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Thursday, October 14, 2021

Plus One Physics Chapter 1 Physical World Question and Answers PDF Download

Plus One Physics Chapter 1 Physical World Question and Answers PDF Download: Students of Standard 11 can now download Plus One Physics Chapter 1 Physical World question and answers pdf from the links provided below in this article. Plus One Physics Chapter 1 Physical World Question and Answer pdf will help the students prepare thoroughly for the upcoming Plus One Physics Chapter 1 Physical World exams.


Plus One Physics Chapter 1 Physical World Question and Answers

Plus One Physics Chapter 1 Physical World question and answers consists of questions asked in the previous exams along with the solutions for each question. To help them get a grasp of chapters, frequent practice is vital. Practising these questions and answers regularly will help the reading and writing skills of students. Moreover, they will get an idea on how to answer the questions during examinations. So, let them solve Plus One Physics Chapter 1 Physical World questions and answers to help them secure good marks in class tests and exams.


Board

Kerala Board

Study Materials

Question and Answers

For Year

2021

Class

11

Subject

Hindi

Chapters

Physics Chapter 1 Physical World

Format

PDF

Provider

Spandanam Blog


How to check Plus One Physics Chapter 1 Physical World Question and Answers?

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  2. Click on the 'Plus One Question and Answers'.
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Plus One Physics Chapter 1 Physical World Question and Answers PDF Download

We have provided below the question and answers of Plus One Physics Chapter 1 Physical World study material which can be downloaded by you for free. These Plus One Physics Chapter 1 Physical World Question and answers will contain important questions and answers and have been designed based on the latest Plus One Physics Chapter 1 Physical World, books and syllabus. You can click on the links below to download the Plus One Physics Chapter 1 Physical World Question and Answers PDF. 

Question 1.
The word ‘Science’ originated from a Latin verb. Which is that verb?
Answer:
‘Scientia’ means ‘to know’.

Question 2.
The word physics comes from a Greek word______.
Answer:
‘Fusis’ means ‘nature’.

Question 3.
Name the branch of science that deals with

  1. Study of stars
  2. Study of earth

Answer:

  1. Astronomy
  2. Geology

Plus One Physics Physical World Three Mark Questions and Answers

Question 1.
Fill in the blanks

Answer:


Plus One Physics All Chapters Question and Answers


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The Plus One Physics Chapter 1 Physical World Question and Answers PDF that has been provided above is extremely helpful for all students because of the way it has been drafted. It is designed by teachers who have over 10 years of experience in the field of education. These teachers use the help of all the past years’ question papers to create the perfect Plus One Physics Chapter 1 Physical World Question and Answers PDF.


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Where can I download Plus One Physics Chapter 1 Physical World Question and Answers PDF?

You can download Plus One Physics Chapter 1 Physical World Question and Answers PDF for the latest 2021 session.

Can I download Plus One All subjects Question and Answers PDF?

Yes - You can click on the links above and download subject wise question papers in PDF

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Plus One Physics Chapter 5 Law of Motion Question and Answers PDF Download

Plus One Physics Chapter 5 Law of Motion Question and Answers PDF Download: Students of Standard 11 can now download Plus One Physics Chapter 5 Law of Motion question and answers pdf from the links provided below in this article. Plus One Physics Chapter 5 Law of Motion Question and Answer pdf will help the students prepare thoroughly for the upcoming Plus One Physics Chapter 5 Law of Motion exams.


Plus One Physics Chapter 5 Law of Motion Question and Answers

Plus One Physics Chapter 5 Law of Motion question and answers consists of questions asked in the previous exams along with the solutions for each question. To help them get a grasp of chapters, frequent practice is vital. Practising these questions and answers regularly will help the reading and writing skills of students. Moreover, they will get an idea on how to answer the questions during examinations. So, let them solve Plus One Physics Chapter 5 Law of Motion questions and answers to help them secure good marks in class tests and exams.


Board

Kerala Board

Study Materials

Question and Answers

For Year

2021

Class

11

Subject

Hindi

Chapters

Physics Chapter 5 Law of Motion

Format

PDF

Provider

Spandanam Blog


How to check Plus One Physics Chapter 5 Law of Motion Question and Answers?

  1. Visit our website - https://spandanamblog.com
  2. Click on the 'Plus One Question and Answers'.
  3. Look for your 'Plus One Physics Chapter 5 Law of Motion Question and Answers'.
  4. Now download or read the 'Class 11 Physics Chapter 5 Law of Motion Question and Answers'.

Plus One Physics Chapter 5 Law of Motion Question and Answers PDF Download

We have provided below the question and answers of Plus One Physics Chapter 5 Law of Motion study material which can be downloaded by you for free. These Plus One Physics Chapter 5 Law of Motion Question and answers will contain important questions and answers and have been designed based on the latest Plus One Physics Chapter 5 Law of Motion, books and syllabus. You can click on the links below to download the Plus One Physics Chapter 5 Law of Motion Question and Answers PDF. 

Question 1.
Which one of the following is not a force?
(a) Impulse
(b) Tension
(c) Thrust
(d) Weight
Answer:
(a) Impulse
Tension, thrust, weight are all common forces in mechanics whereas impulse is not a force.
Impulse = Force × Time duration.

Question 2.
A passenger getting down from a moving bus, falls in the direction of the motion of the bus. This is an example for
(a) Inertia of motion
(b) Second law of motion
(c) Third law of motion
(d) Inertia of rest
Answer:
(a) Inertia of motion
A passenger getting down from a moving bus, falls in the direction of the motion of the bus. This is because his feet come to rest on touching the ground and the remaining body continues to move due to inertia of motion.

Question 3.
Which one of the following is not a contact force?
(a) Viscous force
(b) Magnetic force
(c) Friction
(d) Buoyant force
Answer:
(b) Magnetic force

Question 4.
A jet engine works on the principle of
(a) Conservation of linear momentum
(b) Conservation of mass
(c) Conservation of energy
(d) Conservation of angular momentum
Answer:
(a) Conservation of linear momentum
A jet engine works on the principle of linear momentum.

