell hits the plane. The situation is shown in the given figure.
Muzzle velocity of the gun, u = 600 m/s
Time taken by the shell to hit the plane = t
Horizontal distance travelled by the shell = uxt
Distance travelled by the plane = vt
The shell hits the plane. Hence, these two distances must be equal.
uxt = vt
In order to avoid being hit by the shell, the pilot must fly the plane at an altitude (H) higher than the maximum height achieved by the shell.
Ans: 31
0.86 m/s2; 54.46° with the direction of velocity
Speed of the cyclist,
Radius of the circular turn, r = 80 m
Centripetal acceleration is given as:
The situation is shown in the given figure:
Suppose the cyclist begins cycling from point P and moves toward point Q. At point Q, he applies the breaks and decelerates the speed of the bicycle by 0.5 m/s2.
This acceleration is along the tangent at Q and opposite to the direction of motion of the cyclist.
Since the angle between
is 90°, the resultant acceleration a is given by:
Ans: 32
(a) Let
and 
respectively be the initial components of the velocity of the projectile along horizontal (x) and vertical (y) directions.
Let
and 
respectively be the horizontal and vertical components of velocity at a point P.
Time taken by the projectile to reach point P = t
Applying the first equation of motion along the vertical and horizontal directions, we get:
(b) Maximum vertical height,
Horizontal range,
Solving equations (i) and (ii), we get:
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.
Ans: 2
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.
Ans: 3
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.
Ans: 4
Answer:
(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.
Ans: 5
Answer:
(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.
Ans: 6
(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:
In a triangle, each side is smaller than the sum of the other two sides.
Therefore, in ΔOMN, we have:
ON < (OM + MN)
If the two vectors
and act along a straight line in the same direction, then we can write:Combining equations (iv) and (v), we get:
(b) Let two vectors
and be represented by the adjacent sides of a parallelogram OMNP, as shown in the given figure.Here, we have:
In a triangle, each side is smaller than the sum of the other two sides.
Therefore, in ΔOMN, we have:
… (iv)If the two vectors
and act along a straight line in the same direction, then we can write: … (v)Combining equations (iv) and (v), we get:
(c) Let two vectors
and be represented by the adjacent sides of a parallelogram PORS, as shown in the given figure.Here we have:
In a triangle, each side is smaller than the sum of the other two sides. Therefore, in ΔOPS, we have:
If the two vectors act in a straight line but in opposite directions, then we can write:
… (iv)Combining equations (iii) and (iv), we get:
(d) Let two vectors
and be represented by the adjacent sides of a parallelogram PORS, as shown in the given figure.The following relations can be written for the given parallelogram.
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:
If the two vectors act in a straight line but in the opposite directions, then we can write:
Combining equations (iv) and (v), we get:
Ans: 7
Answer:(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.
Answer:(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).
Answer:(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.
Answer:(d) Correct
For a + b + c + d = 0
a + (b + c) + d = 0
The resultant sum of the three vectors a, (b + c), and dcan be zero only if (b + c) lie in a plane containing aand d, assuming that these three vectors are represented by the three sides of a triangle.
If aand 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.
Ans: 8
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.
Ans: 9
(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
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 = OP + PQ + QO
Time taken = 10 min
∴Average speed
Ans: 10
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.
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 coresponding to the required turns is shown in the given table
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 coresponding 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
|
Ans: 11
a) Total distance travelled = 23 km
Total time taken = 28 min
∴Average speed of the taxi
(b) Distance between the hotel and the station = 10 km = Displacement of the car
∴Average velocity
Therefore, the two physical quantities (averge speed and average velocity) are not equal.
Ans: 12
Thedescribed 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.
Hence, the woman must hold the umbrella toward the south, at an angle of nearly 18° with the vertical.
Ans: 13
Speed of the man, vm= 4 km/h
Width of the river = 1 km
Time taken to cross the river
Speed of the river, vr= 3 km/h
Distancecovered with flow of the river = vr× t
Ans: 14
Velocity of the boat, vb= 51 km/h
Velocity of the wind, vw= 72 km/h
Theflag 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 vwand (–vb) = 90° + 45°
Anglewith respect to the east direction = 45.11°– 45° = 0.11°
Hence, the flag will flutter almost due east.
