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KSEEB Class 11 Physics Solutions For Chapter 5 Motion Very Short Type Questions And Answers

Question 1. Why are spokes provided in a bicycle wheel?
Answer:

The spokes of the cycle wheel increase its moment of inertia. The greater the moment of inertia, the greater the opposition to any change in uniform rotational motion.

As a result, the cycle runs smoother and steadier. If the cycle wheel had no spokes, the cycle would be driven with jerks and hence unsafe.

Read and Learn More KSEEB Class 11 Physics Solutions

Question 2. What is inertia? What gives the measure of inertia?
Answer:

The inability of a body to change its state by itself is known as inertia. The mass of a body is a measure of its Inertia.

Types Of Inertia

1. Inertia of rest
2. Inertia of motion
3. Inertia of direction.

Question 3. According to Newton’s third law, every force is accompanied by an equal and opposite force. How can a movement ever take place?
Answer:

From Newton’s third law action = -reaction. But action and reaction are not working on the same system. So they will not cancel each other. Hence, motion is possible.

Question 4. When a bullet is fired from a gun, the gun gives a kick in the backward direction. Explain.
Answer:

The firing of a gun is due to Internal form. Terminal forces do not change The momentum of the system. Before firing m₁u₁ + m₂u₂ = 0.

Since the system is at rest after firing m₁v₁ + m₂v₂ = 0(or) m₁v₁ = – m₂v₂. So gun and bullet will move in opposite directions to satisfy the law of conservation of linear momentum.

Question 5. Why does a heavy rifle not recoil as strongly as a light rifle using the same cartridges?
Answer:

Velocity (or) recoil \(v=\frac{m v}{M}\) i.e., the ratio of momentum of the bullet to the mass of the gun. If the mass of the gun is high then the velocity of the recoil is less with the same cartridge.

KSEEB Physics Class 11 Laws Of Motion Notes

Question 6. If a bomb at rest explodes into two pieces, the pieces must travel in opposite directions. Explain.
Answer:

Explosion is due to internal forces. From the law of conservation of linear momentum, internal forces cannot change the momentum of the system. So after explosion m₁v₁ + m₂v₂ = 0 or m₁v₁ = – m₂v₂. According to the law of conservation of linear momentum, they will fly in opposite directions.

Question 7. Define force. What are the basic forces in nature? Answer:

Force is that which changes (or) tries to change the state of a body.

The Basic Forces In Nature Are:

1. Gravitational forces,
2. Electromagnetic forces,
3. Nuclear forces.

Question 8. Can the coefficient of friction be greater than one?
Answer:

Yes. Generally coefficient of friction between the surfaces is always less than one. But under some special conditions like on extremely rough surfaces coefficient of friction may be greater than one.

Question 9. Why does the car with a flat tire stop sooner than the one with inflated tires?
Answer:

Due to the flatting of tires, the frictional force increases. Because rolling frictional force between the surfaces Is proportional to the area of contact. Area of contact Increases for flattened tires. So rolling frictional force Increases and the car will be stopped quickly.

Question 10. A horse has to pull harder during the start of the motion than later. Explain.
Answer:

To start motion in a body we must apply force to overcome static friction(F_s = μ_smg). When once motion is started between the bodies then kinetic frictional force comes into act.

Kinetic friction (F_k = μ_kmg) is always less than static friction. So it is tougher to start a body from rest than to keep it in. motion.

Laws Of Motion Question And Answers In KSEEB Physics

Question 11. What happens to the coefficient of friction if the weight of the body is doubled?
Answer:

When the weight of the body is doubled still then there is no change in coefficient of friction. Because frictional force oc normal reaction.

So when the weight of a body is doubled then frictional force and normal reaction will also become doubled and the coefficient of friction remains constant.

Laws Of Motion Solutions KSEEB Class 11 Physics Short Type Questions And Answers

Question 1. A stone of mass 0.1 kg is thrown vertically upwards. Give the magnitude and direction of the net force on the stone

1. During its upward motion,
2. During its downward motion,
3. At the highest point, where it momentarily comes to rest.

Answer:

Mass of stone, m = 0. 1 kg.

1. During upward motion force acts downwards due to acceleration due to gravity.

Magnitude of force F = mg = 0.1 x 9.8 = 0.98 N(↓)

During downward motion force acts downward.

Magnitude of force F = mg = 0.1 x 9.8 = 0.98N(↓)

3. At the highest point velocity v = 0. Hut still g will act on it only In downward motion so resultant forceF = 0.98 N. downward.

Note: In the entire Journey of the body force due to gravitational pull acts only In a downward direction.

4. If the body is thrown with an angle of 30° with the horizontal then the vertical component of gravitational force does not change, hence in this case downward force F = mg = 0.1 x 9.8 = 0. 98 newton.

Question 2. Define the terms momentum and impulse. , State and explain the law of conservation of linear momentum. Give examples.
Answer:

Momentum \((\overline{\mathbf{p}})\): It is the product of mass and velocity of a body.

Momentum \((\overline{\mathbf{p}})\) = mass(m) x velocity(v)

∴ \((\overline{\mathbf{p}})\) = m \((\overline{\mathbf{v}})\)

Impulse (J): When a large force(F) acts on a body for a small time(t) then the product of force and time is called Impulse.

Impulse(J) = Force(F) x time(t)

∴ Impulse(J) = F x t

KSEEB Class 11 Physics Laws of Motion 

Law Of Conservation Of Linear Momentum: There is no external force acting on the system. The total linear momentum of the isolated system remains constant.

Proof: Let two bodies of masses say A and B are moving with initial momenta PA and PB collided with each other.

During collision they are in contact for a small time say Δt. During this time of contact, they will exchange their momenta. Let the final momenta of the bodies be P_A¹ and P_B¹. Let force applied by A on B is F_AB and force applied by B on A is F_BA.

From Newton’s 3rd Law F_AB = F_BA or F_AB Δt = F_BA Δt

From 2nd Law \(F_{A B} \Delta t=P_A^1-P_A\) change in momentum of A.

\(F_{B A} \Delta t=P_{11}^{\prime}-P_B\) change in momentum of B,

i.e., the sum of momentum before collision Is equal to the sum of momentum after collision.

Question 3. Why are shock absorbers used in motorcycles and cars?
Answer:

1. When vehicles are passing over the vertices and depressions of a rough road they will collide with them for a very short period. This causes an impulse effect.
2. Due to the large mass and high speed of the vehicles, the magnitude of impulse is also high. Impulse may cause damage to the car or even to the passengers in it. \(\mathrm{F} \propto \frac{1}{\Delta \mathrm{t}}\)
3. The bad effects of impulse are less if the time of contact is longer. Impulse J = F.t. For the same magnitude of impulse(change in momentum) if the time of contact is high force acting on the vehicle is less.

Shock absorbers will absorb the impulse and release the same force slowly. This is due to the large time constant of the springs.

So shock absorbers are used in vehicles to reduce impulse effects.

Question 4. Explain the terms limiting friction, dynamic friction, and rolling friction.
Answer:

Limiting Friction: Frictional forces always oppose relative motion between the bodies. These forces are self-adjusting forces. Their magnitude will increase to some extent with the value of applied force.

The maximum frictional force between the bodies at rest is called”limiting friction”.

Dynamic (Or) Kinetic Friction: When the applied force is equal to or greater than limiting friction then the body will move. When once motion is started then frictional force will abruptly fall to a minimum value.

The frictional force between moving bodies Is called dynamic (or) kinetic friction. Kinetic friction Is always less than limiting friction.

Rolling Friction: The resistance encountered by a rolling body on a surface is called rolling friction.

Practice Questions For Laws Of Motion KSEEB Physics

Question 5. Explain the advantages and disadvantages of friction.
Answer:

Advantages Of Friction:

1. We are able to walk because of friction.
2. It is impossible for a car to move on a slippery road.
3. The braking system of vehicles works with the help of friction.
4. Friction between roads and tires provides the necessary external force to accelerate the car.
5. Transmission of power to various parts of a machine through belts is possible by friction.

Disadvantages Of Friction:

1. In many cases, we will try to reduce friction because it dissipates energy into heat.
2. It causes wear and tear to machine parts which causes frequent replacement of machine parts.

Question 6. Explain Friction. Mention the methods used to decrease friction.
Answer:

Friction: It is a contact force parallel to the surfaces in contact. Friction will always oppose relative motion between the bodies.

Methods To Reduce Friction:

1. Polishing: Friction is caused due to surface irregularities. So by polishing friction can be reduced to some extent.
2. Lubricants: By using lubricants friction can be reduced. Lubricants will spread as an ultra-thin layer between the surfaces in contact and friction decreases.
3. Stream Elining: By making a front portion of vehicles in a curved shape friction due to air can be reduced.
4. Ball Bearings: Ball bearings are used to reduce friction between machine parts. Ball bearing will convert sliding motion into rolling motion. As a result, friction is reduced.

Question 7. State the laws of rolling friction.
Answer:

When a body is wiling over the other, then friction between the bodies is known as rolling friction.

Rolling friction coefficient, \(\mu_{\mathrm{r}}=\frac{\text { Rolling friction }}{\text { Normal reaction }}=\frac{\mathrm{Fr}}{\mathrm{N} \cdot \mathrm{R}}\)

Laws Of Rolling Friction:

1. Rolling friction will develop a point of contact between the surface and the rolling sphere. For objects like wheels line of contact will develop.
2. Rolling friction(f_r) has the least value for a given normal reaction when compared with static friction(f_s) or kinetic friction (f_k).
3. Rolling friction is directly proportional to F = ma from it. to normal reaction, f_r ∝ N.
4. In rolling friction, the surfaces in contact will get momentarily deformed a little.
5. Rolling friction depends on the area of contact. Due to this reason, friction increases when air pressure is less in tires (Flattened tires).
6. Rolling friction is inversely proportional to the radius of the rolling body \(\mu_r \propto \frac{1}{r}\).

Practice Questions For Laws Of Motion KSEEB Physics

Question 8. Why is pulling the lawn roller preferred to pushing it?
Answer:

Let a lawn roller be pulled by means of a force F with some angle θ to the horizontal. By resolving the force into two components.

1. The horizontal component F cos θ is useful. to pull the body.
2. The vertical component F sin θ opposes the weight

So N.R. = mg – F sin θ

But frictional force μ N.R.

∴ Frictional force [μ(mg- F sin θ)] decreases,

So it is easier to pull the body.

When the lawn roller is pushed by a force, the vertical component F sin θ causes the apparent increase in the weight of the object.

So the normal reaction N.R. = mg + F sin θ

∴ Frictional force [μ(mg + F sinθ)] increases and it will be difficult to pull the body.

KSEEB Class 11 Physics Chapter 5 Laws Of Motion Long Type Questions And Answers

Question 1.

1. State Newton’s second law of motion. Hence, derive the equation of motion F = ma from it.
2. A body is moving along a circular path such that its speed always remains constant. Should there be a force acting on the body?

Answer:

1. Newton’s 2nd law: The rate of change of momentum of a body is proportional to external force and acts along the direction of force applied.

i.e., \(\frac{d p}{d t} \propto F\)

Derivation Of Equation F = ma: According to Newton’s 2nd law.

We know \(\frac{\mathrm{dp}}{\mathrm{dt}} \propto \vec{\mathrm{F}} \Rightarrow \mathrm{F}=\mathrm{k} \cdot \frac{\mathrm{d} \overline{\mathrm{p}}}{\mathrm{dt}}\)…..(1)

But  \(\vec{p}\) = momentum of the body = m \(\bar{v}\)

∴ F = \(\mathrm{k} \cdot \frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{m} \overline{\mathrm{v}}) \Rightarrow \mathrm{km} \frac{\mathrm{d} \overline{\mathrm{v}}}{\mathrm{dt}}\)……(2)

But \(\frac{\mathrm{d} \overline{\mathrm{v}}}{\mathrm{dt}}\) = Rate of change in velocity = acceleration ‘a’.

∴ F = ma…… (3)

Here, k = constant.

The proportional constant is made equal to one, by properly selecting the unit of force.

∴ F = ma.

2. Force On A Body Moving In A Circular Path: Let a body of mass ‘m’ be moving in a circular path of radius ‘r’ with constant speed.

The velocity of the body is given by the tangent drawn at that point. Since velocity is changing continuously the body will have acceleration.

So the body will experience some acceleration. This is called normal acceleration (or) centripetal acceleration.

Question 2. Define the Angle of friction and Angle of repose. Show that the angle of friction is equal to the angle of repose for a rough inclined plane.