Question 5.
Newton’s second and third laws of motion lead to the conservation of
(a) linear momentum
(b) angular momentum
(c) potential energy
(d) kinetic energy
Answer:
(a) linear momentum
Newton’s second and third laws lead to the conservation of linear momentum.

Question 6.
A large force is acting on a body for a short time. The impulse imparted is equal to the change in
(a) acceleration
(b) momentum
(c) energy
(d) velocity
Answer:
(b) momentum
If a large force F acts for a short time dt, the impulse imparted is
I = F.dt, = \(\frac{d p}{d t}\).dt
I = dp = change in momentum.

Question 7.
When a shell explodes, the fragments fly apart though no external force is acting on it. Does this violate Newton’s first law of motion?
Answer:
No. The explosion takes place due to internal force. The internal force does not change the position of centre of mass.

Question 8.
In taking a catch, a cricket player moves his hands backward on holding the ball. Why?
Answer:
We know F = \(\frac{\Delta P}{\Delta t}\)
When ∆t increases, the force acting on hand decreases.

Question 9.
Name the factor on which inertia depends.
Answer:
Mass

Question 10.
Why does a swimmer push the water backwards?
Answer:
A swimmer pushes the water backward in order to be pushed forward (Newton’s third law).

Question 11.
Rocket works on the principle of conservation of_______.
Answer:
Momentum

Question 12.
A man experience a backward jerk, while firing bullet from gun. Which law is applicable here? Answer:
Conservation of momentum.

Question 13.
If you jerk a piece of paper under a book quick enough, the book will not move. Why?
Answer:
This is due to inertia of rest.

Question 14.
Why it is difficult to walk on a slipper road?
Answer:
We will not get required reaction from slippery road.

Question 15.
A stone, when thrown on a glass window, smashes the window pan to pieces. But a bullet fired from the gun passes through it making a hole why?
Answer:
This is due to inertia of rest of glass window.

Question 16.
Why an athlete runs some distance before taking a jump?
Answer:
An athletic runs some distance before taking a jump to gain some initial momentum. It helps the athlete to jump more.

Question 17.
Why a horse can not pull a cart and run in empty space?
Answer:
The horse-cart system moves forward due to reaction of ground on the feet of horse. In free space, there is no reaction. So it can not pull cart.

Question 18.
Why parachute descends slowly?
Answer:
Parachute has large surface area. This increases fluid friction and slows down the motion of parachute.

Question 19.
Sand is thrown on tracks with snow. Why?
Answer:
The presence of snow on tracks reduces friction and driving is not safe. If sand is thrown, friction will be increased and driving becomes safe.

Question 20.
It is difficult to move a cycle along a road with its brakes on. Explain.
Answer:
When the cycle is moved with its brakes on, wheels can only skid. There will be sliding friction. The sliding friction is more compared to rolling friction. Hence it is difficult to move a cycle with its brakes on.

Plus One Physics Law of Motion Two Mark Questions and Answers

Question 1.
Two masses are in the ratio 1:5

  1. What is inertia?
  2. What is the ratio of inertia of the above case?

Answer:

  1. The inability of a body to change it’s state of rest or uniform motion is called inertia.
  2. Mass is a measure of inertia. Hence ratio of inertia is 1:5.

Question 2.
More force is required to push a body than pull to get same speed on a ground with some friction. Why?
Answer:
When we push, the action on the surface and normal reaction on the body increases. (Friction is directly proportional to normal reaction).

As a result more force is required to push the body. When we pull, normal reaction decreases. Hence friction decreases. Hence less force is required to pull the body.

Question 3.
A lift in a multistoried building is moving from ground floor to third floor. What will happen to weight of a person sitting in side of the lift.

  1. A When starts to move up from ground floor.
  2. When the lift moves with constant speed.

Answer:

  1. A weight increases weight w = mg + ma
  2. weight is constant ie. w = mg

Question 4.
Why it is advisable to hold a gun tight to one’s shoulder when it is being fired?
Answer:
The recoiling gun can hurt the shoulder. If gun is held tightly against the shoulder, the body and gun act a system. This will reduce recoil velocity as it is inversly proportional to mass of system.

Question 5.
Why shockers are used in vehicles?
Answer:
When there is a jerk or jump, the time for which force acts (∆t) increases. As the product of force and time for which force acts (F∆t) remains constant, increase in At will reduce the force. This provide smooth motion.

Plus One Physics Law of Motion Three Mark Questions and Answers

Question 1.
Give the magnitude and direction of net force on

  1. a drop of rain falling down with a constant velocity.
  2. a stone of mass 0.1 kg just after it dropped from the window of a tram accelerating at 1 ms-2.

Answer:
1. Net force is zero

2. When stone is dropped, gravitational force will act on the stone.
Gravitational force F = mg
= 0.1 × 10
= 1 N downward.

Question 2.
An external force is always required to break the inertia of a body which is either in the state of rest or state of uniform motion.

  1. Which law governs this statement?
  2. Can all forces produce acceleration? Why?
  3. A boy holding a spring balance in his hand suspend a mass 2kg from it. If the balance slips from his hand and falls down, find the reading of the balance while it is in the air.

Answer:

  1. Newtons first law of motion.
  2. No. If resultant force acting on the body is zero, the body will move with constant velocity or remain at rest.
  3. Zero

Question 3.
A man weighs 70 kg. He stands on a weighing scale in a lift which is moving.

  1. upward with a uniform speed of 10 m/s.
  2. downward with an uniform acceleration of 5 m/s2.
  3. upward with an uniform accelerate of 5 m/s2. (Take g = 10m/s2). Find weight in each case.

Answer:
1. Weight W = mg
= 70 × 10 = 700 N.

2. W = mg – ma
= 70 × 10 – 70 × 5
= 700 – 350
= 350 N

3. W = mg + ma
= 70 × 10 + 70 × 5
= 700 + 350
= 1050N.

Question 4.
A body of mass ‘m’ is placed on a rough inclined plane having coefficient of friction µs. The inclination of plane is given as ‘θ’.

  1. Which component of weight brings the body towards the bottom along the plane.
  2. Find how much force is required to pull the body along the plane.

Answer:

  1. mg Sinθ brings the body downwards
  2. When the body moves upwards the frictional force (Fs) acts downwards
    Total pulling Force = mg Sinθ + Frictional force (Fs) (u, mgCosθ).

