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Test: Motion in a Plane - 1 - NEET MCQ


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30 Questions MCQ Test Physics Class 11 - Test: Motion in a Plane - 1

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Test: Motion in a Plane - 1 - Question 1

Which is a constant for a freely falling object?

Detailed Solution for Test: Motion in a Plane - 1 - Question 1

For a freely falling object under the influence of gravity alone, the acceleration remains constant and is equal to the acceleration due to gravity (g). On Earth, g has a standard value of approximately 9.8 m/s2.

Displacement changes continuously as the object falls.

Velocity increases linearly with time due to constant acceleration.

Speed, which is the magnitude of velocity, also increases as the object falls.

Thus, acceleration is the only constant in this scenario.

Test: Motion in a Plane - 1 - Question 2

Which of the following statements not true?

Detailed Solution for Test: Motion in a Plane - 1 - Question 2

Option A: True. The velocity vector of a particle at any point on its path is always tangent to the path at that point, as velocity represents the direction of motion.

Option B: True. For a particle in uniform circular motion, the centripetal acceleration points towards the center. Over one complete cycle, the acceleration vectors cancel out, resulting in an average acceleration of zero (null vector).

Option C: True. The net acceleration in uniform circular motion is centripetal acceleration, which always points towards the center of the circle along the radius.

Option D: Not true. This statement is incorrect because it implies that the net acceleration in all types of circular motion is always radial. While in uniform circular motion, the acceleration is purely radial, in non-uniform circular motion, there is an additional tangential component of acceleration due to the change in speed. Thus, the net acceleration in such cases is not always directed towards the center.

Test: Motion in a Plane - 1 - Question 3

When a ball is thrown upwards, as it rises, the vertical component of its velocity

Detailed Solution for Test: Motion in a Plane - 1 - Question 3

When a ball is thrown upwards, the vertical component of its velocity decreases as it rises because of the acceleration due to gravity (g), which acts downward.

  • Gravity opposes the upward motion, reducing the ball's vertical velocity until it becomes zero at the highest point.
  • After reaching the highest point, the ball starts descending, and the vertical velocity increases in the downward direction.

Thus, as the ball rises, the vertical component of its velocity decreases.

Test: Motion in a Plane - 1 - Question 4

Vectors can be added by

Detailed Solution for Test: Motion in a Plane - 1 - Question 4

Explanation:Parallelogram law of vector addition is,If two vectors are considered to be the adjacent sides of a Parallelogram, then the resultant of two vectors is given by the vector which is a diagonal passing through the point of contact of two vectors.

Test: Motion in a Plane - 1 - Question 5

The path of a projectile is 

Detailed Solution for Test: Motion in a Plane - 1 - Question 5

The path of a projectile under the influence of gravity is a parabola. This is derived from the kinematic equations of motion in two dimensions:

  • The horizontal motion is uniform (constant velocity).
  • The vertical motion is uniformly accelerated (due to gravity).

Combining these motions, the equation of the trajectory is:

This equation represents a parabolic path, where θ is the angle of projection, u is the initial velocity, and g is the acceleration due to gravity.

Thus, the trajectory of a projectile is a parabola.

Test: Motion in a Plane - 1 - Question 6

The reason why cyclists bank when taking a sharp turn is

Detailed Solution for Test: Motion in a Plane - 1 - Question 6

Explanation:In order to take a safe turn, the cyclist has to bend a little from his vertical position. In this case, a component of the reaction provides the required centripetal force.If q is angle made by the cyclist with the vertical then

In actual practice, the value of q is slightly less because the force of friction also contributes towards the centripetal force.

Test: Motion in a Plane - 1 - Question 7

Magnitude of displacement of a particle is

Detailed Solution for Test: Motion in a Plane - 1 - Question 7

The magnitude of displacement is the shortest straight-line distance between the initial and final positions of the particle.