A block of mass 4 kg is resting on a rough horizontal plane and is about to move when a horizontal force of 30 N is applied to it. If g = 10 m/s². Find the total contact force exerted by the plane on the block.

Answer:

Angle Of Friction: The angle made by the resultant of the Normal reaction and the limiting friction with the Normal reaction is called the angle of friction (ø).

Angle Of Repose: Let a body of mass m is placed on a rough inclined plane. Let the angle with the horizontal ‘θ’ be gradually increased then for a particular angle of inclination (say a) the body will just slide down without acceleration.

This angle θ = α is called the angle of repose. At this stage, the forces acting on the body are in equilibrium

Equation For Angle Of Repose: Force acting on the body in a vertically downward direction = W = mg.

By resolving this force into two components.

1. Force acting along the inclined plane in a downward direction = mg sinθ.
   • This component is responsible for downward motion.
2. The component mg cos θ. which is balanced by the normal reaction.

If the body slides down without acceleration resultant force on the body is zero, then

mg sin θ = Frictional force (f_k)

mg cos θ = Normal reaction (N.R.)

But coefficient friction \(\mu_{\mathrm{k}}=\frac{\mathrm{f}_{\mathrm{k}}}{\mathrm{N} \cdot \mathrm{R}}=\frac{\mathrm{mg} \sin \theta}{\mathrm{mg} \cos \theta}=\tan \theta\)

Hence θ = α is called the angle of repose.

∴ μ_k = tan α

Hence tangent of the angle of repose (tan θ) is equal to the coefficient of friction (f_k) between the bodies.

2. Given m = 4 kg

g = 10 m/s²

Normal reaction N = mg

N = 4 x 10 = 40 N

Horizontal frictional force f = 30 N

Total contact force \(\mathrm{F}=\sqrt{\mathrm{f}^2+\mathrm{N}^2}\)

F = \(\sqrt{30^2+40^2}\)

F = \(\sqrt{900+1600}\)

F = \(\sqrt{2500}\)

∴ F = 50 N

Chapter 5 Solutions For Laws Of Motion KSEEB Physics Problems

Question 1. The linear momentum of a particle as a function of time is given by, p = a + bt, where a and b are positive constants. What is the force acting on the particle?
Solution:

Linear momentum of a particle, p = a + bt

We know that force acting on a particle is equal to the rate of change of linear momentum.

i.e., \(\vec{F}=\frac{d p}{d t} \Rightarrow \vec{F}=\frac{d}{d t}(a+b t)\)

⇒ \(\vec{F}=\frac{d}{d t}(b t)=b \frac{d}{d t}=b \times 1=b\)

Question 2. Calculate the time needed for a net force of 5 N to change the velocity of a 10 kg mass by 2 m/s.
Solution:

Force, F = 5N

Change in velocity, v – u = 2ms^-1

Mass, m = 10 kg

From Newton’s second law of motion,

F = \(m a=m \frac{(v-u)}{t}\)

∴ t = \(\frac{\mathrm{m}(\mathrm{v}-\mathrm{u})}{\mathrm{F}}=\frac{10 \times 2}{5}=4 \mathrm{~s}\)

Question 3. A ball of mass ‘m’ is thrown vertically upward from the ground and reaches a height ‘h’ before momentarily coming to rest, If ‘g’ is the acceleration due to gravity. What is the impulse received by the ball due to gravity force during its flight? (neglect air resistance)
Solution:

Impulse, J = force x time

⇒ J = \(\mathrm{ma} \times \frac{2 \mathrm{u}}{\mathrm{g}}\);

Here, a = g and \(\mathrm{u}=\sqrt{2 \mathrm{gh}}\)

Now \(\mathrm{J}=\mathrm{mg} \times \frac{2 \times \sqrt{2 \mathrm{gh}}}{\mathrm{g}} \Rightarrow \mathrm{J}=2 \mathrm{~m} \sqrt{2 \mathrm{gh}}\)

∴ Impulse \(\mathrm{J}=\sqrt{8 \mathrm{~m}^2 \mathrm{gh}}\)

KSEEB Physics Laws of Motion Questions And Answers

Question 4. A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 m s^-1 to 3.5 m s^-1 in 25 s. The direction of motion of the body remains unchanged. What is the magnitude and direction of the force?
Answer:

Mass of the body, m = 3.0 kg

The initial velocity of the body, u = 2.0 ms^-1

The final velocity of the body, v = 3.5 ms^-1

Time, t = 25 s

From Newton’s second law of motion,

F = \(\mathrm{ma}=\frac{\mathrm{m}(\mathrm{v}-\mathrm{u})}{\mathrm{t}}=\frac{3.0(3.5-2.0)}{25}=0.18 \mathrm{~N}\)

∴ Magnitude of force acting on the body, F = 0.18 N. The direction of force acting on the body is along the direction of motion of the body because force is positive.

Question 5. A man in a lift feels an apparent weight W when the lift is moving up with a uniform acceleration of 1/3rd of the acceleration due to gravity. If the same man was in the same lift now moving down with a uniform acceleration that is 1/2 of the acceleration due to gravity, then what is his| apparent weight?
Solution:

Case (1): When Lift Is Moving Upwards:

The apparent weight of the man = W

Acceleration, a = g/3

The apparent weight of the man, when the lift is moving upwards, is,

W = \(\mathrm{m}(\mathrm{a}+\mathrm{g})=\mathrm{m}\left(\frac{\mathrm{g}}{3}+\mathrm{g}\right) \)

= \(\frac{4}{3} \mathrm{mg}=\frac{4}{3} \mathrm{~N}\) (because \(\mathrm{N}=\mathrm{mg}\))

∴ \(\mathrm{N}=\frac{3}{4} \mathrm{~W}\)

Case (2): When the lift is moving downwards: Let W’ be (lie apparent weight of the man Acceleration, a = g/2

The apparent weight of the man when the lift is moving downwards is, W” = m(g – a) = m (g – g/2)

= \(\frac{1}{2}\) mg = \(\frac{N}{2}\) [N = mg]

= \(\frac{1}{2}\)(\(\frac{3}{4}\) W) = \(\frac{3}{8}\) W

Question 6. A container of mass 200 kg rests on the back of an open truck. If the truck accelerates at 1.5 m/s², what is the minimum coefficient of static friction between the container and the bed of the truck required to prevent the container from sliding off the back of the truck?
Solution:

Mass of the container, m = 200 kg

Acceleration of truck, a = 1.5 ms^-2

Coefficient of static friction, u_s = \(\frac{a}{g}\)

= \(\frac{1.5}{9.8}\) = 0.153

Question 7. A bomb initially at rest at a height of 40 m above the ground suddenly explodes into two identical fragments. One of them starts moving vertically downwards with an initial speed of 10 m/s. If acceleration due to gravity is 10m/s², what is the separation between the fragments 2s after the explosion?
Solution:

Case (1): (for a downward-moving fragment)

Initial velocity, u = 10 ms^-1

Acceleration, a = +g = 10 ms^-1

Time, t = 2s

From the equation of motion, s = ut + \(\frac{1}{2}\) at² the distance moved in downward direction is,

s₁ = 10 x 2 + \(\frac{1}{2}\) x 10 x (2)² = 40

Case (2) (For Upward Moving Fragment) Given that two fragments are identical hence, after explosion, the fragments move in opposite directions.

Here the first fragment moves in a downward direction, hence, the second fragment moves upward direction.

Again from s = ut + \(\frac{1}{2}\) at² we can write,

s₂ = – 10 x 2+ \(\frac{1}{2}\) x 10 x 4 =-20 + 20 = 0 m

Separation between the fragments 2s after the explosion = s₁ – s₂ = 40 – 0 = 40m

Question 8. A fixed pulley with a smooth grove has a light string passing over it with a 4 kg attached on one side and a 3 kg on the other side. Another 3 kg is hung from the other 3 kg as shown with another light string. If the system is released horn rest, find the common acceleration. (g = 10 m/s²)

Solution:

Here, m₁ = 3 + 3 = 6 kg; m₂ = 4 kg; g = 10 ms^-2

Acceleration of the system, \(a=\left(\frac{m_1-m_2}{m_1+m_2}\right) g \Rightarrow a=\left(\frac{6-4}{6+4}\right) \times 10=2 \mathrm{~ms}^{-2}\)

Question 9. A block of mass of 2 kg slides on an inclined plane that makes an angle of 30° with the horizontal. The coefficient of friction between the block and the surface is √3/2

1. What force should be applied to the block so that it moves down without any acceleration?
2. What force should be applied to the block so that it moves up without any acceleration?

Solution:

Mass of the block, m = 2kg

The angle of Inclination, θ = 30°

Coefficient of friction between the block and the surface, \(\mu=\frac{\sqrt{3}}{2}\)

1. The required force to move the block down without acceleration is,

F = \(\mathrm{mg}\left(\sin \theta-\mu_{\mathrm{k}} \cos \theta\right)\)

= \(2 \times 9.8\left(\sin 30^{\circ}-\frac{\sqrt{3}}{2} \times \cos 30^{\circ}\right)\)

= \(2 \times 9.8\left(\frac{1}{2}-\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}\right)=4.9 \mathrm{~N}\)

2. The required force to move the block up without any acceleration is,

F = \(\mathrm{mg}\left(\sin \theta+\mu_{\mathrm{k}} \cos \theta\right)\)

= \(2 \times 9.8\left(\sin 30^{\circ}+\frac{\sqrt{3}}{2} \cos 30^{\circ}\right)\)

= \(2 \times 9.8\left(\frac{1}{2}+\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}\right)=24.5 \mathrm{~N}\)

KSEEB Class 11 Physics Chapter 5 Laws Of Motion 

Question 10. A block is placed on a ramp of parabolic shape given by the equation y = x²/20, see Figure.

If μ_s = 0.5, what is the maximum height I above the ground at which the block can be placed without slipping? \(\left(\tan \theta=\mu_{\mathrm{s}}=\frac{d y}{d x}\right)\)

Solution:

For the body not to drop

mg cos θ = μ mg sin θ

⇒ tan θ = μ given μ = 0.5
.
But tan θ = \(\frac{dy}{dx}\) slope of parabolic region

⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}=\mu=0.5\)…..(1)

given \(\mathrm{y}=\frac{\mathrm{x}^2}{20}\)

∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2 \mathrm{x}}{20}=\cdot \frac{\mathrm{x}}{10}\)…..(2)

∴ slope \(\frac{\mathrm{y}}{\mathrm{x}}=0.5 \Rightarrow 0.5=\frac{\mathrm{x}}{10} \Rightarrow \mathrm{x}=5\)…..(3)

From eq. (1) and (3) vertical height, y = \(\frac{x^2}{20}=\frac{5 \times 5}{20}=1.25 \mathrm{~m}\)

Question 11. A block of metal of mass 2 kg on a horizo¬ntal table is attached to a mass of 0.45 kg by a light string passing over a frictionless pulley at the edge of the table. The block is subjected to a horizontal force by allowing the 0.45 kg mass to fall. The coefficient of sliding friction between the block and the table is 0.2. Calculate

1. The initial acceleration,
2. The tension in the string,
3. The distance the block would continue to move if, after 2 s of motion, the string should break.