Question 5.
Four person sitting in the back seat of a car at rest, is pushing on the front seat.

  1. Does the car move. Why?
  2. State the law which help you to answer above question.
  3. Long jumpers take a long run before the jump. Why?

Answer:

  1. No. Action and reaction cancel each other.
  2. Newtons third law of motion.
  3. To get large inertia of motion.

Question 6.
A Cricket player lowers his hands while catching a Cricket ball to avoid injury.

  1. What do you mean by impulsive force?
  2. Prove impulse-momentum theorem.

Answer:
1. The forces which acton bodies for short time are called impulsive forces.
Example:

  • In hitting a ball with a bat
  • In firing a gun

2. F = \(\frac{d p}{d t}\)
F∆t = dp
impulse = change in momentum.

Plus One Physics Law of Motion Four Mark Questions and Answers

Question 1.
A bead sliding on a wire A moves to C through B as shown in the figure. The bead at A has a speed of200cms

  1. what is speed at B?
  2. To what height will it rise before it returns?
  3. Why the ball moves up even after reaching the bottom-most point B?

Answer:
1. mgh = 1/2 mv2
m × 10 × 0.8 = 1/2 mv2
V2 = 2 × 10 × 0.8
V = \(\sqrt{2 \times 10 \times 0.8}\)
V = 4 m/s.

2. 80 cm (if friction is neglected).

3. when the ball reaches at B, the potential energy is converted into kinetic energy. Due to this kinetic energy the ball raises to the point c.

Question 2.
Figure shows a block (mass m1) on a smooth horizontal surface, connected by a thin cord that passes over a pulley to a second block (m2), which hangs vertically.

  1. Obtain formula for the acceleration of the system and tension in the cord.
  2. If m1 and m2 interchanges its position, will it affect the tension of the string?
  3. What is the acceleration of the system if m1 = 5 kg and m2 = 2kg?

Answer:
1. When the body m2 moves in down ward direction.
m2g – T = m1 a
T = m2g – m1a.

2. New tension can be found from the relation
m1g – T = m2a
T = m1g – m2 a.

3. Acceleration of system, a

Question 3.
The collision of two ice hockey players are shown in figure.

Analyse the data given in the figure and answer the following questions.

  1. Which conservation law is applicable in this case.
  2. In which direction and at what speed do they travel after they stick together.
    [Hint – towards right can be taken us +ve direc¬tion and vice versa]
  3. If we assume the friction of playing ground is zero, predict the nature of motion and the point at which they come to rest.

Answer:
1. Conservation of linear momentum.

2. Total momentum before collision = Total momentum after collision.
110 × 4 + 90 × -6 = (110 + 90)v
v = 0.5 m/s
-ve direction, (in the direction of man mass 90 kg).

3. Uniform motion They will not stop.

Question 4.
A circular track of radius 300m is kept with outside of track raised to make 5 degree with the horizontal.

  1. Name the process in which outside of the road is raised little above the inner.
  2. Obtain an expression for the optimum speed to avoid skidding (considering to friction)

Answer:
1. Banking of roqd

2.


Consider a vehicle along a curved road with angle of banking θ. Then the normal reaction on the ground will be inclined at an angle θ with the vertical.
The vertical component can be divided into N Cosθ (vertical component) and N sinθ (horizontal component). The frictional force can be divided into two components. Fcosθ (horizontal component) and F sinθ (vertical component).
From the figure
N cos θ = F sinθ + mg
N cosθ – F sinθ = mg ______(1)
The component Nsin0 and Fsinθ provide centripetal force. Hence
N sinθ + F cos θ = \(\frac{\mathrm{mv}^{2}}{\mathrm{R}}\) ______(2)
eq (1) by eq (2)

Dividing both numerator and denominator of L.H.S by N cosθ. We get

This is the maximum speed at which vehicle can move over a banked curved road.
Optimum speed:
Optimum speed is the speed at which a vehicle can move over a curved banked road without using unnecessary friction. Putting µ = 0 in the above equation we get
v0 = \(\sqrt{\mathrm{Rg} \tan \theta}\).

Question 5.
A circular track of radius 400m is kept with outer side of track raised to make 5° with the horizontal (coefficient of friction 0.2)
(a) Name such track?
(b) What is optimum speed to avoid wear and tear of type?
(c) What is the maximum permissible speed to avoid skidding?
Answer:
(a) Banking.

Plus One Physics Law of Motion Five Mark Questions and Answers

Question 1.
A horse pulls a cart with a constant force so that the cart moves at a constant speed.

  1. Does it violate Newton’s second law of motion?
  2. If not, how will you account for the non-acceleration of the cart?
  3. Will the speed of the cart increase, decrease, or remain the same if the horse applied more force?
  4. A body of mass 5kg is acted upon by two perpendicular forces 8N and 6N. Give the magnitude and direction of the acceleration of the body.

Answer:
1. No.

2. The force applied by the car is balanced by the frictional force. Hence the cart moves with constant velocity.

3. If the horse is applied more force, the speed of the cart increases.

4. The resultant force,

F = 10N
We know, F = ma
10 = 5 × a
acceleration, a = \(\frac{5}{10}\) = 2 m / sec2

The angle of resultant force,

θ = tan-1 6/8
θ = 36°521
The angle of acceleration θ = 36°521.

Question 2.

  1. Friction is the force which opposes the relative motion between two surfaces in contact with each other. What is limiting static friction? State the laws related to this.
  2. Show that the coefficient of friction is equal to the tan of the angle between the resultant and normal reactions.
  3. For a body of mass 5kg on a plane at limiting static friction of 30 degrees. What is the force of friction?

Answer:
1. The maximum value of static friction is called limiting static friction.

  • The magnitude of the limiting friction is independent of the area of contact between the surfaces.
  • The limiting static friction is directly proportional to the normal reaction R.

ie f α R
fs = µsR.

2. Angle of friction is the angle whose tangent gives the coefficient of friction.

Consider a body placed on a surface. Let N be the normal reaction and limit is the limiting friction. Let ‘θ’ be the angle between Resultant vector and normal reaction. From the triangle OBC,

∴ tanθ = µ.