  • If the particle moves in a straight line without changing direction, the displacement is equal to the path length.
  • If the particle follows a curved or non-linear path, the displacement is less than the path length.

Thus, the magnitude of displacement is always either less than or equal to the path length of the particle between two points.

Test: Motion in a Plane - 1 - Question 8

A motorcycle stunt rider rides off the edge of a cliff. Just at the edge his velocity is horizontal, with magnitude 9.0 m/s. Find the magnitude of the motorcycle’s position vector after 0.50s it leaves the edge of the cliff.

Detailed Solution for Test: Motion in a Plane - 1 - Question 8

Explanation:The motorcycle’s x- and  y-coordinates at t=0.50 s are

The negative value of y shows that the motorcycle is below its starting point.

The motorcycle’s distance from the origin at t=0.50 s is 

Test: Motion in a Plane - 1 - Question 9

Two vectors are equal if

Detailed Solution for Test: Motion in a Plane - 1 - Question 9

Explanation:Equal vectors are vectors that have the same magnitude and the same direction. Equal vectors may start at different positions.

Test: Motion in a Plane - 1 - Question 10

A sports car has a “lateral acceleration” of 0.96g = 9.4 m s−2. This is the maximum centripetal acceleration the car can sustain without skidding out of a curved path. If the car is traveling at a constant 40 m/s on level ground, what is the radius R of the tightest unbanked curve it can negotiate?

Detailed Solution for Test: Motion in a Plane - 1 - Question 10

Explanation:

The car is in uniform circular motion because it’s moving at a constant speed along a curve that is a segment of a circle. Hence we know 

This is the minimum turning radius because arad is the maximum centripetal acceleration.

Test: Motion in a Plane - 1 - Question 11

A cyclist moves along a circular path of radius 70m. If he completes one round in 11s, calculate total length of path.

Detailed Solution for Test: Motion in a Plane - 1 - Question 11

Radius of the circular path, r = 70m
Time takes to complete one round, t = 11s
Total length of the path, s = 2πr = 2 × 22/7 × 70 = 440m

Test: Motion in a Plane - 1 - Question 12

In a harbor, wind is blowing at the speed of 72 km/h and the flag on the mast of a boat anchored in the harbor 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?

Detailed Solution for Test: Motion in a Plane - 1 - Question 12

Explanation: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.

Hence, the flag will flutter approximately due east.

Test: Motion in a Plane - 1 - 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?

Detailed Solution for Test: Motion in a Plane - 1 - Question 13

Explanation:

Speed of the man, vm = 4 km/h
Width of the river = 1 km

Test: Motion in a Plane - 1 - Question 14

A ball rolls off top of a staircase with a horizontal velocity u m/s. If the steps are h metre high and b mere wide, the ball will just hit the edge of nth step if n equals to

Detailed Solution for Test: Motion in a Plane - 1 - Question 14

Horizontal Motion: The horizontal distance traveled by the ball after time t is: x = u ⋅ t

Vertical Motion: The vertical displacement after time t under gravity is:

Conditions for Hitting the n-th Step:

  1. Horizontal distance x must be equal to n⋅b (the horizontal distance to the n-th step):

    From this, the time t is:
  2. Vertical distance y must be equal to n⋅h (the vertical distance to the n-th step):
    y = n ⋅ h



Expand and simplify:

Test: Motion in a Plane - 1 - Question 15

Which of the following statements false?

Detailed Solution for Test: Motion in a Plane - 1 - Question 15

Option A: False. The total path length is the actual distance traveled by the particle, whereas the magnitude of the displacement vector is the straight-line distance between the initial and final positions. These two are only equal if the particle moves in a straight line without changing direction. Otherwise, the path length is greater than the magnitude of displacement.

Option B: True. The magnitude of a vector is always a scalar (it has only magnitude, no direction).

Option C: True. Three vectors not lying in the same plane cannot form a closed triangle (null vector condition), as they cannot satisfy vector addition due to being in different planes.