Solution:

Mass of the first block, m₁ = 0.45 kg

Mass of the second block, m₂ = 2kg

coefficient of sliding friction between the block and table, μ = 0.2

1. Initial acceleration, \(a=\left(\frac{m_1-\mu m_2}{m_1+m_2}\right) g\)

⇒ a = \(\left(\frac{0.45-0.2 \times 2}{0.45+2}\right) \times 9.8\)

⇒ a = \(0.2 \mathrm{~ms}^{-2}\)

2. Tension in the string

T = \(\frac{\left(\mathrm{m}_1+\mu \mathrm{m}_2\right) g+\left(\mathrm{m}_2-\mathrm{m}_1\right) \mathrm{a}}{2}\)

T = \(\frac{(0.45+0.2 \times 2) 9.8+(2-0.45) 0.2}{2}\)

= \(\frac{(0.85 \times 9.8)+(1.55 \times 0.2)}{2}\)

T = \(\frac{8.33+0.31}{2}=\frac{8.64}{2}=4.32\) Newton

3. Velocity of string after 2 sec=u in this case; \(\mathrm{u}^{\prime}=0\)

∴ \(\mathrm{u}=\mathrm{u}^{\prime}+\mathrm{at}=0+0.2 \times 2=0.4 \mathrm{~m} / \mathrm{s}\)

Stopping distance, \(\mathrm{s}=\frac{\mathrm{u}^2}{2 \mu \mathrm{g}}\)

= \(\frac{0.4 \times 0.4}{2 \times 0.2 \times 9.8}=\frac{0.4}{9.8}=0.0408 \mathrm{~m}\)

or \(\mathrm{s} \simeq 4.1 \mathrm{~cm}\)

KSEEB Class 11 Physics Solutions For Chapter 5 Laws Of Motion

Question 12. On a smooth horizontal surface a block A of mass 10 kg is kept. On this block, a second block B of mass 5 kg is kept. The coefficient of friction between the two blocks is 0.4. A horizontal force of 30 N is applied on the lower block as shown. The force of friction between the blocks is (take g = 10 m/s²)

Solution:

The mass of block ‘A’ is m_A = 10 kg

The mass of block ‘B’ is m_B = 5 kg

Applied horizontal force, F = 30 N

Coefficient of friction between two blocks,  μ = 0.4

The frictional force of block ’B’  Is f = μ mg

⇒ f = 0,4 x 5 x 10 = 20N

∴ The frictional force acting between the two blocks, = F – f = 30 – 20 = 10N

Question 13. A batsman hits back a ball straight in the direction of the bowler without changing its initial speed of 12 ms^-1. If the mass of the ball is 0.15 kg, determine the impulse imparted to the ball. (Assume linear motion of the ball).
Solution:

Impulse = change in momentum = (0.15 x 12)-(-0.15 x 12) = 3.6 NS .

In the direction from the batsman to the bowler.

Question 14. A force 2\(\bar{i}\) + \(\bar{j}\) — \(\bar{k}\) Newton acts on a body which is initially at rest. At the end of 20 seconds the velocity of the body is 4\(\bar{i}\) + 2\(\bar{j}\) – 2\(\bar{k}\) m/s. What is the mass of the body?
Solution:

Force F = 2\(\bar{i}\) + \(\bar{j}\) – \(\bar{k}\) = 20sec

Initial velocity u₀ = 0.

Final velocity U = 4\(\bar{i}\) + 2\(\bar{j}\) – 2\(\bar{k}\)

Mass of the body m = \(\frac{F}{a}\)

But acceleration \(a=\frac{U-U_0}{t}\)

∴ \(\mathrm{m}=\frac{\text { F.t }}{\mathrm{U}-\mathrm{U}_0}=\frac{20 \cdot|\mathrm{F}|}{\left|\mathrm{U}-\mathrm{U}_0\right|}\)

= \(\frac{20 \sqrt{2^2+1+1}}{\sqrt{4^2+2^2+2^2}-0}=\frac{20 \sqrt{6}}{\sqrt{24}}\)

∴ \(\mathrm{m}=\frac{20 \sqrt{6}}{\sqrt{4 \times 6}}=\frac{20 \sqrt{6}}{2 \sqrt{6}}=10 \mathrm{~kg} \text {. }\)

Question 15. A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 ms^-1. How long does the body take to stop?
Solution:

Here, F = – 50N, m = 20 kg

u = 15 ms^-1, v = 0, t = ?

From F = \(m a, a=\frac{F}{m}=\frac{-50}{20}=-2.5 \mathrm{~ms}^{-2}\)

From \(\mathrm{v}=\mathrm{u}+\mathrm{at} ; 0=15-2.5 \mathrm{t}\)

t = \(\frac{15}{2.5}=6 \mathrm{~s}\)

Question 16. A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6N. Give the magnitude and direction of the acceleration of the body.
Solution:

Here, m = 5 kg, \(\vec{a}\) = ?

⇒ \(\vec{F}_1=\overrightarrow{O A}=8 \mathrm{~N} ; \vec{F}_2=\overrightarrow{O B}=6 \mathrm{~N}\)

Resultant force \(\overrightarrow{\mathrm{F}}=\overrightarrow{\mathrm{OC}}=\sqrt{\mathrm{F}_1^2+\mathrm{F}_2^2}\)

= \(\sqrt{8^2+6^2}=10 \mathrm{~N}\)

If \(\angle A O C=\theta, \tan \theta=\frac{\mathrm{AC}}{\mathrm{OA}}=\frac{\mathrm{OB}}{\mathrm{OA}}=\frac{6}{8}=0.75\)

∴ \(\theta=36^{\circ} 52^{\prime}\)

This is the direction of the resultant force and hence the direction of acceleration of the body as shown.

Also, \(\mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}=\frac{10}{5}=2 \mathrm{~ms}^{-2}\)

Laws Of Motion Solutions KSEEB Class 11 Physics

Question 17. The driver of a three-wheeler moving with a speed of 36 km /h sees a child standing in the middle of the road and brings his vehicle to rest in 4.0 seconds just in time to save the child. What is the average retarding force on the vehicle? The mass of the three-wheeler is 400 kg and the mass of the driver is 65 kg.
Solution:

Here, u = 36 km/h = 10 m/s, v = 0, t = 4s

m = 400 + 65 = 465 kg

Retarding force = \(\mathrm{F}=\mathrm{ma}=\frac{\mathrm{m}(\mathrm{v}-\mathrm{u})}{\mathrm{t}}\)

= \(\frac{465(0-10)}{4}=-1162.5 \mathrm{~N}\)

Question 18. A rocket with a lift-off mass of 20,000 kg Is blasted upwards with an initial acceleration of 5.0 ms^-2. Calculate the initial thrust (force) of the blast.
Solution:

Here, m = 20000 kg = 2 x 10⁴ kg

Initial acceleration, a = 5 ms^-2;

Thrust, F =?

Clearly, the thrust should be such that it overcomes the force of gravity besides giving it an upward acceleration of 5 ms^-2.

Thus the force should produce a net acceleration of 9.8 + 5.0 = 14.8 ms^-2.

As thrust = force = mass x acceleration

∴ F = 2 X 10⁴ X 14.8 = 2.96 X 10⁵  N

Question 19. A man of mass 70 kg stands on a weighing scale in a lift which is moving

1. Upwards with a uniform speed of 10 ms^-1,
2. Downwards with a uniform acceleration of 5 ms^-2,
3. Upwards with a uniform acceleration of 5 m s^-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?

Solution:

Here, m = 70 kg, g = 10 m/s²

The weighing machine in each case measures the reaction R i.e., the apparent weight.

1. When the lift moves upwards with a uniform speed, its acceleration is zero.

∴  R = mg = 70 x 10 = 700 N

2. When the lift moves downwards with a = 5 ms^-2

∴ R = m(g- a) = 70 (10 – 5) = 350 N

3. When the lift moves upwards with a = 5 ms^-2

∴ R = m (g + a) = 70 (10 + 5) = 1050 N

4. If the lift were to come down freely under gravity, downward acceleration. a = g

∴ R = m (g – a) = m (g – g) = Zero.

Question 20. Two bodies of masses 10 kg and 20 kg respectively kept on a smooth, horizontal surface arc tied to the ends of a light string, a horizontal force F = 000 N Is applied to

1. A,
2. B along the direction of hiring.

What is the tension In the string in each ease?

Solution:

Here, F = 600 N m₁ = 10 kg, m₂ = 20 kg

Let T be the tension in the string and a be the acceleration of the system, in the direction of force applied.

∴ a = \(\frac{F}{m_1+m_2}=\frac{600}{10+20}=20 \mathrm{~m} / \mathrm{s}^2\).

1. When force is applied on lighter block A, T = m₂a = 20 x 20 N = 400 N
2. When force is applied on heavier block B, T = m₁a = 10 x 20 N T = 200 N

Which is different from the value of T in case (1).

Hence our answer depends on which mass end, the force is applied.

KSEEB Class 11 Physics Chapter 5 Laws Of Motion

Question 21. Two masses 8 kg and 12 kg are connected at the two ends of a light in an extensible string that goes over a frictionless pulley. Find the acceleration of the masses, and the tension in the string when the masses are released.

Solution:

Here, m₂ = 8 kg, ; m₁ = 12kg

As a = \(\frac{\left(m_1-m_2\right) g}{m_1+m_2}=\frac{(12-8) 9.8}{12+8}=\frac{39.2}{20}\) = \(1.96 \mathrm{~ms}^{-2}\)

Again, \(T=\frac{2 m_1 m_2 g}{m_1+m_2}=\frac{2 \times 12 \times 8 \times 9.8}{(12+8)}\)

= 94.1 N

Question 22. A nucleus is at rest In the laboratory frame of reference. Show that If It disintegrates Into two smaller nuclei the products must move in opposite directions.
Solution:

Let m₁, m₂ be the masses of products and \(\overrightarrow{v_1}, \overrightarrow{v_2}\) be their respective velocities. Therfore, total linear momentum after disinte-gration = \(m_1 \overrightarrow{v_1}+m_2 \overrightarrow{v_2}\).

Before disintegration, the nucleus is at rest. Therefore, its linear momentum before disintegration is zero.

According to the principle of conservation of linear momentum, \(m_1 \overline{v_1}+m_2 \overline{v_2}=0 \text { or } \overline{v_2}=\frac{-m_1 \overline{v_1}}{m_2}\)

Negative sign shows that \(\overrightarrow{\mathrm{v}_1} \text { and } \overrightarrow{\mathrm{v}_2}\) are in opposite directions.

Question 23. Two billiard balls each of mass 0.05 kg moving in opposite directions with a speed of 6 ms^-1 collide and rebound with the same speed. What is the impulse imparted to each ball due to the other?
Solution:

Here, initial momentum of the ball A = 0.05 (6) = 0.3 kg ms^-1

As the speed is reversed on collision, the final momentum of the ball A = 0.05 (-6) = – 0.3 kg ms^-1

Impulse imparted to ball A = change in momentum of ball A = final momentum – initial momentum = – 0.3 – 0.3 = – 0.6 kg ms^-1.

Question 24. A shell of mass 0.02 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 ms^-1, what is the recoil speed of the gun?
Solution:

Here, the mass of the shell, m = 0.02 kg.

mass of gun, M = 100 kg

muzzle speed of shell, V = 80 ms^-1

recoil speed of gun, v =?

According to the principle of conservation of linear momentum, mV + Mν = 0

or, \(\mathrm{v}=-\frac{\mathrm{mV}}{\mathrm{M}}=\frac{-0.02 \times 80}{100}=0.016 \mathrm{~ms}^{-1}\)

Solutions For Laws of Motion KSEEB Physics Chapter 5 

Question 25. A slone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 ni with a speed of 40 rev./min in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N?
Solution:

Here, m = 0.25 kg, r = 1.5 m;

n = \(40 \mathrm{rpm}=\frac{40}{60} \mathrm{rps}=\frac{2}{3} \mathrm{rps}, \mathrm{T}=?\)

T = \(\mathrm{mr} \omega^2=\mathrm{mr}(2 \pi \mathrm{n})^2=4 \pi^2 \mathrm{mr} \mathrm{r}^2\)

T = \(4 \times \frac{22}{7} \times \frac{22}{7} \times 0.25 \times 1.5 \times\left(\frac{2}{3}\right)^2=6.6 \mathrm{~N}\)

If \(\mathrm{T}_{\max }=200 \mathrm{~N}\), then from \(\mathrm{T}_{\max }=\frac{\mathrm{mv}_{\max }^2}{\mathrm{r}}\)

⇒ \(\mathrm{v}_{\max }^2=\frac{\mathrm{T}_{\max } \times \mathrm{r}}{\mathrm{m}}=\frac{200 \times 1.5}{0.25}=1200\)

⇒ \(\mathrm{v}_{\max }=\sqrt{1200}=34.6 \mathrm{~m} / \mathrm{s}\)

Question 26. Explain why

1. A horse cannot pull a cart and run in empty space,
2. Passengers are thrown forward from their seats when a speeding bus stops suddenly,
3. It is easier to pull a lawn mover than to push it,
4. A cricketer moves his hands backward while holding a catch.

Solution:

1. While trying to pull a cart, a horse pushes the ground backward with a certain force at an angle. The ground offers an equal reaction in the opposite direction, on the feel of the horse. The forward component of this reaction is responsible for the motion of the cart. In empty space, there Is no reaction, and hence, a horse cannot pull the cart and run.
2. This is due to the “inertia of motion”. When the speeding bus stops suddenly, the lower part of the body in contact with the seats stops. The upper part of the bodies of the passengers tends to maintain a uniform motion. Hence, the passengers are thrown forward.
3. While pulling a lawn mover, force is applied upwards along the handle. The vertical component of this force is upwards and reduces the effective weight of the mover. While pushing a lawn mover, force is applied downwards along the handle. The vertical component of this force is downwards and increases the effective weight of the mover. As the effective weight is lesser in the case of pulling than in the case of pushing, therefore, “pulling is easier than pushing”.
4. While holding a catch, the impulse received by the hands, F x t = change in linear momentum of the ball is constant. By moving his hands backward, the cricketer increases the time of impact (t) to complete the catch. As t increases, F decreases and as a reaction, his hands are not hurt severely.