3. Tangent of the angle cient of friction.
µs = tanθ
µs = tan 30
µs = \(1 / \sqrt{3}\)
Friction F = µsmg
= \(1 / \sqrt{3}\) × 5 × 10
F = \(\frac{50}{\sqrt{3}}\)N.

Question 3.
The rate of change of linear momentum of a body is directly proportional to the external force applied on it, and takes place always in the direction of force applied.

  1. Name this law.
  2. Using this law obtain the expression for force.
  3. The motion of a particle of mass m is described by y = ut + \(\frac{1}{2}\) gt2. Find the force acting on it.

Answer:
1. Newton’s Second Law.

2. Consider a body of mass ‘m’ moving with a momentum \(\vec{p}\). Let \(\vec{F}\) be the force acting on it for time internal ∆t. Due to this force the momentum is changed from \(\vec{p}\) to p + ∆p. Then according to Newtons second law, we can write

Where K is a constant pf proportionality. When we take the limit ∆t → 0, we can write

3.

Hence force F = mg.

Question 4.
Recoil of gun is based on the principle of conservation of momentum.

  1. State the principle of conservation of momentum.
  2. Explain the reoil velocity of gun.
  3. A bullet of mas 100g is fired from a rife of mass 200 kg with a spped of 50 m/s. Calculate the recoil velocity of the rife.

Answer:
1. According to law of conservation of linear momentum, if the external force acting on a body is zero, total linear momentum remains constant. According to Newton’s second law.
F = \(\frac{d p}{d t}\)
If F = 0, \(\frac{d p}{d t}\) = 0 i.e; P is constant.

2. Let M, m be the mass of gun and bullet respectively. Let V and ν be the velocities of gun and bullet after firing.
According to consevation of momentum
Total momentum before firing = Total momentum after firing
∴ O = MV + m ν
-MV = mν
The above equation shows that when bullet moves in forward direction, the gun moves in back direction. This motion of gun is called recoil of gun.

3. M = 200kg, m = 100g = 0.1kg
ν = 50 m/s, V = ?
MV = mν
200 × V = 0.1 × 50
V = \(\frac{0.1 \times 50}{200}\)m/s.

Question 5.
While firing a bullet, the gun must be held tight to the shoulder.

  1. Which conservation law helps you to explain this
  2. “In the firing process, the speed of the gun is very low compared to the speed of the bullet.” Substantiate the above statement using mathematical expressions.
  3. A shell of 20kg moving at 50m/s bursts in to two parts of masses 15kg and 5kg. If the larger part continues to move in the same direction at 70 m/s. What is the velocity and direction of motion of the other piece.

Answer:
1. Conservation of momentum.

2. Total momentum is conserved
∴ mu + MV = 0
V = \(\frac{-m u}{M}\) M is very large. Hence v is small

3. MV = m1 u1 + m2 u2
20 × 50 = 5u1 + 15 × 70
5u1 = 50
u1 = 10m/s.

Question 6.
While firing a bullet, the gun must be held tight to the shoulder.

  1. This is a consequence of______
  2. Show that recoil velocity is opposite to the muzzle velocity of the bullet.
  3. A gun of mass 5 kg fire a bullet of mass 5g, vertically upwards to a height of 100m. Calculate the recoil velocity of gun.

Answer:
1. Conservation of linear momentum.

2. Let M be the mass of gun and m be the mass of bullet. When gun fires, the gun and bullet acquire velocities V and v respectively.
According to conservation of momentum.
Total momentum before firing = Total momentum afterfiring
m × o + M × o = mu + MV
O = mv + MV
ie. – MV = mv
V = \(\frac{-m v}{M}\)

3. M = 5kg, m = 5 × 10-3 kg, h = 100m
v2 = u2 + 2as
0 = u2 + 2 × 10 × 100
Velocity of bullet,

Question 7.
A standing passenger falls backwards when the bus starts suddenly.

  1. Explain why this happens?
  2. Which Newtons law gives the above concept. State the law.
  3. Obtain an expression for force using Newtons law.

Answer:
1. Due to inertia of rest, the body continues in the state of rest.

2. Newtons first law:
Everybody continues in its state of rest or of uniform motion along a straight line unless it is compelled by an external unbalanced force to change that state:

3. Consider a body of mass ‘m’ moving with a momentum \(\vec{p}\). Let \(\vec{F}\) be the force acting on it for time internal ∆t. Due to this force the momentum is changed from \(\vec{p}\) to p + ∆p. Then according to Newtons second law, we can write

Where K is a constant pf proportionality. When we take the limit ∆t → 0, we can write

Question 8.
According to Newton’s law of motion rate of change of momentum is directly proportional to applied force.
a. Impulse has the unit similarto that of

  1. Momentum
  2. force
  3. time
  4. Energy

b. A man falling from certain height receives more injuries when he falls on a marble floor than when he falls on a heap of sand. Explain. Why?
c.

Force – time graph for a body starting from rest is shown in the figure. What is the velocity of the body at the end of 12 second? (Mass of the body is 5 kg)
Answer:
a. 1. Momentum.

b. When a man falls on a marble floor, the momentum is reduced to zero in lesser time. Due to this, the rate of change of momentum is large. So greater force acts on a man falls on marble floor.

c. The area of force – time graph gives change in momentum.
ie. change in momentum,
mv = 1/2 × (12 – 4) × (20 -10)
mv = 40

Plus One Physics Law of Motion NCERT Questions and Answers

Question 1.
Give the magnitude and direction of the net force acting on
(a) a drop of rain falling down with a constant speed.
(b) a cork of mass 10g floating on water
(c) a kite skillfully held stationary in the sky
(d) a car moving with a constant velocity of 30km h-1 on a rough road
(e) a high – speed electron in space far from all material objects, and free of electric and magnetic fields.
Answer:
Applying Newton’s first law of motion, we find that no net force acts in any of the situations, (a) to (d). Again, no force in situation (e). This is because electron is far away from all material agencies producing electromagnetic and gravitational forces.