Option D: True. The average speed is defined as the total path length divided by the time taken, while average velocity is the magnitude of the displacement divided by time. Since the total path length is always greater than or equal to the magnitude of displacement, the average speed is greater than or equal to the magnitude of average velocity.

Thus, Option A is the false statement.

Test: Motion in a Plane - 1 - Question 16

A man standing on the roof of a house of height h throws one particle vertically downwards and another particle horizontally with the same velocity u. The ratio of their velocities when they reach the earth’s surface will be

Detailed Solution for Test: Motion in a Plane - 1 - Question 16

Particle 1 (Thrown Vertically Downward):

Initial velocity (u) is downward.

Final velocity is affected by gravity. Using the equation of motion:

Particle 2 (Thrown Horizontally):

Horizontal velocity remains constant (u).

Vertical velocity due to free fall can be calculated as:

The resultant velocity (v2) at the Earth's surface is given by combining the horizontal and vertical velocities using the Pythagorean theorem:

Velocity Ratio:
The magnitude of both velocities is the same:

Thus, the ratio of their velocities is: 1 : 1

Test: Motion in a Plane - 1 - Question 17

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?

Detailed Solution for Test: Motion in a Plane - 1 - Question 17

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.

Hence, the woman must hold the umbrella toward the south, at an angle of nearly 18° with the vertical.

Test: Motion in a Plane - 1 - Question 18

A man weighing 80 kg is standing in a trolley weighing 320 kg. The trolley is resting on frictionless horizontal rails. If the man starts walking on the trolley with a speed of 1 m/s, then after 4 sec his displacement relative to the ground will be

Detailed Solution for Test: Motion in a Plane - 1 - Question 18

Since there is no external horizontal force acting on the system (man + trolley), the total momentum of the system remains conserved.

Let vm ​ = velocity of the man relative to the ground.

vt ​ = velocity of the trolley relative to the ground.
Initially, both the man and the trolley are at rest. So:
Initial Momentum = 0

When the man walks with velocity 1m/s relative to the trolley:

Momentum of the man relative to the ground:
pm ​= 80⋅vm

Momentum of the trolley relative to the ground:
pt = 320⋅vt

From conservation of momentum:


The man’s velocity relative to the trolley is 1m/s:
= vm ​− vt ​= 1
Substitute 


Thus, the velocity of the trolley:

Man's velocity relative to the ground:

Displacement of the man relative to the ground after 4s:

The correct displacement is: 3.2 m

Test: Motion in a Plane - 1 - Question 19

Null vector or a zero vector has a magnitude

Detailed Solution for Test: Motion in a Plane - 1 - Question 19

A null vector (or zero vector) is a vector with:

  • Zero magnitude.
  • No specific direction.

Mathematically, a null vector is represented as , and its magnitude is given by . It plays a fundamental role in vector algebra, serving as the additive identity in vector addition.

Thus, the magnitude of a null vector is equal to zero.

Test: Motion in a Plane - 1 - Question 20

A particle is projected at an angle θ with horizontal with an initital speed u. When it makes an angle α with horizontal, its speed v is-

Detailed Solution for Test: Motion in a Plane - 1 - Question 20

The horizontal velocity remains constant throughout the motion because there is no acceleration in the horizontal direction:
vx = u cosθ

The vertical velocity changes due to the acceleration due to gravity g. Let the vertical velocity at the point where the angle is α be vy.

 At the given point, the angle α is related to the velocity components by:
 ​
From this, we find:
vy = vx​ tan α = u cos θ tan α

 The speed v of the particle at any point is given by:

Substituting 

  1.  ​Factor out (u cos θ)2 :
  2. Use the trigonometric identity 1 + tan2 α = sec2 α:
  3. Simplify:


Test: Motion in a Plane - 1 - Question 21

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?