Question 27. A stream of water flowing horizontally with a speed of 15 m s^-1 pushes out of a tube of cross-sectional area 10^-2 m², and hits a vertical wall nearby. What is the force exerted on the wall by the impact of water, assuming it does not rebound?
Solution:

Here, v = 15 ms^-1

Area of cross-section, a = 10^-2 m²

The volume of water pushing out/sec = a x v = 10^-2 x 15 m³ s^-1

As the density of water is 10³ kg/m³, therefore, mass of water strikes the wall per sec.

m = (15 x 10^-2) x 10³ = 150 kg/s.

As \(\mathrm{F}=\frac{\text { change in linear momentum }}{\text { time }}\)

∴ \(\mathrm{F}=\frac{\mathrm{m} \times \mathrm{v}}{\mathrm{t}}=\frac{150 \times 15}{1}=2250 \mathrm{~N}\)

KSEEB Physics Class 11 Laws Of Motion Notes

Question 28. Ten one-rupee coins are put on top of each other on a table. Each coin has a mass of m. Give the magnitude and direction of

1. The force on the 7th coin (counted from the bottom) due to all the coins on its top,
2. The force on the 7th coin by the eighth coin,
3. The reaction of the 6th coin on the 7th coin.

Solution:

1. The force on the 7th coin is due to the weight of the three coins lying above it. Therefore, F = (3 m) kgf = (3 mg) N

where g is the acceleration due to gravity. This force acts vertically downwards.

2. The eighth coin is already under the weight of two coins above it and it has its own weight too. Hence force on 7th coin due to 8th coin is sum of the two forces i.e. F = 2 m + m = (3 m) kgf = (3 mg) N

The force acts vertically downwards.

3. The sixth coin is under the weight of four coins above it.

Reaction, r = -F = -4 m (kgf) =- (4 mg) N

The Minus sign indicates that the reaction acts vertically upwards, opposite to the weight.

Question 29. An aircraft executes a horizontal loop at a speed of 720 km/h with its wings banked at 15°. What is the radius of the loop?
Solution:

Here θ = 15°

v \(=720 \mathrm{~km} / \mathrm{h}=\frac{720 \times 1000}{60 \times 60}=200 \mathrm{~ms}^{-1}\);

g = \(9.8 \mathrm{~ms}^{-2} \text {; from } \tan \theta=\frac{\mathrm{v}^2}{\mathrm{rg}} \)

⇒ \(\mathrm{v}^2=\mathrm{rg} \tan \theta\)

∴ r \(=\frac{\mathrm{v}^2}{\mathrm{~g} \tan \theta}=\frac{(200)^2}{9.8 \times \tan 15^{\circ}}\)

= \(15232. \mathrm{m}=15.232 \mathrm{k} \cdot \mathrm{m} \text {. }\)

Question 30. A train runs along an unbanked circular track of radius 30 m at a speed of 54 km/h The mass of the train is 10⁶ kg. What provides the centripetal force required for, this purpose-The engine or the rails? What is the angle of banking required to prevent wearing out of the rail?
Solution:

The centripetal force is provided by the lateral thrust exerted by the rails on the wheels. By Newton’s 3rd law, the train exerts an equal and opposite thrust on the rails causing its wear and tear.

Obviously, the outer rail will wear out faster due to the larger force exerted by the train on it.

Here, \(v=54 \mathrm{~km} / \mathrm{h}=\frac{54 \times 1000}{60 \times 60}=15 \mathrm{~m} / \mathrm{s}\);

g = \(9.8 \mathrm{~ms}^{-2}\)

As \(\tan \theta=\frac{\mathrm{v}^2}{\mathrm{rg}}=\frac{15 \times 15}{30 \times 9.8}=0.76 ;\)

∴ \(\theta=\tan ^{-1} 0.76=37.4^{\circ}\)

Question 31. A block of mass 25 kg Is raised by a 50 kg man in two different ways as shown. What Is the action on the floor by the man in the two cases? If the floor yields to a normal force of 700 N, which mode should the man adopt to lift the block without the floor yielding?

Solution:

Here, the mass of the block, m = 25 kg

Mass of man, M = 50 kg

Force applied to lift the block

F = mg = 25 x 9.8 = 245 N

Weight of man W = Mg = 50 x 9.8 = 490 N

1. When a block is raised by the man as shown, force is applied by the man in the upward direction. This increases the apparent weight of the man. Hence action on the floor.

W’ = W + F = 490 + 245 = 735 N

When a block is raised by a man as shown, force is applied by the man in the downward direction. This decreases the apparent weight of the man. Hence, action on the floor in this case would be W’ = W – F = 490 – 245 = 245 N.

As the floor yields to a normal force of 700 N, mode (2) has to be adopted by the man to lift the block.

Question 32. A monkey of mass 40 kg climbs on a rope that can stand a maximum tension of 600 N. In which of the following cases will the rope break the monkey

1. Climbs up with an acceleration of 6 ms^-2
2. Climbs down with an acceleration of 4 ms^-2
3. Climbs up with a uniform speed of 5 ms^-1
4. Fell down the rope nearly freely under gravity?

(Ignore the mass of the rope).

Solution:

Here, the mass of the monkey, m = 40 kg.

Maximum tension the rope can stand, T = 600 N.

In each case, the actual tension in the rope will be equal to the apparent weight of the monkey (R),

The rope will break when R exceeds T.

1. When monkey climbs up with a = 6 ms^-2, R = m (g + a) = 40 (10 + 6) = 640 N (which is greater than T).

Hence the rope will break.

2. When the monkey climbs down with a = 4 ms^-2

R = m (g – a) = 40 (10 – 4) = 240 N, which is less than T

∴ The rope will not break.

3. When the monkey climbs up with a uniform speed v = 5 ms^-1,

its acceleration a = 0

R = mg = 40 x 10 = 400 N, which is less than T

∴ The rope will not break.

4. When the monkey falls down the rope nearly freely under gravity, a = g

R = m (g – a) = m (g – g) = 0 (Zero.)

Hence the rope will not break.

Question 33. A 70 kg man stands In contact against the Inner wall of a hollow cylindrical drum of radius 3 m rotating about its vertical avis with 200 rev/mln. The coefficient of friction between the wall and his clothing is 0.15. What is the minimum rotational speed of the cylinder to enable the man to remain stuck to the wall (without falling) when the floor is suddenly removed?
Solution:

Here, m = 70 kg, r = 3 m

n = 200, rpm = \(\frac{200}{60}\) rps, μ = 0.15, ω= ?

The horizontal force N by the wall on the man provides the necessary centripetal force = m r ω². The frictional force (f) in this case is vertically upwards opposing the weight (mg) of the man.

After the floor is removed, the man will remain stuck to the wall, when mg = f < μ N, i.e. mg < μ m r ω² or g < μ r ω²

∴ The minimum angular speed of rotation of the cylinder is \(\omega=\sqrt{\frac{g}{\mu r}}=\sqrt{\frac{10}{0.15 \times 3}}\)

= \(4.7 \mathrm{rad} / \mathrm{s}\)

Question 34. A thin circular loop of radius R rotates about its vertical diameter with an angular frequency ω. Show that a small bead on the wire loop remains at its lowermost point for \(\omega \leq \sqrt{g / R},\). What is the angle made by the radius vector joining the center to the bead with the vertically downward direction form = \(\omega \leq \sqrt{2g / R},\) Neglect friction?
Solution:

In Figure, we have shown that the radius vector joining the bead to the center of the wire makes an angle 0 with the verticle downward direction. If N is a normal reaction, then as is clear from the figure,

⇒ \(\mathrm{mg}=\mathrm{N} \cos \theta\)….(1)

⇒ \(\mathrm{mr} \omega^2=\mathrm{N} \sin \theta\)…..(2)

or \(\mathrm{m}(\mathrm{R} \sin \theta) \omega^2=\mathrm{N} \sin \theta \text { or } \mathrm{m} \mathrm{R}^2=\mathrm{N}\)

from (1), \(\mathrm{mg}=\mathrm{m} R \omega^2 \cos \theta\)

or \(\cos \theta=\frac{\mathrm{g}}{\mathrm{R \omega}^2}\) …..(3)

As |cos θ| ≤ 1, therefore, the bead will remain at its

lowermost point for \(\frac{g}{R \omega^2} \leq 1 \text {, or } \omega \leq \sqrt{\frac{g}{R}}\)

When \(\omega=\sqrt{\frac{2 g}{R}}\) from (3), \(\cos\theta=\frac{g}{R}\left(\frac{R}{2 g}\right)=\frac{1}{2}\)

∴ \(\theta=60^{\circ}\)

KSEEB Class 11 Physics Solutions For Chapter 4 Motion in a Plane Very Short Answer Questions

Question 1. Write the equation for the horizontal – range covered by a projectile and specify
when it will be maximum.
Answer:

Range of a projectile (R) = \(\frac{u^2 \sin 2 \theta}{g}\)

When θ = 45° Range is maximum. (sin 90° = 1)

Maximum Range (R_max) = \(\frac{u^2}{g}\)

Read and Learn More KSEEB Class 11 Physics Solutions

Question 2. The vertical component of a vector is equal to its horizontal component. What is the angle made by the vector with the X-axis?
Answer:

Let R be a vector.

Vertical component = R sin θ;

Horizontal component = R cos θ

∴ R sin θ = R cos θ

So sin θ = cos θ ⇒ θ = 45°

Question 3. A vector V makes an angle θ with the horizontal. The vector is rotated through an angle α. Does this rotation change the vector V?
Answer:

Magnitude of vector = V ;

Let the initial angle with horizontal = θ

Angle rotated = α

So new angle with horizontal = θ + α

Now horizontal component, V_α = V cos (θ + α)

Vertical component, V_y = V sin (θ + α)

Magnitude of vector, V = \(\sqrt{V_x^2+V_y^2}=V\)

So rotating the vector does not change its magnitude.

KSEEB Class 11 Physics Motion In A Plane

Question 4. Two forces of magnitude 3 units and 5 units act at 60 with each other, What is the magnitude of their resultant?
Answer:

Given \(\bar{P}\) = 3 units, \(\bar{Q}\) = 5 units and θ = 60°

∴ R = \(\sqrt{P^2+Q^2+2 P Q \cos \theta}\)

= \(\sqrt{3^2+5^2+2(3)(5) \cos 60^{\circ}}\)

= \(\sqrt{9+25+30 \times \frac{1}{2}}=\sqrt{49}=7 \text { units. }\)

Question 5. A = \(\vec{i}\) + \(\vec{j}\). What is the angle between the vector and the X-axis?
Answer:

Given that, \(\vec{A}\) = \(\vec{i}\) + \(\vec{j}\)

|\(\vec{A}\)| = \(\sqrt{(1)^2+(1)^2}=\sqrt{2}\)

If ‘θ’ is the angle made by the vector with X-axis then, \(\cos \theta=\frac{A_x}{|\vec{A}|} \Rightarrow \cos \theta=\frac{1}{\sqrt{2}} \Rightarrow \theta=45^{\circ}\)

Question 6. When two right-angled vectors of magnitude 7 units and 24 units combine, what is the magnitude of their resultant?
Answer:

Given \(\vec{P}\) = 7 units; \(\vec{Q}\) = 24 units;  θ = 90°

∴ \(\sqrt{\mathrm{P}^2+\mathrm{Q}^2+2 \mathrm{PQ} \cos \theta}=\sqrt{\mathrm{P}^2+\mathrm{Q}^2}\)

∴ \(\mathrm{R}=\sqrt{\mathrm{P}^2+\mathrm{Q}^2}=\sqrt{7^2+24^2}=25\) units.