Question 2.
A constant retarding force of 50 N is applied to a body of mass 20kg moving initially with a speed of 15ms-1. How long does the body take to stop?
Answer:
Acceleration, a = –\(\frac{50 \mathrm{N}}{20 \mathrm{kg}}\) = -2.5ms-2
[Negative sign indicates retardation]
u = 15ms-1, v = 0, t = ?
v = u + at
0 = 15 – 2.5t or 2.5t = 15 or
t = \(\frac{15}{2.5}\)s = 6.0s.

Question 3.
A constant force acting on a body of mass 3.0kg changes its speed from 2.0ms-1 to 3.5 ms-1 in 25s. The direction of motion of the body remains unchanged. What is the magnitude and direction of the force?
Answer:
m = 3kg; u = 2ms-1; v = 3.5 ms-1;
t = 25s ; F = ?
v = u + at
3.5 = 2 + 25a or a = 0.06 ms-2
F = ma = 3kg × 0.06 ms-2 = 0.18N.
The direction of force is along the direction of motion.

Question 4.
A bob of mass 0.1 kg hung from the ceiling of a room by a string 2 m long is set into oscillation. The speed of the bob at its mean position is 1 ms-1. What is the trajectory of the bob if the string is cut when the bob is

  1. at one of its extreme positions?
  2. at its mean position.

Answer:

  1. At the extreme position, the speed of the bob is zero. If the string is cut, it will fall vertically downwards.
  2. At the mean position, the bob has a horizontal velocity. If the string is cut, it will fall along a parabolic path.

Question 5.
A man of mass 70kg stands on a weighing scale in a lift which is moving

  1. upwards with a uniform speed of 10ms-1
  2. downwards with a uniform acceleration of 5ms-2
  3. upwards with a uniform acceleration of 5ms-2 What would be the readings on the scale in each case?
  4. What would be the reading if the lift mechanism failed and it hurtled down freely under gravity?

Answer:

  1. a = 0, R = mg = 70 × 10 N = 700N
  2. mg – R = , ma ; R – mg – ma = (g – a)
    = 70(10 – 5) N = 350N
  3. R – mg = ma or R = m(g + a)
    = 70(10 + 5)N = 1056 N
  4. In the event of free fall, it is a condition of weightlessness.

Question 6.
A nucleus is at rest in the laboratory frame of reference. Show that if it dist integrates into two smaller nuclei, the products must move in opposite directions.
Answer:
Applying principle of conservation of momentum,

The negative sign indicates that the products move in opposite directions.

Question 7.
A shell of mass 0.020 kg is fired by a gun of mass 100kg. If the muzzle speed of the shell is 80ms-1, what is the recoil speed of the gun?
Answer:
m = 0.02kg, M = 100kg, v = 80ms-1, V = ?

= -0.016ms-1 = -1.6cm s-1
Negative sign indicates that gun moves in a direction opposite to the direction of motion of the bullet.


Plus One Physics All Chapters Question and Answers


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Plus One Physics Chapter 4 Motion in a Plane Question and Answers PDF Download

Plus One Physics Chapter 4 Motion in a Plane Question and Answers PDF Download: Students of Standard 11 can now download Plus One Physics Chapter 4 Motion in a Plane question and answers pdf from the links provided below in this article. Plus One Physics Chapter 4 Motion in a Plane Question and Answer pdf will help the students prepare thoroughly for the upcoming Plus One Physics Chapter 4 Motion in a Plane exams.


Plus One Physics Chapter 4 Motion in a Plane Question and Answers

Plus One Physics Chapter 4 Motion in a Plane question and answers consists of questions asked in the previous exams along with the solutions for each question. To help them get a grasp of chapters, frequent practice is vital. Practising these questions and answers regularly will help the reading and writing skills of students. Moreover, they will get an idea on how to answer the questions during examinations. So, let them solve Plus One Physics Chapter 4 Motion in a Plane questions and answers to help them secure good marks in class tests and exams.


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Question 1.
A cricketer can throw a ball to a maximum horizontal distance of 100m. With the same speed how much high above the ground can the cricketer throw the same ball?
(a) 200m
(b) 150m
(c) 100m
(d) 50m
Answer:
(d) 50m
Rmax = \(\frac{u^{2}}{g}\)
100 = \(\frac{u^{2}}{g}\) or u2 = 100g
Using, v2 = u2 + 2as
0 = (100g) + 2(-g)h or h = 50m.

Question 2.
If vectors \(\hat{\mathrm{i}}-3 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}\) and \(\hat{\mathrm{i}}-3 \hat{\mathrm{j}}-\mathrm{ak}\) are equal vectors, then the value of a is
(a) 5
(b) 2
(c) -3
(d) -5
Answer:
(d) Comparing vector, we get = \(+5 \hat{k}=-a \hat{k}\)
a = -5.

Question 3.
State which of the following algebraic operations are not meaningful
(a) Addition of a scalar to a vector
(b) Multiplication of any two scalars.
(c) Multiplication of vector by scalar
(d) Division of a vector by scalar Addition of a scalar to a vector
Answer:
(a) Addition of a scalar to a vector

Question 4.
What is the acceleration of train travelling at 40ms-1 as it goes round a curve of 160m radius?
Answer:
a = \(\frac{v^{2}}{r}=\frac{40 \times 40}{160}\) = ms-2.

Question 5.
What provides centripetal force in the following cases.

  1. Electron revolving around nucleus.
  2. Earth revolving around sun

Answer:

  1. Electrostatic force
  2. Gravitational force

Question 6.
Why a cyclist has to bend inwards while going on a circular track?
Answer:
The cyclist bends inwards to provide required centripetal force.

Question 7.
A body executing uniform circular motion has constant
(i) velocity
(ii) acceleration
(iii) speed
(iv) angular velocity
Answer:
(iii) speed

Question 8.
Name a quantity which remains unchanged during projectile motion.
Answer:
Horizontal component.

Question 9.
What is the effect of air resistance in time of flight and horizontal range?
Answer:
The effect of air resistance is to increase time of flight and decrease horizontal range.

Question 10.
What is the angle between directions of velocity and acceleration at the highest point of trajectory of projectile?
Answer:
At the highest point velocity is horizontal and acceleration is vertical. So angle is 90°.

Question 11.
Can a body have constant velocity and still have a varying speed?
Answer:
No.
If velocity is constant, speed also will be constant.