Detailed Solution for Test: Motion in a Plane - 1 - Question 21

(b) Distance between the hostel and the station = 10 km = Displacement of the car

∴ Average velocity = 

Test: Motion in a Plane - 1 - Question 22

A particle is moving along the path y = x2 from x = 0 m to x = 2 m. Then the distance traveled by the particle is:

Detailed Solution for Test: Motion in a Plane - 1 - Question 22

Formula for Arc Length:

The distance traveled by the particle is given by the arc length formula:

Here:

  • y = x2, so dy/dx = 2x
  • The limits are a = 0 and b = 2.

Step-by-Step Solution:

  1. Differentiate y = x2 :
    dy/dx = 2x
  2. Square dx/dy :​
  3. Substitute into the arc length formula:
  4. Approximation: The integral  cannot be solved exactly using elementary functions, but its value is slightly greater than ​ because ​ grows faster than  ​ over the interval [0, 2].

Why Option c is Correct:

  • The exact value of the integral lies just above  , making Option 3 ( > ) the correct choice.
Test: Motion in a Plane - 1 - Question 23

A projectile is fired a velocity of 150 meters per second at an angle of 30 degrees with the horizontal. What is the magnitude of the vertical component of the velocity at the time the projectile is fired?

Detailed Solution for Test: Motion in a Plane - 1 - Question 23

Explanation:

Initial velocity u = 150 m/s

Test: Motion in a Plane - 1 - Question 24

Six particles situated at the corners of a regular hexagon of side a move at constant speed v. Each particle maintains a direction towards the particle at the next. The time which the particles will take to meet each other is:

Detailed Solution for Test: Motion in a Plane - 1 - Question 24

1. Symmetry of Motion:

  • Each particle moves towards the next particle, but due to the symmetry of the hexagon, their motion results in a gradual spiral towards the center of the hexagon.
  • At any instant, the direction of motion is along the line joining the current position of the particle to the position of the next particle.

2. Relative Velocity Towards the Center:

  • The relative motion of the particles ensures that the effective velocity component toward the center is responsible for reducing the distance between them.
  • At any moment, the velocity of a particle can be resolved into two components:
    One along the line joining it to the next particle.
    The other perpendicular to it.

The component of velocity along the line joining the particle to the next one is:
veffective = vcos 
where θ = 60 (the interior angle of a regular hexagon).
Using cos ⁡60 = 1/2
veffective ​ = v/2

3. Initial Distance to be Covered:

  • Initially, the distance between two adjacent particles is the side length of the hexagon, a.
  • Due to the symmetry, the particles spiral inward and meet at the center of the hexagon. This meeting point is equidistant from all the initial positions, and the total distance covered by each particle is effectively along its initial direction.

Time to Meet:

The time it takes for the particles to meet can be calculated by dividing the total initial distance a by the effective velocity veffective :

Test: Motion in a Plane - 1 - Question 25

A body is projected with velocity 20√3 m/s with an angle of projection 60° with horizontal. Calculate velocity on that point where body makes an angle 30° with the horizontal.

Detailed Solution for Test: Motion in a Plane - 1 - Question 25

The velocity of the body at any point during its motion is composed of:

  1. Horizontal Component (vx): This remains constant throughout the motion because there’s no horizontal acceleration.

    Using cos ⁡60 = 1/2
  2. Vertical Component (vy): This changes with time due to the acceleration due to gravity (g = 9.8 m/s2). At the point where the angle with the horizontal is 30, the relation between the components of velocity is:
     ​ ​
    Using tan ⁡ 
     ​ ​
    Solve for vy​ :
  3. Magnitude of the Velocity (v): The total velocity is the vector sum of vx and vy​ :

    Substitute

Final Answer:
The speed of the body at the point where it makes an angle of 30 with the horizontal is: 20 m/s

Test: Motion in a Plane - 1 - Question 26

We can define the difference of two vectors A and B as the

Detailed Solution for Test: Motion in a Plane - 1 - Question 26

Explanation:

Vector subtraction is defined in the following way.