KSEEB Physics Class 11 Motion In A Plane Notes

Question 7. If \(\vec{P}\) = 2i + 4j + 14k and \(\vec{Q}\) = 4i + 4j + 10k, find the magnitude of \(\vec{P}\) + \(\vec{Q}\).
Answer:

Given \(\overline{\mathrm{P}}=2 \overline{\mathrm{i}}+4 \overline{\mathrm{j}}+14 \overline{\mathrm{k}}\) and

⇒ \(\bar{Q}=4 \overline{\mathrm{i}}+4 \overline{\mathrm{j}}+10 \overline{\mathrm{k}}\)

⇒ \(\overline{\mathrm{P}}+\overline{\mathrm{Q}}=6 \overline{\mathrm{i}}+8 \overline{\mathrm{j}}+24 \overline{\mathrm{k}}\)

Magnitude of \(|\overline{\mathrm{P}}+\overline{\mathrm{Q}}|=\sqrt{6^2+8^2+24^2}\)

= \(\sqrt{36+64+576}=\sqrt{676}=26\) units.

Question 8. Can a vector of magnitude zero have non-zero components?
Answer:

A vector with zero magnitude cannot have non-zero components, because the magnitude of a given vector \(\bar{V}=\sqrt{V_x^2+V_y^2}\) must be zero. This is possible only when \(V_x^2\) and \(V_y^2\) are zero.

Question 9. What is the acceleration of a projectile at the top of its trajectory?
Answer:

At Highest point acceleration, a = g. In projectile motion acceleration will always act towards the centre of the earth. It is irrespective of its position, whether it is at the highest point or somewhere.

Question 10. Can two vectors of unequal magnitude add up to give the zero vector? Can three unequal vectors add up to give the zero vector?
Answer:

No. Two unequal vectors can never give zero vector by addition. But three unequal vectors when added may give zero vector.

Motion In A Plane Solutions KSEEB Class 11 Physics Chapter 4 Short Answer Questions

Question 1. State parallelogram law of vectors. Derive an expression for the magnitude and direction of the resultant vector.
Answer:

Parallelogram Law: If two vectors are represented by the two adjacent sides of a parallelogram, then the diagonal passing through the intersection of given vectors represents their resultant both in direction and magnitude.

Proof: Let \(\overline{\mathrm{P}}\) and \(\overline{\mathrm{Q}}\) be two adjacent vectors ‘θ’ be the angle between them. Construct a parallelogram OACB as shown in the figure.

Extend the line OA and draw a normal D from C. The diagonal OC = the resultant \(\overline{\mathrm{R}}\) both in direction and magnitude

In figure OCD = right angle triangle ⇒ OC² = OD² + DC²

But OD = OA + AD = \(\overline{\mathrm{P}}\) + \(\overline{\mathrm{Q}}\) cos θ and CD = Q sin θ and \(\overrightarrow{\mathrm{OC}}=R\)

∴ \(R^2=[P+Q \cos \theta]^2+[Q \sin \theta]^2\)

⇒ \(\overline{\mathrm{R}}^2=\overline{\mathrm{P}}^2+\overline{\mathrm{Q}}^2 \cos ^2 \theta+2 \overline{\mathrm{PQ}} \cos \theta+\overline{\mathrm{Q}}^2 \sin ^2 \theta\)

∴ \(\overline{\mathrm{R}}^2=\overline{\mathrm{P}}^2+\overline{\mathrm{Q}}^2\left(\sin ^2 \theta+\cos ^2 \theta\right)+2 \overline{\mathrm{PQ}} \cos \theta\)

But \(\sin ^2 \theta+\cos ^2 \theta=1\)

∴ \(\mathrm{R}^2=[\mathrm{P}^2+\mathrm{Q}^2+2 \mathrm{PQ} \cos \theta\).

∴ Resultant vector, \(\overline{\mathrm{R}}=\sqrt{\mathrm{P}^2+\mathrm{Q}^2+2 \mathrm{PQ}}\)

The angle of resultant with adjacent side ‘α’

In triangle OCD, \(\tan \alpha=\frac{C D}{O D}\) (OD = OA + AD)

⇒ \(\text{Tan} \alpha=\frac{C D}{O A+A D}=\frac{Q \sin \theta}{P+Q \cos \theta}\)

(because A D = \(Q \cos \theta\))

Angle of \(\overline{\mathrm{R}}\) with adjacent side, \(\alpha=\tan ^{-1}\left[\frac{Q \sin \theta}{P+Q \cos \theta}\right]\)

Question 2. What is relative motion? Explain it.
Answer:

Relative velocity is the velocity of a body with respect to another moving body.

Relative velocity in two-dimensional motion

Let two bodies A and B are moving with velocities \(\overrightarrow{\mathrm{V}}_{\mathrm{A}}\) and \(\overrightarrow{\mathrm{V}}_{\mathrm{B}}\) then relative velocity of A with respect to B is \(\overrightarrow{\mathrm{V}}_{\mathrm{AB}}=\overrightarrow{\mathrm{V}}_{\mathrm{A}}-\overrightarrow{\mathrm{V}}_{\mathrm{B}}\)

The relative velocity of B with respect to A is \(\overrightarrow{\mathrm{V}}_{\mathrm{BA}}=\overrightarrow{\mathrm{V}}_B-\overrightarrow{\mathrm{V}}_{\mathrm{A}}\)

Procedure To Find Resultant: To find relative velocity in two-dimensional motion use vectorial subtraction of V_A or V_B. Generally to find relative velocity one vector \(\overrightarrow{\mathrm{V}}_{\mathrm{A}}\) or \(\overrightarrow{\mathrm{V}}_{\mathrm{B}}\) is reversed (as the case may be) and parallelogram is constructed. Now resultant of that parallelogram is equal to \(\overrightarrow{\mathrm{V}}_{\mathrm{A}}\) – \(\overrightarrow{\mathrm{V}}_{\mathrm{B}}\) or \(\overrightarrow{\mathrm{V}}_{\mathrm{B}}\) – \(\overrightarrow{\mathrm{V}}_{\mathrm{A}}\) (one vector is reversed VA is taken as –\(\overrightarrow{\mathrm{V}}_{\mathrm{A}}\) or \(\overrightarrow{\mathrm{V}}_{\mathrm{B}}\) is taken as –\(\overrightarrow{\mathrm{V}}_{\mathrm{B}}\))

In the figure relative velocity of B with respect to A is V_BA = V_R = V_B -V_A

Question 3. Show that a boat must move at an angle with respect to river water in order to cross the river in minimum time.
Answer:

The motion of a boat in a river: Let a boat travel with a speed of V_bE in still water with respect to earth. It is used to cross a river which flows with a speed of V_WE with respect to earth. Let the width of the river be W.

We can cross the river in two different ways.

1. In shortest path
2. In the shortest time.

To cross the river in the shortest time:

Shortest time, t = \(\mathrm{t}=\frac{\text { width of river }}{\text { velocity of boat }}=\frac{W}{V_{\mathrm{bE}}}\)

To cross the river in the shortest time boat must be rowed along the width of the river i.e., the boat must be rowed perpendicular to the bank or 90° with the flow of water.

Because the width of the river is the shortest distance. So velocity must be taken in that direction to obtain the shortest time.

In this case, V_bE and V_WE are perpendicular and the boat will travel along AC. The distance BC is called drift. So to cross the river in the shortest time angle with the flow of water = 90°.

KSEEB Class 11 Physics Chapter 4 Solved Examples

Question 4. Define unit vector, null vector and position vector.
Answer:

Unit Vector: A vector whose magnitude is one unit is called a unit vector.

Let \(\overline{\mathrm{a}}\) is a given vector then the unit vector of \(\bar{a}=\frac{\bar{a}}{|\bar{a}|}=\hat{a}\)

When \(\bar{a}\) ≠ 0 or \(\frac{\bar{a}}{|\bar{a}|}\) = unit vector of \(\bar{a}\). It is denoted by a \(\hat{a}\)

Null Vector: A vector whose magnitude is zero is called a null vector. But it has direction.

For a null vector, the origin and terminal point are the same.

Example: Let \(\overline{\mathrm{A}} \times \overline{\mathrm{B}}=\overline{0}\). Here magnitude of \(\overline{\mathrm{A}} \times \overline{\mathrm{B}}=\overline{0}\). But still, it has direction perpendicular to the plane of \(\overline{\mathrm{A}}\) and \(\overline{\mathrm{A}}\).

Position Vector: Any vector in space can be represented by the linear combination of \(\overline{\mathrm{i}}, \overline{\mathrm{j}}\) and \(\overline{\mathrm{k}} \text {. }\).

Let ‘O’ is the origin then \(\overline{\mathrm{OP}}\) is represented as \(\overline{\mathrm{OP}}=x \bar{i}+y \bar{j}+z \bar{k}\) where x, y and z are magnitudes of \(\overline{\mathrm{OP}}\) along \(\bar{i}, \bar{j}\) and \(\overline{\mathrm{k}}\) axis.

Magnitude of \(\overrightarrow{\mathrm{OP}}=\sqrt{\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2}\)

Question 5. If \(|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}|\), prove that the angle between \(\vec{a}\) and \(\vec{b}\) is 90°.
Answer:

Let \(\vec{a}\), \(\vec{b}\) are the two vectors.

Sum of vectors = \(\bar{a}+\bar{b}=\sqrt{a^2+b^2+2 a b \cos \theta}\)

Difference of vectors = \(\bar{a}-\bar{b}=\sqrt{a^2+b^2-2 a b \cos \theta}\)

Given \(|\bar{a}+\bar{b}|=|\bar{a}-\bar{b}|\)

⇒ \(\sqrt{a^2+b^2+2 a b \cos \theta}\)

= \(\sqrt{a^2+b^2-2 a b \cos \theta}\) by squaring on both sides, \(\mathrm{a}^2+\mathrm{b}^2+2 \mathrm{ab} \cos \theta=\mathrm{a}^2+\mathrm{b}^2-2 \mathrm{ab} \cos \theta\)

∴ \(4 \mathrm{ab} \cos \theta=0\) or \(\theta=90^{\circ}\)

So if \(|\bar{a}+\bar{b}|=|\bar{a}-\bar{b}|\) then angle between \(\bar{a}\) and \(\bar{b}\) is \(90^{\circ}\).

Motion In A Plane Solutions KSEEB Class 11 Physics

Question 6. Show that the trajectory of an object thrown at a certain angle with the horizontal is a parabola.
Answer:

Projectile: A body thrown into the air same angle as the horizontal, (other than 90°) its motion under the influence of gravity is called a projectile. The path followed by it is called a trajectory.

Let a body be projected from point O, with velocity ‘u’ at an angle θ with horizontal. The velocity ‘u’ can be resolved into two rectangular components u_x and u_y along X-axis and Y-axis.

u_x = u cos θ and u_y = u sin θ

After time t, Horizontal distance travelled x = u cos θ t……(1)

After a time sec; vertical displacement y = u sinθ • t – \(\frac{1}{2}\) gt²

y = \(u \sin \theta \cdot t-\frac{1}{2} g t^2\)

From (1) t = \(\frac{x}{u \cos \theta}\)

y = \(\frac{u \sin \theta \cdot x}{u \cos \theta}-\frac{1}{2} g \cdot \frac{x^2}{u^2 \cos ^2 \theta}\)

⇒ \(y=x \tan \theta-\frac{g x^2}{2 u^2 \cos ^2 \theta}\)

Let \(\tan \theta=\mathrm{A}\) and \(\frac{\mathrm{g}}{2 \mathrm{u}^2 \cos ^2 \theta}=\mathrm{B}\) then y = \(A x-B x^2\)

Let \(\tan \theta=\mathrm{A}\) and \(\frac{\mathrm{g}}{2 \mathrm{u}^2 \cos ^2 \theta}=\mathrm{B}\) then \(y=A x-B x^2\)

The above equation represents a “parabola”. Hence the path of a projectile is a parabola.

Question 7. Explain the terms average velocity and instantaneous velocity. When are they equal?
Answer:

Average Velocity: It is the ratio of total displacement to total time taken.

Average velocity = \(\frac{\text { total displacement }}{\text { total time taken }}\)

= \(\frac{x_2-x_1}{t_2-t_1}\)

Average velocity is independent of the path followed by the particle. It just deals with the initial and final positions of the body.

Instantaneous Velocity: The velocity of a body at any particular instant of time is defined as instantaneous velocity.

Mathematically \(\frac{\mathrm{dx}}{\mathrm{dt}}=\underset{\Delta t \rightarrow 0}{\text{Lt}} \frac{\Delta \mathrm{x}}{\Delta \mathrm{t}}\) is defined as instantaneous velocity.

For a body moving with uniform velocity its average velocity = Instantaneous velocity.

Question 8. Show that the maximum height and range of a projectile are \(\frac{\mathrm{U}^2 \sin ^2 \theta}{2 \mathrm{~g}}\) and \(\frac{\mathrm{U}^2 \sin 2 \theta}{\mathrm{g}}\) respectively where the terms have their regular meanings.
Answer:

Let a body be projected with an initial velocity ’u’ and with an angle θ to the horizontal.