Question 12.
Can a body have zero velocity, still accelerating?
Answer:
Yes.
When a body is at highest point of motion, its velocity is zero but acceleration is equal to acceleration due to gravity.

Question 13.
A quantity has both magnitude and direction. Is it necessarily a vector? Give an example.
Answer:
No. The given quantity will be a vector only if it obeys laws of vector addition.
Example: Current.

Question 14.
What is the angle between \(\vec{A} \times \vec{B}\) and \(\vec{B} \times \vec{A}\) ?
Answer:
These two vectors will be antiparallel. Hence θ = 180°.

Plus One Physics Motion in a Plane Two Mark Questions and Answers

Question 1.
A particle is projected with a velocity u so that its horizontal range is twice the greatest hieght attained. The horizontal range is
(a) \(\frac{u^{2}}{g}\)
(b) \(\frac{2 u^{2}}{g}\)
(c) \(\frac{4 u^{2}}{5g}\)
(d) None of these
Answer:
(c) Horizontal range, R = \(\frac{\mathrm{u}^{2} \sin 2 \theta}{\mathrm{g}}\);
Maximum height, H = \(\frac{u^{2} \sin ^{2} \theta}{2 g}\)
As per question, R = 2H

or sin2θ = sin2θ
or 2sinθcosθ = sin2θ or tanθ = 2
Hence, sinθ = \(\left(\frac{2}{\sqrt{5}}\right)\) and cosθ = \(\left(\frac{1}{\sqrt{5}}\right)\)
Horizontal range, R = \(\frac{2 u^{2} \sin \theta \cos \theta}{g}\)

Question 2.

Answer:

Question 3.
Choose the correct alternative given below. A Particle executing uniform circular motion. Then its
(a) Velocity and acceleration are radial.
(b) Velocity and acceleration are tangential.
(c) Velocity is tangential, acceleration is radial.
(d) Velocity is radial, acceleration is tangential.
(e) In a circus, a rider rides in a circular track of radius Y in a vertical plane. The minimum velocity at the highest point of the track will be

  • \(\sqrt{2 g r}\)
  • \(\sqrt{ g r}\)
  • \(\sqrt{3 g r}\)
  • 0

Answer:
(c) Velocity is tangential, acceleration is radial.
(e) \(\sqrt{ g r}\)

Question 4.
Two non-zero vectors \(\bar{A}\) and \(\bar{B}\) are such that \(|\bar{A}+\bar{B}|=|\bar{A}-\bar{B}|\). Find the angle between them.
Answer:
\(|\bar{A}+\bar{B}|=|\bar{A}-\bar{B}|\)
|A2 + B2 + 2ABcosθ| = |A2 + B2 – 2ABcosθ|
|4ABcosθ| = 0
Since A and B are non-zero we get, cos θ = 0 or θ = 90°.

Question 5.
Consider a particle moving along the circumference of a circle of radius R with constant speed with a time period T.

  1. During T, what is the distance coverd and displacement?
  2. What is the direction of the velocity at each point?

Answer:

  1. Distance = 2πR. Displacement = 0
  2. Tangent to the circle at every point.

Question 6.
Classify into scalars and vectors. Frequency, velocity gradient, instantaneous velocity, Area.
Answer:

Scalars Vectors Frequency Instantaneous velocity Velocity gradient Ares

Question 7.
A body is projected so that it has maximum range R. What is the maximum height reached during the fight?
Answer:
At maximum range, θ = 45°

Plus One Physics Motion in a Plane Three Mark Questions and Answers

Question 1.
An electron of mass ‘m’ moves with a uniform speed v around the nucleus along a circular radius Y.

  1. Derive an expression for the acceleration of the electron.
  2. Explain why the speed of electron does not increase even though it is accelerated by the above acceleration.

Answer:
1.


The acceleration is directed towards the centre of the circle and is called centripetal acceleration.
a0 = \(\frac{v^{2}}{R}\)
But V = Rω
Substituting we get
ac = Rω2

2. The direction of centripetal force is towards the centre. The angle between force and displacement is 90°. Hence the work done by the centripetal force is zero. So speed does not increase.

Question 2.
A boy pulls his friend in a home made trolley by means of a rope inclined at 30° to the horizontal. If the tension in the rope is 400N.

  1. Draw the vertical and horizontal components of tension in the rope.
  2. Find the effective force pulling the trolley along the ground.
  3. Find the force tending to lift the trolley off the ground.

Answer:
1.

2. Effective horizontal force = T Cos30°
= 400 × Cos 30°

3. Vertical force = T Sin 30°
= 400 × Sin 30°

Question 3.
A stone tied to the end of a string is whirled in a horizontal circle with constant speed.

  1. Name the acceleration experienced by the stone.
  2. Arrive at an equation for magnitude of acceleration experienced by the stone.

Answer:
1. Centripetal acceleration

2.


The acceleration is directed towards the centre of the circle and is called centripetal acceleration.
a0 = \(\frac{v^{2}}{R}\)
But V = Rω
Substituting we get
ac = Rω2

Plus One Physics Motion in a Plane Four Mark Questions and Answers

Question 1.
Two balls are released simultaneously from a certain height, one is allowed to fall freely and other thrown with some horizontal velocity.

  1. Will they hit the ground together?
  2. At any time during the fall will the velocities of the balls are same?
  3. How does the path of the balls appear to a person standing on the ground?

Answer:

  1. Both balls will reach at same time.
  2. Total velocity of first body and second body is different. First ball has only downward velocity but second ball has both downward and horizontal velocity.
  3. The path of first ball appears to be straight line and that of second ball appears to be parabola.

Question 2.
A ball is thrown straight up.

  1. Obtain a mathematical expression for the height to which it travels.
  2. What js its velocity and acceleration at the top?
  3. Draw the velocity-time graph for the ball showing its motion up and down.

Answer:
1. u = u, v = o, a = -g, h = ?
We can find maximum height using the equation
v2 = u2 + 2as
0 = u2 + 2 × – g × H
2gh = u2
H = \(\frac{u^{2}}{2 g}\)

2. Velocity is zero, but it has a acceleration and its value g = 9.8 m/s2.

3.

Question 3.