  • The difference of two vectors, A - B , is a vector C that is, C = A - B
  • The addition of two vector such that C = A + (-B). B has been taken in opposite direction.

Thus vector subtraction can be represented as a vector addition.

Test: Motion in a Plane - 1 - Question 27

In a uniform circular motion, which of the following quantity is not constant

Detailed Solution for Test: Motion in a Plane - 1 - Question 27

Uniform Circular Motion:

  • In uniform circular motion, an object moves along a circular path with a constant speed.
  • The direction of motion (and hence the velocity vector) changes continuously due to centripetal acceleration, even though the magnitude of speed remains constant.
  1. Angular Momentum:
    • Angular momentum (L) is given by: L = m ⋅ v ⋅ r
      where m is the mass,v is the speed, and r is the radius of the circular path.
    • Both v and r remain constant, so angular momentum is constant.
  2. Speed:
    • By definition of uniform circular motion, speed remains constant.
  3. Kinetic Energy:
    • Kinetic energy (KE) is given by:

      Since m and v are constant, KE is also constant.
  4. Momentum:
    • Momentum  is a vector quantity given by:

      While the magnitude of momentum (related to speed) remains constant, the direction of  changes continuously as the object moves along the circular path.
    • Therefore, momentum is not constant in uniform circular motion.

Final Answer: The correct answer is: Momentum

Test: Motion in a Plane - 1 - Question 28

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 Figure. What is the magnitude of the displacement vector for each ? For which girl is this equal to the actual length of path skate ?

Detailed Solution for Test: Motion in a Plane - 1 - Question 28

Explanation:

(i) PQ = diameter = displacement for each girl = 2r = 2 x 200 = 400 m

Since, displacement vector does not depend upon the actual path length and it is the shortest distance between initial and final position, so in the case of each girl the displacement is 400 m.

(ii)This is equal to the actual length of the path skated by girl B.

Test: Motion in a Plane - 1 - Question 29

A particle is moving with veocity ; where k is constant. The general equation for the path is:

Detailed Solution for Test: Motion in a Plane - 1 - Question 29

 From the given velocity vector:
vx = ky(velocity in the x-direction),
vy = kx(velocity in the y-direction).

 From the definitions of vx and vy, we have:

 Divide the two equations to eliminate t:

 Rearrange the equation:
ydy = xdx.
Integrate both sides:

where C is the constant of integration.
y2 = x2 + constant.

 The general equation of the path is: y2 = x2 + constant.
This matches Option b.

Test: Motion in a Plane - 1 - Question 30

A particle is projected with a velocity u making an angle θ with the horizontal. At any instant, its velocity v is at right angles to its initial velocity u; then v is:

Detailed Solution for Test: Motion in a Plane - 1 - Question 30

Key Concepts:

  1. Initial Velocity Components: The particle's initial velocity u has two components:
    • Horizontal velocity: ux ​= u cos θ,
    • Vertical velocity: uy ​= u sin θ.
  2. Velocity at Any Instant: At any time t, the horizontal and vertical velocities of the particle are:
    • Horizontal: vx = u cos θ (remains constant),
    • Vertical: vy = u sin θ−gt, where g is the acceleration due to gravity.
  3. Condition for Perpendicularity: At the instant when the velocity vector  becomes perpendicular to the initial velocity  their dot product is zero:

    The dot product is:

    Substituting the components:
  4. Solve for t: Expand the equation:

    Use the trigonometric identity  ⁡

    Solve for t:
  5. Vertical Velocity at This Instant: At time , the vertical velocity is:

    Substitute

    Simplify:
  6. Horizontal Velocity at This Instant: The horizontal velocity remains constant:
    vx ​= u cos θ.
  7. Resultant Velocity v: Since vy = 0, the resultant velocity is equal to the horizontal velocity:
    v = vx ​= u cosθ.

 At the point where the particle's velocity is perpendicular to its initial velocity, the magnitude of the velocity is: v = u cot θ, which matches Option c.

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