Ans: 15
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:
sin2 θ = 0.30625
sin θ = 0.5534
∴θ = sin–1(0.5534) = 33.60°
Horizontal range, R
Ans: 16
Maximum horizontal distance,R = 100 m
Thecricketer 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:
The ball will achieve the maximum height when it is thrown vertically upward. For such motion, the final velocity vis zero at the maximum height H.
Acceleration,a = –g
Usingthe third equation of motion:
Ans: 17
Length of the string, l = 80 cm = 0.8 m
Number of revolutions = 14
Time taken = 25 s
Frequency,
Angular frequency, ω =2πν
Centripetal acceleration,
The direction of centripetal acceleration is always directed along the string, toward the centre, at all points.
Ans: 18
Radius of the loop, r= 1 km = 1000 m
Speed of the aircraft, v= 900 km/h
Centripetal acceleration,
Accelerationdue to gravity, g = 9.8 m/s2
Ans: 19
(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
Inuniform 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.
Ans: 20
(a)
The position of the particle is given by:
Velocity
, of the particle is given as:Acceleration
, of the particle is given as:(b) 8.54 m/s, 69.45°below the x-axis
The magnitude of velocity is given by:
The negative sign indicates thatthe direction of velocity is below the x-axis.
Ans: 21
Velocity of the particle,
Acceleration of the particle
Also,
But,
Integrating both sides:
Where,
= Velocity vector of the particle at t = 0= Velocity vector of the particle at time tIntegrating the equations with the conditions: at t = 0; r = 0 andat t = t; r = r
Since the motion of the particle is confined to the x-y plane, on equating the coefficients of
, we get:(a) Whenx = 16 m:
∴y = 10 × 2 + (2)2 = 24 m
(b) Velocity of the particle is given by:
Ans: 22
Consider a vector
, given as:On comparing the components on both sides, we get:
Hence,the magnitude of the vector
is.Let
be the angle made by the vector, with the x-axis, as shown in the following figure.Hence,the magnitude of the vector
is.Let
be the angle made by the vector, with the x- axis, as shown in the following figure.It is given that:
On comparing the coefficients of
, we have: Let
make an angle with the x-axis, as shown in the following figure.Angle between the vectors
Component of vector
, along the direction of , making an angle Let
be the angle between the vectors.Component of vector
, along the direction of, making an angle Ans: 23
Answer:(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.
Ans: 24
(a) False
Despitebeing a scalar quantity, energy is not conserved in inelastic collisions.
(b) False
Despitebeing 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.
Ans: 25
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:
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
Ans: 26
Answer:No; Yes; 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.
Ans: 28
Answer: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.
Ans: 29
Answer: No
Range,R = 3 km
Angle of projection, θ= 30°
Acceleration due to gravity, g = 9.8 m/s2
Horizontal range for the projection velocity u0, is given by the relation:
The maximum range (Rmax) is achieved by the bullet when it is fired at an angle of 45°with the horizontal, that is,
On comparing equations (i) and (ii), we get:
Hence, the bullet will not hit a target 5 km away.
Ans: 30
Height of the fighter plane = 1.5 km = 1500 m
Speed of the fighter plane, v = 720 km/h = 200 m/s
Let θ be the angle with the vertical so that the shell hits the plane. The situation is shown in the given figure.
Muzzle velocity of the gun, u = 600 m/s
Time taken by the shell to hit the plane = t
Horizontal distance travelled by the shell = uxt
Distance travelled by the plane = vt
The shell hits the plane. Hence, these two distances must be equal.
uxt = vt
In order to avoid being hit by the shell, the pilot must fly the plane at an altitude (H) higher than the maximum height achieved by the shell.
Ans: 31
0.86 m/s2; 54.46° with the direction of velocity
Speed of the cyclist,
Radius of the circular turn, r = 80 m
Centripetal acceleration is given as:
The situation is shown in the given figure:
Suppose the cyclist begins cycling from point P and moves toward point Q. At point Q, he applies the breaks and decelerates the speed of the bicycle by 0.5 m/s2.
This acceleration is along the tangent at Q and opposite to the direction of motion of the cyclist.
Since the angle between
is 90°, the resultant acceleration a is given by:Ans: 32
(a) Let
and respectively be the initial components of the velocity of the projectile along horizontal (x) and vertical (y) directions.Let
and respectively be the horizontal and vertical components of velocity at a point P.Time taken by the projectile to reach point P = t
Applying the first equation of motion along the vertical and horizontal directions, we get:
(b) Maximum vertical height,
Horizontal range,
Solving equations (i) and (ii), we get:
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