  1. Parallelogram law helps to find the magnitude and direction of the resultant of two forces. State the law.
  2. For two vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) are acting at a point with an angle a between them, find the magnitude and direction of the resultant vector.
  3. What will be the angle between two vectors of equal magnitude for their resultant to have the same magnitude as one of the vectors?

Answer:
1. Law of parallelogram of vectors:
If two vectors acting simultaneously at a point are represented in magnitude and direction by the two adjacent sides of a parallelogram then the diagonal of the parallelogram passing through that point represents the resultant in magnitude and direction.

2.

Consider two vectors \(\vec{A}(=\overrightarrow{O P}) \text { and } \vec{B}(=\overrightarrow{O Q})\) making an angle θ. Using the parallelogram method of vectors, the resultant vector R can be written as,

SN is normal to OP and PM is normal to OS. From the geometry of the figure
OS2 = ON2 + SN2
but ON = OP + PN
ie. OS2 = (OP+PN)2 + SN2 ______(1)
From the triangle SPN, we get
PN = Bcosθ and SN = Bsinθ
Substituting these values in eq.(1), we get
OS2 = (OP + Bcosθ)2 + (Bsinθ)2
But OS = R and OP = A
= A2 + 2ABcosθ + B2cos2θ + B2sin2θ
2 = A2 +2 ABcosθ + B2
R = \(\sqrt{A^{2}+2 A B \cos \theta+B^{2}}\)
The resultant vector \(\overrightarrow{\mathrm{R}}\) make an angle a with \(\overrightarrow{\mathrm{R}}\). From the right angled triangle OSN,

But SN = Bsinθ and PN = Bcosθ

3. 120°

Question 4.
A ball of mass m is projected at an angle with the ground and it is found that its kinetic energy at the highest point is 75% of that at the point of projection.

  1. Is it a one-dimensional or a two-dimensional motion? Why?
  2. Find the angle of projection
  3. Determine another angle of projection which produces the same range.

Answer:
1. Two-dimensional motion. The projectile has two dimensions.

2. The K.E at highest point
Ek = E cos2θ,
where E = initial K.E
0.75 E = E cos2θ
cos2θ = 0.75 = 3/4
cosθ = \(\sqrt{3} / 2\)
θ = 60°

3. When an object is projected with velocity ‘u’ making an angle θ to with the horizontal direction, the horizontal range will be
R1 = \(\frac{u^{2} \sin 2 \theta}{g}\) ____(1)
when an object is projected with velocity u making an angle (90° – θ) with the horizontal direction, then horizontal range will be

From eq (1) eq (2), we get R1 = R2, which means that we get same horizontal range for two angles θ and (90 – θ).

Question 5.
“The graphical method of adding vectors helps us in visualizing the vectors and the resultant vector. But, sometimes, it is tedious and has limited accuracy”.

  1. Name the alternative method of vector addition.
  2. Write a mathematical expression to find resultant of two vectors.
  3. A particle is moving eastward with a velocity of 5m/s. |f in 10s, the velocity changes by 5 m/s northwards, what is the average acceleration in this time.

Answer:
1. Analytical method of vector addition.

2. R = \(\sqrt{A^{2}+B^{2}+2 A B \cos \theta}\)

3. Change in velocity

Average acceleration

Question 6.

Answer:

Question 7.
A projectile is an object projected into air with a velocity V so that it is moving under the influence of gravity.

  1. What is the shape of the path of projectile?
  2. As a projectile moves in its path, is there any point along the path where the velocity and acceleration vectors are perpendicular to each other
  3. If E is energy with a projectile is projected.

(i) What is the Kinetic energy at the highest point.
(ii) What is P.E at highest point?

Answer:
1. Parabola

2. Yes, highest point

3. Energy with a projectile is projected:
(i) Kinetic energy Ek = 1/2mv2
Velocity at the highest point = Vcosθ
∴ K.E. at highest point = 1/2m(Vcosθ)2
= 1/2 mv2cos2θ
K.E. = Ek cos2θ

(ii) Potential energy at highest point,
P.E. = mgh
= mg\(\frac{v^{2} \sin ^{2} \theta}{2 g}\) = 1/2mv2sin2θ
P.E. = Ek sin2θ.

Plus One Physics Motion in a Plane Five Mark Questions and Answers

Question 1.
An object is projected with velocity U at an angle θ to the horizontal.

  1. Obtain a mathematical expression for the range in the horizontal plane.
  2. What are the conditions to obtain maximum horizontal range?
  3. Find the maximum height of the object when its path makes an angle of 30° with the horizontal (velocity of projection = 8 ms-1)

Answer:
1. If we neglect the air resistance, the horizontal velocity (ucosθ) of projectile will be a constant. Hence the horizontal distance (R) can be found as
R = horizontal velocity × time of flight

2. when θ = 45° we get maximum horizontal range

3. Height H

Question 2.
The path of projectile from A is shown in the figure. M is the mass of the particle.

When the particle moves from A to M.

  1. a) What is the change in vertical velocity?
  2. b) What is the change in speed?
  3. What is the change in linear momentum?
  4. The ceiling of a long hall is 25m high. What is the maximum horizontal distance that a ball thrown with a speed of 40 ms-1 can go without hitting the ceiling of the hall?

Answer:
1.

When the particle reaches at M, the vertical component of velocity becomes zero.
change in vertical velocity = u sinθ – 0
= u sinθ

2. change in speed = u cosθ – u

3. change in momentum = mu sinθ – 0
= mu sinθ

4.

θ = 33°481
The horizontal range of the ball is

Question 3.
A particle moving uniformly along a circle, experiences a force directed towards the centre and an equal and opposite force directed away from the centre.

  1. Name the two forces directed towards and away from the centre.
  2. Obtain an expression for the force directed towards the centre.
  3. An aircraft executes a horizontal loop at a speed of 720 km hr-1 with its wings banked at 15°. What is the radius of the loop?

Answer:
1. centripetal force and centrifugal force.

2.


force which produces this centripetal acceleration is called centripetal force.
Centripetal force can be written as F = ma.
F = m\(\frac{V^{2}}{R}\)

3. Speed of aircraft, 720 × \(\frac{5}{18}\) = 200 m/s
The velocity of aircraft, ν = \(\sqrt{r g \tan \theta}\)

Question 4.
A body is projected with a velocity ‘u’ in a direction making an angle θ with the horizontal.