Initial velocity along x direction, u_x = u cos θ

Initial velocity along y direction, u_y = u sin θ

Maximum Height: The vertical distance covered by the projectile until its vertical component becomes zero.

from equation v² – u² = 2as ……..(1)

at maximum height v = 0

u = u_y = u sin θ

a = -g

S = H

From equation (1)

0² – u² sin²θ = 2 (- g) H

– u² sin²θ = – 2 gH

∴ H = \(\frac{u^2 \sin ^2 \theta}{2 g}\)

KSEEB Class 11 Physics Chapter 4 Motion In A Plane

Horizontal Range: It is the distance covered by the projectile along the horizontal between the point of projection to the point on the ground, where the projectile returns again.

It is denoted by R. The horizontal distance covered by the projectile at the time of flight is called horizontal range.

Therefore, R = u cos θ x t.

R = \(\mathrm{u} \cos \theta \times \frac{2 \mathrm{u} \sin \theta}{\mathrm{g}}=\frac{\mathrm{u}^2(2 \sin \theta \cos \theta)}{\mathrm{g}}\),

But \(2 \sin \theta \cos \theta=\sin 2 \theta\),

∴ Range(R) = \(\frac{u^2 \sin 2 \theta}{g}\)

Angle Of Projection For Maximum Range: For a given velocity of projection, the horizontal range will be maximum, when sin 2θ = 1.

∴ Angle of projection for maximum range is 2θ = 90° or θ = 45°

∴ \(R_{\max }=\frac{u^2}{g}\)

Question 9. If the trajectory of a body IN parabolic In one reference frame, can It be parabolic in another reference frame that moves at constant velocity with respect to the first reference frame? If the trajectory can be other than parabolic, what else can it be?
Answer:

Yes. According to Newton’s first law, a body at rest or a body moving with uniform velocity is treated as the same. Both of them belong to an inertial frame of reference.

If a frame (say 1) is moving with uniform velocity with respect to others, then that second frame must be at rest or it maintains a constant velocity concerning the first.

So both frames are inertial frames. So if the trajectory of a body in one frame is a parabola, then the trajectory of that body in another frame is also a parabola.

Question 10. A force 2i + j – k Newton acts on a body which is initially at rest. At the end of 20 seconds, the velocity of the body is 4i + 2j – 2k ms^-1. What is the mass of the body?
Answer:

Force, F = 2i + j – k

time, t = 20

Initial velocity, u = 0

Final velocity, v = 4i + 2j – 2k = 2(2i + j – k)

Acceleration, \(\mathrm{a}=\frac{\mathrm{v}-\mathrm{u}}{\mathrm{t}}\)

= \(\frac{2(2 i+j-k)-0}{20}=\frac{2 i+j-k}{10}\)

Mass of the body, m = \(\frac{F}{a}\)

= \(\frac{2 i+j-k}{(2 i+j-k) / 10}=10 \mathrm{~kg}\)

Chapter 4 Motion In-Plane Problems In KSEEB Physics Class 11

Question 1. Ship A is 10 km due west of ship B. Ship A is heading directly north at a speed of 30 km/h, while ship B is heading in a direction 60° west of north at a speed of 20 km/h.

1. Determine the magnitude of the velocity of ship B relative to ship A.
2. What will be their distance of closest approach?

Answer:

The velocity of A 30 kmph due North

∴ V_A=30 \(\hat{j}\)

Velocity of B = 20 kmph 60° west of North

∴ \(V_B=-20 \sin 60^{\circ}+20 \cos 60^{\circ}=10 \sqrt{3} \hat{\mathrm{i}}+10 \hat{\mathrm{j}}\)

Velocity of B with respect to A = V_B – V_A

= \(-10 \sqrt{3} \hat{\mathrm{i}}+10 \hat{\mathrm{j}}-30 \hat{\mathrm{j}}=-10 \sqrt{3} \hat{\mathrm{i}}-20 \hat{\mathrm{j}}\)

∴ \(\mathrm{V}_R=\left|\mathrm{V}_B-\mathrm{V}_A\right|=\sqrt{100 \times 3+400}\)

= \(10 \sqrt{7} \mathrm{kmph}\)

Shortest Distance: In Δ le ANB shortest distance, AN = AB sin θ

But distance, AB = 10 km

∴ AN = \(10 \times \frac{20}{10 \sqrt{7}}=\frac{20}{\sqrt{7}}=7.56 \mathrm{~km}\)

Motion In A Plane KSEEB Physics

Question 2. If θ is the angle of projection, R is the range, h is the maximum height, and T is the time of flight, then show that

1. tan θ = 4h/R and
2. h = gT²/8

Answer:

1. Given angle of projection = η,

Range, \(R=\frac{u^2 \sin 2 \theta}{g}\)

h = \(\mathrm{h}_{\max }=\frac{\mathrm{u}^2 \sin ^2 \theta}{2 \mathrm{~g}} \text {. }\)

Time of flight \(\mathrm{T}=\frac{2 \mathrm{u} \sin \theta}{\mathrm{g}}\)

= \(\frac{\sin ^2 \theta}{2 \sin 2 \theta}=\frac{\sin \theta \sin \theta}{2 \times 2 \sin \theta \cos \theta}\)

∴ \(\frac{h}{R}=\frac{\tan \theta}{4} \Rightarrow \tan \theta=\frac{4 h}{R}\)

2. \(\mathrm{h}=\frac{\mathrm{u}^2 \sin ^2 \theta}{2 \mathrm{~g}}\)

But \(\mathrm{T}=\frac{2 \mathrm{u} \sin \theta}{\mathrm{g}} \Rightarrow \frac{\mathrm{u} \sin \theta}{\mathrm{g}}=\mathrm{T}/2\).

multiply with (g/g)

h = \(\frac{u^2 \sin ^2 \theta}{2 g}=\frac{u^2 \sin ^2 \theta}{2 \cdot g^2} \cdot g\)

= \(\left(\frac{T}{2}\right)^2 \times \frac{g}{2}=\frac{g T^2}{8}\)

∴ \(\mathrm{h}=\frac{\mathrm{gT}^2}{8}\) is proved.

Question 3. A projectile is fired at an angle of 60° to the horizontal with an initial velocity of 800 m/s:

1. Find the time of flight of the projectile before it hits the ground.
2. Find the distance it travels before it hits the ground (range).
3. Find the time of flight for the projectile to reach its maximum height.

Answer:

Angle of projection, θ = 60°,

Initial velocity, u = 800 m/s

1. Time of flight, \(\mathrm{T}=\frac{2 \mathrm{u} \sin \theta}{\mathrm{g}}\)

= \(\frac{2 \times 800}{9.8} \times \frac{\sqrt{3}}{2}=\frac{800 \times 1.732}{9.8}=141.4 \mathrm{sec}\)

2. Range, \(R =\frac{u^2 \sin 2 \theta}{g}\)

= \(\frac{800 \times 800 \times \sin \left(2 \times 60^{\circ}\right)^{\prime}}{9.8}\)

R = \(\frac{800 \times 800}{9.8} \frac{\sqrt{3}}{2}\)

= \(\frac{800 \times 800}{9.8} \times 0.8660=56.56 \mathrm{~km}\)

3. Time of flight to reach maximum height = \(\frac{T}{2}\)

= \(\frac{800 \times 0.8660}{9.80}=70.7 \mathrm{sec}\)

Motion In A Plane Problems In KSEEB Physics

Question 4. For a particle projected slantwise from the ground, the magnitude of its position vector with respect to the point of projection, when it is at the highest point of the path is found to be √2 times the maximum height reached by it. Show that the angle of projection is tan^-1 √2.
Answer:

The position vector of h (max point) from 0, is \(\frac{\mathrm{R}}{2}=\frac{1}{2} \frac{\mathrm{u}^2 \sin 2 \theta}{\mathrm{g}}\)

But R/2 = √2 h_max(given)

∴ \(\frac{1}{2} \frac{u^2 \sin 2 \theta}{g}=\frac{\sqrt{2} \mathrm{u}^2 \sin ^2 \theta}{2 g} \)

∴ \(2 \sin \theta \cos \theta=\sqrt{2} \sin \theta \cdot \sin \dot{ }\)

⇒ \(\frac{2}{\sqrt{2}}=\frac{\sin \theta}{\cos \theta} \Rightarrow \tan \theta=\sqrt{2}\)

∴ Angle of projection, \(\theta=\tan ^{-1} \sqrt{2}\)

Question 5. An object is launched from a cliff 20 m above the ground at an angle of 30° above the horizontal with an initial speed of 30 m/s. How far horizontally does the object travel before landing on the ground? (g = 10 m/s²).
Answer:

Height of cliff = 20m

The angle of projection, θ = 30°

Velocity of projection, u = 30 m/s

Total horizontal distance travelled = OC = OB’+ B’C

But OB’ = AB = Range R = \(\frac{u^2 \sin 2 \theta}{\mathrm{g}}\)

∴ R = \(\frac{30 \times 30 \sin 60^{\circ}}{10}=90 \frac{\sqrt{3}}{2}=45 \sqrt{3} \rightarrow(1)\)

2. Distance B’C = Range of a horizontal projectile.

∴ Range = u cos θ t

u \(\cos \theta=30 \cdot \frac{\sqrt{3}}{2}=15 \sqrt{3}\)

Time taken to reach the ground, t =?

Given \(S_y=20, \quad u_y=u \sin \theta=30 \sin 30^{\circ}\)

= \(15 \mathrm{~m} / \mathrm{s}\)

from \(\mathrm{s}=\mathrm{ut}+\frac{1}{2} a t^2\)

∴ \(\mathrm{s}_{\mathrm{y}}=20=15 \mathrm{t}+\frac{10}{2} \mathrm{t}^2 \Rightarrow 5 \mathrm{t}^2+15 \mathrm{t}-20=0\)

or \(t^2+3 t-4=0\) or \((t+4)(t-1)=0\)

∴ \(\mathrm{t}=-4\) or \(\mathrm{t}=1\)

∴ \(\mathrm{t}\) is Not -ve use t=1

∴ Range = \(\mathrm{u} \cdot \cos \theta \cdot \mathrm{t}=15 \sqrt{3} \times 1 \rightarrow(2)\)

Total distance travelled before reaching the equation (1) + (2) ground = 45√3 + 15√3 =60√3 m.

KSEEB Class 11 Physics Motion In A Plane Questions And Answers

Question 6. ‘O’ is a point on the ground chosen as origin. A body first suffers a displacement of 10√2 mm North-East, next 10 m North and finally 10√2 North-West. How far it is from its origin?
Answer:

1. 10√2 m North-East
2. 10 m North
3. 10√2 m North West

From the figure total displacement from the origin, ‘O’ is OC

But OC = OA’ + A’B’ + B’C = 10 + 10 + 10 = 30 m.

Question 7. From a point on the ground, a particle is projected with initial velocity u, such that its horizontal range is maximum. Find the magnitude of average velocity during its ascent.
Answer:

Velocity of projection = u.

Range is maximum ⇒ θ = 45°

During time of ascent ⇒ when h = h_max ⇒ u_x = v_x= u · cos θ

Average velocity, \(V_A=\sqrt{V_x^2+V_y^2}\)

V_x = Average velocity along X-axis = \(\frac{\mathrm{u} \cdot \cos \theta+\mathrm{u} \cos \theta}{2}\)

⇒ \(\mathrm{v}_{\mathrm{x}}=\mathrm{u} \cdot \cos \theta=\mathrm{u} \cdot \cos 45^{\circ}=\mathrm{u} / \sqrt{2} \longrightarrow(1)\)

Average velocity along \(\mathrm{Y} \text {-axis }=\frac{\mathrm{u}_{\mathrm{y}}+\mathrm{v}_{\mathrm{y}}}{2}\)

⇒ \(\mathrm{u}_{\mathrm{y}}=\mathrm{u} \sin \theta=\mathrm{u} / \sqrt{2}, \mathrm{v}_{\mathrm{y}}=0\left(\text { at }_{\mathrm{h}_{\max }}\right)\)

∴ \(\mathrm{v}_{\mathrm{y}}=\frac{\mathrm{u} / \sqrt{2}}{2}=\frac{\mathrm{u}}{2 \sqrt{2}} \longrightarrow(2)\)

Average velocity during the time of ascent

= \(\sqrt{\left(\frac{u}{\sqrt{2}}\right)^2+\left(\frac{u}{2 \sqrt{2}}\right)^2}\)

Average velocity

= \(\sqrt{\frac{u^2}{2}+\frac{u^2}{4 \times 2}}=\sqrt{\frac{4 u^2+u^2}{8}}=\frac{u \sqrt{5}}{2 \sqrt{2}}\)

Question 8. A particle is projected from the ground with some initial velocity making an angle of 45° with the horizontal. It reaches a height of 7.5 m above the ground while it travels a horizontal distance of 10 m from the point of projection. Find the initial speed of projection (g = 10 m/s²).
Answer:

Angle of projection = 45°

Vertical height, h_y = 7.5 m

Horizontal distance, h_x = 10 m

From figure \(\tan \theta=\frac{y}{x}=\frac{7.5}{10}=3 / 4, \cos \theta=\frac{4}{5}\)

⇒ \(\mathrm{s}_{\mathrm{x}}=\mathrm{u} \cdot \cos \theta \cdot \mathrm{t} \Rightarrow 10=\mathrm{u} \cdot \frac{1}{\sqrt{2}} \cdot \mathrm{t}\)

t = \(\frac{10 \sqrt{2}}{\mathrm{u}}\)…..(1)

⇒ \(\mathrm{s}_{\mathrm{y}}=7.5=(\mathrm{u} \sin \theta) \mathrm{t}-\frac{1}{2} g \mathrm{t}^2\)

7.5 = \(\frac{\mathrm{u}}{\sqrt{2}} \frac{10 \sqrt{2}}{\mathrm{u}}-\frac{1}{2} \times \frac{10 \times 10 \times 10 \times 2}{\mathrm{u}^2}\)

7.5 = \(10-\frac{1000}{\mathrm{u}^2} \Rightarrow-2.5=\frac{-1000}{\mathrm{u}^2}\)

∴ \(\mathrm{u}^2=\frac{1000}{2.5}=400\)

∴ \(u^2=400 \Rightarrow u=20 \mathrm{~m} / \mathrm{s}\)

Question 9. The wind is blowing from the south at 5 ms^-1. To a cyclist, It appears to be blowing from the cast at 5 ms^-1. Show that the velocity of the cyclist is ms^-1 towards the north-cast.
Answer:

The direction of wind is South to North 5 m/s.

The apparent direction is from East to West at 5 m/s.

This is relative velocity.

To find the velocity of the cyclist reverse the direction of the resultant vector OB and find the resultant

∴ Velocity of cyclist = \(\sqrt{5^2+5^2+0}\)

= \(5 \sqrt{2} \mathrm{~m}\)

Motion In A Plane KSEEB Physics

Question 10. A person walking at 4 m/s finds raindrops falling slantwise into his face at a speed of 4 m/s at an angle of 30° with the vertical. Show that the actual speech of the raindrops is 4 m/s.
Answer:

Velocity of man = 4 m/sec

The apparent velocity of raindrop = 4 m/sec with θ = 30° with vertical.

This is relative velocity V_B.

1. Velocity of man = 4 \(\hat{i}\)

Velocity of rain = 4 m/s, 30° with vertical

∴ It is represented by = -4 sin 30 i – 4.cos 30° \(\hat{j}\)

= \(-4 \cdot \frac{1}{2} \hat{\mathrm{i}}+4 \frac{\sqrt{3}}{2} \text { say } V_R=-2 \hat{\mathrm{i}}+2 \sqrt{3} \hat{\mathrm{j}}\)

But V_R = V_B – V_A where V_B is the actual velocity of rain

⇒ \(V_B-V_A+V_A=V_B\)

= \(-2 \hat{i}+2 \sqrt{3} \hat{j}+4 \hat{\mathrm{i}}=2 \hat{\mathrm{i}}+2 \sqrt{3} \hat{\mathrm{j}}\)

∴ \(\left|V_B\right|=\sqrt{4+4 \times 3}=\sqrt{16}=4 \mathrm{~m} / \mathrm{s}\)

Question 11. The ceiling of a long hall is 25 m high. What is the maximum horizontal distance that a ball thrown with a speed of 40 ms_^-1 can go without hitting the ceiling of the hall?
Solution:

Here, u = 40 ms^-1; H = 25m, R = ?

Let θ be the angle of projection with the horizontal direction to have the maximum range, with maximum height = 25 m.

Maximum height, \(\mathrm{H} \Rightarrow 25=\frac{\mathrm{u}^2 \sin ^2 \theta}{2 \mathrm{~g}}\)

= \(\frac{(40)^2 \sin ^2 \theta}{2 \times 9.8}\)

or \(\sin \theta=\left(\frac{25 \times 2 \times 9.8}{40^2}\right)=0.5534\)

= \(\sin 33.6^{\circ}\) or \(\theta=33.6^{\circ}\)

Horizontal range, \(\mathrm{R}=\frac{\mathrm{u}^2 \sin 2 \theta}{\mathrm{g}}\)

= \(\frac{(40)^2 \times \sin \left(2 \times 33.6^{\circ}\right)}{9.8}\)

= \(\frac{1600 \times 0.9219}{9.8}=150.5 \mathrm{~m}\)

Question 12. A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 seconds, what is the magnitude and direction of acceleration of the stone?
Solution:

Here, r = 80 cm = 0.8 m; v = 14/25 s^-1.

∴ \(\omega=2 \pi v=2 \times \frac{22}{7} \times \frac{14}{25}=\frac{88}{25} \text { rad. s } \mathrm{s}^{-1} \text {. }\)

The centripetal acceleration, \(\mathrm{a}=\omega^2 \mathrm{r}=\left(\frac{88}{25}\right)^2 \times 0.80=9.90 \mathrm{~m} / \mathrm{s}^2\)

The direction of centripetal acceleration is along the string directed towards the centre of the circular path.

KSEEB Class 11 Physics Motion In A Plane Key Concepts

Question 13. An aircraft executes a horizontal loop of radius 1 km with a steady speed of 900 km/h. Compare its centripetal acceleration with the acceleration due to gravity.
Solution:

Here, r = 1 km = 1000 m;

v = 900 km h^-1 = 900 x (1000m) x (60 x 60s)^-1 = 250 ms^-1

Centripetal acceleration, \(\mathrm{a}=\frac{\mathrm{v}^2}{\mathrm{r}}=\frac{(250)^2}{1000}\)

Now, \(\frac{\mathrm{a}}{\mathrm{g}}=\frac{(250)^2}{1000} \times \frac{1}{9.8}=6.38\)

Question 14. An aircraft is flying at a height of 3400 m above the ground. If the angle subtended at a ground observation point by the aircraft positions 10.0 s apart is 30°, what is the speed of the aircraft?
Solution:

Figure O is the observation point at the ground. A and B are the positions of aircraft for which ∠AOB = 30°.

Draw a perpendicular OC on AB. Here OC = 3400 m and ∠AOC = ∠COB = 15°. Time taken by aircraft from A to B is 10 s.

In ΔAOC, AC = OC tan 15° = 3400 x 0.2679

AC = 910.86 m.

AB = AC + CB = AC + AC = 2 AC

= 2 x 910.86 m

Speed of the aircraft, v = \(\frac{\text { distance } A B}{\text { time }}\)

= \(\frac{2 \times 910.86}{10}=182.17 \mathrm{~ms}^{-1}=182.2 \mathrm{~ms}^{-1}\).

Question 15. A bullet fired at an angle of 30° with the horizontal hits the ground 3.0 km away. By adjusting its angle of projection, can one hope to hit a target 5.0 km away? Assume the muzzle speed to be fixed and neglect air resistance.
Solution:

Horizontal range, R= \(\frac{u^2 \sin 2 \theta}{g}\)

or, 3 = \(\frac{u^2 \sin 60^{\circ}}{g}\) or \(\frac{u^2}{g}=2 \sqrt{3}\)

3 = \(\frac{u^2}{g} \times \frac{\sqrt{3}}{2}\)

2\(\sqrt{3}=\frac{u^2}{g}\)

∴ \(R_{\max }=2 \sqrt{3}\)

∴ \(R_{\max }=3.464 \mathrm{~m} \).

KSEEB Class 11 Physics Solutions For Motion in a Straight Line Chapter 3 Motion In A Straight Line Very Short Answer Questions

Question 1. The states of motion and rest are relative Explain.
Answer:

Rest: If the position of a body does not change with respect to its surroundings, it is said to be at “rest”.

Motion: If the position of a body changes with respect to its surroundings, it is said to be in “motion”.

By definitions rest and motion are relative with respect to surroundings.

Read and Learn More KSEEB Class 11 Physics Solutions

Question 2. How is average velocity different from instantaneous velocity?
Answer:

Average Velocity: It is the ratio of total displacement to total time taken. It is independent of the path of the body.

∴ Average velocity = \(\frac{\mathrm{s}_2-\mathrm{s}_1}{\mathrm{t}_2-\mathrm{t}_1}(\mathrm{Or}) \mathrm{V}_{\text {avg }}=\frac{\mathrm{ds}}{\mathrm{dt}}\)

The velocity of a particle at a particular instant of time is known as instantaneous velocity. Here time interval is very small.

V = \(\frac{dx}{dt}\)

Only in uniform motion, instantaneous velocity = average velocity. For all other, cases instantaneous velocity may differ from average velocity.

Question 3. Give an example where the velocity of an object is zero but its acceleration is not zero.
Answer:

In the case of a Vertically Projected Body at maximum height its velocity v = 0. But acceleration due to gravity ’g’ is not zero

So even though velocity v = 0 ⇒ acceleration is not zero.

Question 4. A vehicle travels half the distance L with speed v₁ and the other half with speed v₂ What is the average speed?
Answer:

The average speed of a vehicle for the two  equal parts \(\frac{L}{2}\), \(\frac{L}{2}\) is

⇒ \(V_{\text {avg }}=\frac{\text { Total distance }}{\text { Total time }}=\frac{\frac{L}{2}+\frac{L}{2}}{t_1+t_2}\) (because \(t=\frac{L / 2}{V}\))

⇒ \(V_{\text {avg }}=\frac{L}{\left(\frac{\frac{L}{2}}{V_1}+\frac{\frac{L}{2}}{V_2}\right)} \)

= \(\frac{L}{\frac{L}{2}\left(\frac{1}{V_1}+\frac{1}{V_2}\right)}\)

= \(\frac{1}{\frac{1}{2}\left(\frac{V_1+V_2}{V_1 V_2}\right)}\)

∴ \(V_{\text {avg }}=\frac{2 V_1 V_2}{V_1+V_2}\)

Motion In A Straight Line Solutions KSEEB Class 11 Physics

Question 5. A lift coming down is just about to reach the ground floor. Taking the ground floor as the origin and positive direction upwards for all quantities, which one of the following is correct?

1. x < 0, v < 0, a > 0
2. x > 0, v < 0, a < 0
3. x > 0, v < 0, a > 0
4. x > 0, v > 0, a > 0

Answer:

As the lift is coming down, the value of x becomes less hence negative, i.e., x < 0.

Velocity is downwards (i.e., negative). So v < 0.

Just before reaching the ground floor, the lift is retarded, i.e., acceleration is upwards. Hence a > 0.

We can conclude that x < 0, v < 0 and a > 0.

∴ (1) is correct.

Question 6. A uniformly moving cricket ball is hit with a bat for a very short time and is turned back. Show the variation of its acceleration with time taking the acceleration in the backward direction as positive.
Answer:

For a ball moving with uniform velocity acceleration is zero. But during the time of contact between the ball and bat acceleration is applied in the opposite direction. The shape of the acceleration – time graph is as shown.

Question 7. Give an example of one-dimensional motion where a particle moving along the positive x-direction comes to rest periodically and moves forward.
Answer:

When the length of the pendulum is high and the amplitude is less then its motion is along a straight line. The pendulum will come to a stop at the extreme position and move back in the forward direction (‘x’ + ve) periodically.

Question 8. An object falling through a fluid is observed to have an acceleration given by a = g – bv where g is the gravitational acceleration and b is a constant. After a long time, it is observed to fall with a constant velocity. What would be the value of this constant velocity?
Answer:

Acceleration, a = g – bv when moving with constant velocity, a = 0 ⇒ 0 = g – bv

∴ Constant velocity, v = \(\frac{g}{b}\) m/sec.

Question 9. If the trajectory of a body is parabolic in one frame, can it be parabolic in another frame that moves with a constant velocity with respect to the first frame? If not, what can it be?
Answer:

If the trajectory of a body is parabolic with reference frames one and two then those two frames are of rest or moving with uniform velocity.