  1. Derive the mathematical equation of the path followed.
  2. Draw the velocity-time graphs for the horizontal and vertical components of velocity of the projectile.
  3. Obtain an expression for the time of flight of the projectile.

Answer:
1. The vertical displacement of projectile at any time t, can be found using the formula.
S = ut + 1/2at2
y = usinθt – 1/2gt2
But we know horizontal displacement,
x = ucosθxt

In this equation g, θ and u are constants. Hence eq.(4) can be written in the form
y = ax + bx2
where a and b are constants. This is the equation of parabola, ie. the path of the projectile is a parabola.

2.

3. The time taken by the projectile to cover the horizontal range is called the time of flight.
Time of flight of projectile is decided by usinθ. The time of flight can be found using the formula s = ut + 1/2 at2
Taking vertical displacement s = 0, a = -g and initial vertical velocity = usinθ, we get
0 = usinθt – 1/2gt2
1/2 gt2 = usinθt
t = \(\frac{2 u \sin \theta}{g}\).

Question 5.
An object projected into air with a velocity is called a projectile.

  1. What will be the range when the angle of projections are zero degrees and ninety degrees?
  2. Show that fora projectile, the upward time of flight is equal to the downward time of flight.
  3. At what angles will a projectile have the same range fora velocity?

Answer:
1. When θ = 0°
R = 0
When θ = 90°

2. Upward motion
u = u sinθ, a = -g, v = 0
V = u + at
0 = u sinθ + gt
ta = \(\frac{u \sin \theta}{g}\) ____(1)
Downward motion
u = 0, V= u sinθ, a = +g
V = u + at
u sinθ = 0 + gt
ta = \(\frac{u \sin \theta}{g}\) ____(2)
eq (1) and (2), shows up ward time of flight is equal to downward time of flight.

3. θ, 90 – θ.

Question 6.
“An object that is in flight after being thrown or projected is called a projectile”.
1. Which of the following remains constant throughout the motion of the projectile?
(i) Vertical component of velocity
(ii) Horizontal component of velocity
2. Derive an expression for maximum range of a projectile.
3. Show that range of projection of a projectile for two angles of a projection a and (3 is same where α + β = 90°.

Answer:
1. (ii)

2. If we neglect the air resistance, the horizontal velocity (ucosθ) of projectile will be a constant. Hence the horizontal distance (R) can be found as
R = horizontal velocity × time of flight

3. Range of projectile R = \(\frac{\mathrm{u}^{2} \sin 2 \theta}{\mathrm{g}}\)
Case – 1
at angle α,
Rα = \(\frac{\mathrm{u}^{2} \sin 2 \alpha}{\mathrm{g}}\) ____(1)
Case – 2
at an angle β,

From (1) and (2), we get
Rα = Rβ

Question 7.
A bullet is fired with a velocity ‘u’ at an angle ‘θ’ with the horizontal such that it moves under the effect of gravity.

  1. What is the nature of its trajectory.
  2. Arrive at an expression for time of flight of the bullet.
  3. What is the relation between time of ascent and time of decent, when air resistance is neglected.
  4. How the relation is affected when air resistance is considered.

Answer:
1. Parabola

2. The time taken by the projectile to cover the horizontal range is called the time of flight. Time of flight of projectile is decided by usinθ. The time of flight can be found using the formula s = ut + 1/2 at2
Taking vertical displacement s = 0, a = -g and initial vertical velocity = usinθ, we get
0 = usinθt – 1/2gt2
1/2 gt2 = usinθt
t = \(\frac{2 u \sin \theta}{g}\).

3. Time of ascent = time of descent

4. Time of descent > time of ascent

Plus One Physics Motion in a Plane NCERT Questions and Answers

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:

  • Scalars: Volume, mass, speed, density, number of moles and angular frequency.
  • Vectors: 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:
Work, current.

Question 3.
State, with reasons, whether the following algebraic operations with scalar and vector physical quantities are meaningful:

  1. Adding any two scalars.
  2. Adding a scalar to a vector of the same dimensions
  3. Multiplying any vector by any scalar
  4. Multiplying any two scalars
  5. Adding any two vectors
  6. Adding a component of a vector to the same vector.

Answer:

  1. No. Scalars must represent same physical quantity.
  2. No. Vector can be added only to another vector.
  3. Yes. When a vector is multiplied by any scalar, we get a vector.
  4. Yes. In multiplication, scalars may not represent the same physical quantity.
  5. No. The two vectors must represent the same physical quantity.
  6. Yes. The vector and its component must have the same dimensions.

Question 4.
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.

Question 5.
Read each statement below carefully and state with reasons, if it is true or false:

  1. The magnitude of a vector is always a scalar.
  2. Each component of a vector is always a scalar.
  3. The total path length is always equal to the magnitude of the displacement vector of a particle.
  4. 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.
  5. Three vectors not lying in a plane can never add up to give a null vector.

Answer:

  1. True
  2. False
  3. False
  4. True
  5. True.

In order to give a null vector, the third vector should have the same magnitude and opposite direction to the resultant of two vectors.

Question 6.
A cyclist starts from the centre O of a circular park, then cycles along the circumference, and returns to the centre along QO as shown. If the round trip takes 10 minutes, what is the

  1. Net displacement.
  2. Average velocity and
  3. Average speed of the cyclist?


Answer:
1. Since both the initial and final positions are the same therefore the net displacement is zero.

2. Average velocity is the ratio of net displacement and total time taken. Since the net displacement is zero therefore the average velocity is also zero.

3. Average Speed

Question 7.
A passenger arriving in a new town wishes to go from the station to hotel located 10km 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 minute, what is

  1. The average speed of the taxi
  2. The magnitude of average velocity? Are the two equal?

Answer:
Magnitude of displacement = 10km
Total path length = 23km
Time taken = 28 min

Average Speed = \(\frac{23 \mathrm{km}}{\frac{7}{15} \mathrm{h}}\)
= 49.3 kmh-1
Magnitude of average velocity


Plus One Physics All Chapters Question and Answers


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