If they are not parabolic then for that reference frame it may be in a straight line path.

Example: When a body is dropped from a moving plane its path is parabolic for a person outside the plane. But for the pilot in the plane, it is falling vertically downwards.

Question 10. A spring with one end attached to amass and the other to rigid support is stretched and released. When is the magnitude of acceleration a maximum?
Answer:

Maximum restoring force set up in the spring when stretched by a distance ‘r’ is F = – kr

Potential energy of stretched spring = \(\frac{1}{2}\) kx²

As F ∝ r and this force are directed towards the equilibrium position, hence if the mass is left free, it will execute damped SHM due to gravity pull.

Magnitude of acceleration in the mass attached to one end of spring when just released is \(\mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}=\frac{-\mathrm{k}}{\mathrm{m}} \mathrm{r}=(\text { Maximum })\)

∴ The magnitude of the spring’s acceleration will be maximum when it is just released.

KSEEB Class 11 Physics Chapter 3 Solutions

Question 11. Define average velocity and average speed. When does the magnitude of average velocity become equal to the average speed?
Answer:

Average Velocity: It is defined as the ratio of total displacement to total time taken.

Average velocity = \(\frac{\text { total displacement }}{\text { total time taken }}=\frac{\mathrm{x}_2-\mathrm{x}_1}{\mathrm{t}_2-\mathrm{t}_1}\)

Average velocity is independent of the path followed by the particle. It just deals with the initial and final positions of the body.

Average Speed: The ratio of the total path length travelled to the total time taken is known as “average speed”.

Speed and average speed are scalar quantities so no direction for these quantities.

Average speed = \(\frac{\text { Total path length }}{\text { Total time interval }}\)

When the body is along with the straight line its average velocity and average speed are equal.

Solutions For Motion In A Straight Line KSEEB Physics Chapter 3 Short Answer Questions

Question 1. Can the equations of kinematics be used when the acceleration varies with time? If not, what form would these equations take?
Answer:

The equations of motion are

1. v =u + at
2. s = ut + \(\frac{1}{2}\)at² and
3. v² – u² = 2as.

All these three equations are applicable to body moves with uniform acceleration ‘a’.

No, the equations are not applicable when the acceleration varies with time.

Question 2. A particle moves in a straight line with uniform acceleration. Its velocity at time t = 0 is v₁ and at time t₂ = t is v₂. The average velocity of the particle in this time interval is (v₁ + v₂) / 2. Is this correct? Substantiate your answer.
Answer:

t₁ = 0 ⇒ u = v₁; t₂ = t ⇒ v = v₂

Average velocity = \(\frac{u+v}{2}=\frac{v_1+v_2}{2}\)

Question 3. Can the velocity of an object be in a direction other than the direction of acceleration of the object? If so, give an example.
Answer:

Yes, the Velocity of a body and its acceleration may be in different directions.

Explanation:

1. In the case of a vertically projected body ⇒ velocity of the body is in the upward direction and acceleration is in a downward direction.
2. When brakes are applied the velocity of the body before coming to rest is opposite to retarding acceleration.

Question 4. A parachutist flying in an aeroplane jumps when it is at a height of 3 km above the ground. He opens his parachute when he is about 1 km above ground. Describe his motion.
Answer:

Initially, the path is a parabola as seen by an observer on the ground. It is a vertical straight line as seen by the pilot. He opens his parachute, it is moving vertically downwards with decreasing velocity and finally, it reaches the ground.

Question 5. A bird holds n fruit In its beak and flics parallel to the ground. It lets go of the fruit at some height. Describe the trajectory of the fruit as it falls to the ground as seen by

1. The bird
2. A person on the ground.

Answer:

1. As the bird Is flying parallel to the ground, it possesses velocity in the horizontal direction. Hence the fruit also possesses velocity in the horizontal direction and acceleration in a downward direction. Hence the path of the fruit is a straight line with respect to the bird.
2. With respect to a person on the ground, the fruit seems to be on a parabolic path.

Question 6. A man runs across the roof of a tall building and jumps horizontally onto the (lower) roof of an adjacent building. If his speed is 9 ms^-1 the horizontal distance between the buildings is 10 m and the height difference between the roofs is 9 m, will he be able to land on the next building? (take g = 10 ms^-2)
Answer:

Given that,

initial speed, u = 9 ms^-1 ; g = 10m/s²

The height difference between the roofs, h = 9 m

Horizontal distance between two buildings, d = 10 m

Time of flight \(\mathrm{T}=\sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}}=\sqrt{\frac{2 \times 9}{10}}\)

= \(\sqrt{1.8}=1.341 \mathrm{~s}\)

Range of the man = R = u x T = 9 x 1.341 s = 12.069 m

Since R > d, the man will be able to land on the next building.

Question 7. A ball is dropped from the roof of a tall building and simultaneously another ball Is thrown horizontally with some velocity from the same roof. Which ball lands first? Explain your answer.
Answer:

Let ‘h’ be the height of the tall building.

For Dropped Ball: Let ‘t₁‘ be the time taken by the dropped ball to reach the ground.

Initial velocity, u = 0 ; Acceleration, a = + g

Distance travelled, s = h; Time of travel, t = t₁

From the equation of motion, s = ut + \(\frac{1}{2}\) at²

We can write, h = \(0 \times t_1+\frac{1}{2} \times g \times t_1^2 \Rightarrow h=\frac{1}{2} g t_1^2 \Rightarrow t_1^2=\frac{2 h}{g}\)

(or) \(t_1=\sqrt{\frac{2 h}{g}}\)…..(1)

For Horizontally Projected Ball: If the ball is thrown horizontally then its initial velocity along the vertical direction is zero and in this case, let ‘t₂‘ be the time taken by the ball to reach the ground.

Again from the equation of motion,

s = \(\mathrm{ut}+\frac{1}{2} \mathrm{at}^2\) we can write,

⇒ h = \(0 \times \mathrm{t}_2+\frac{1}{2} \mathrm{~g} \times \mathrm{t}_2^2\)

⇒ h = \(\frac{1}{2} \mathrm{gt}_2^2 ;\)

⇒ \(\mathrm{t}_2^2=\frac{2 \mathrm{~h}}{\mathrm{~g}}\) (or)

∴ \(\mathrm{t}_2=\sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}}\)…….(2)

From equation (1) and (2) t₁ = t₂

i.e., both the balls reach the ground at the same time.

Question 8. A ball Is dropped from a building and simultaneously another ball is projected upward with some velocity. Describe the change In relative velocities of the balls as a function of time.
Answer:

1. For a body dropped from building its velocity, v₁ = gt…..(1) (u₁ = 0)
2. For a body thrown up with a velocity ‘u’ its velocity, v² = u – gt….(2)

The two balls are moving in the opposite direction the relative velocity, v_R = v₁ + v₂

∴ v_R = gt + u – gt = u

Here the relative velocity remains constant, but the velocity of one body increases at a rate of ‘g’ m/sec and the velocity of another body decreases at a rate of ‘g’m/sec.

Question 9. A typical raindrop is about 4 mm in diameter. If a raindrop falls from a cloud which is 1 km above the ground, estimate its momentum when it hits the ground.
Answer:

Diameter, D = 4 m

⇒ radius, r = 2mm = 2 x 10^-3 m

mass of raindrop = volume x density = \(\frac{4}{3} \pi r^3 \times 1000 \mathrm{~m}\)

(mass of one m³ of water = 1000 kg)

∴ m = \(\frac{4}{3} \times \frac{22}{7} \times\left[2 \times 10^{-3}\right]^3 \times 1000\)

= \(33.52 \times 10^{-6} \mathrm{~kg}\)

Velocity of raindrop \(v=\sqrt{2 g h}\)

But h = 1 km = 1000 m

∴ \(\mathrm{v}=\sqrt{2 \times 9.8 \times 1000}=\sqrt{19600}\) = 140 m/sec

Momentum of raindrop, \(\vec{p}=m v\)

= 33.52 x 10^-6 x 140 = 4.693 x 10^-3 kg-m

Question 10. Show that the maximum height reached by a projectile launched at an angle of 45° is one-quarter of its range.
Answer:

In projectiles Range, \((R)=\frac{u^2 \sin 2 \theta}{g}\)

Maximum height \((H)=\frac{u^2 \sin ^2 \theta}{2 g}\)

When \(\theta=45^{\circ}\)

⇒ \(\mathrm{R}_1=\mathrm{R}_{\max }=\frac{\mathrm{u}^2 \sin 90^{\circ}}{\mathrm{g}}\)

∴ \(\mathrm{R}_{\max }=\frac{\mathrm{u}^2}{\mathrm{~g}}\)

Maximum height \(H_{\max }=\frac{u^2 \sin ^2\left(45^{\circ}\right)}{2 g}\)

= \(\frac{u^2(1 / \sqrt{2})^2}{2 g}=\frac{u^2}{4 g}\)

∴ \(H_{\max }=\frac{1}{4} R_{\max }\)

∴ When θ = 45° maximum height reached is one-quarter of the maximum range.

Question 11. Derive the equation of motion x = v₀t + \(\frac{1}{2}\) at² using appropriate graph.
Answer:

The velocity-time graph of a body moving with initial velocity ‘u’ and with uniform acceleration ‘a is shown. Let V be the velocity of the body after a time t.

In the v -t graph area of the velocity-time graph = total displacement travelled by it. The area under velocity – time graph = area of OABCD

∴ Area of Rectangular part OACD = Area of OACD + Area of ABC.

A₁ = OA x OD = v₀ t…….(1)

Area of triangle ABC = A₂

A₂ = \(\frac{1}{2}\) Base x height

= \(\frac{1}{2}\) AC x (BC)

= \(\frac{1}{2}\)t(v-v)

But v – v₀ = at

∴ A₂ = \(\frac{1}{2}\)t at = \(\frac{1}{2}\) at²

∴ Total area under graph = s = A₁ + A₂

s(n)= \(v_0 t+\frac{1}{2} a t^2\) is graphically proved.

KSEEB Physics Class 11 Motion in a Straight Line Chapter 3 Problems

Question 1. A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h^-1. Finding the market closed, he instantly turns and walks back home at a speed of 7.5 km h^-1. What is the

1. Magnitude of average velocity and
2. The average speed of the man over the time interval of 0 to 50 minutes?

Solution:

Time taken by man to go from his home to market, \(\mathrm{t}_1=\frac{\text { distance }}{\text { speed }}=\frac{2.5}{7.5}=\frac{1}{3} \mathrm{~h}\)

∴ Total time taken = \(t_1+t_2=\frac{1}{2}+\frac{1}{3}=\frac{5}{6} h\) = 50 min

In time interval 0 to 50 min

Total distance travelled = 2.5 + 2.5 = 5 km.

Total displacement = zero.

1. Average velocity = \(\frac{\text { displacement }}{\text { time }}=0\)
2. Average speed = \(\frac{\text { displacement }}{\text { time }}=\frac{5}{\frac{5}{6}}\) = 6 km/h

Question 2. A stone is dropped from a height of 300 m and at the same time, another stone is projected vertically upwards with a velocity of 100 m/sec. Find when and where the] two stones meet.
Solution:

Height h = 300 m ;

Initial velocity u₀ = 100 m/s

Let the two stones meet at a height ‘x’ above the ground.

For 1st stone \(\mathrm{h}-\mathrm{x}=\frac{1}{2} \mathrm{gt}^2\)…..(1)

For 2nd stone \(\mathrm{x}=\mathrm{u}_0 \mathrm{t}-\frac{1}{2} \mathrm{gt}^2\)

⇒ \(\frac{1}{2} g t^2=u_0 t-x\)….(2)

Since t is the same for the two stones

From equations 1 and 2.

h – x = u₀t – x

⇒ \(\mathrm{u}_0 \mathrm{t}=\mathrm{h} \text { or time } \mathrm{t}=\frac{\mathrm{h}}{\mathrm{u}_{\mathrm{o}}}=\frac{300}{100}=3 \mathrm{sec}\).

⇔ The two stones will meet 3 seconds after the 1st stone is dropped or the 2nd stone is thrown up.

Question 3. A car travels the first third of a distance with a speed of 10 kmph, the second third at 20 kmph and the last third at 60 kmph. What is its mean speed over the entire distance?
Solution:

Total distance = s

distance travelled, \(\mathbf{s}_1=\frac{\mathbf{s}}{3}\)

velocity, v₁ = 10 kmph

distance, \(\mathbf{s}_2=\frac{\mathbf{s}}{3}\)

velocity, v₂ = 20 kmph

distance, \(\mathbf{s}_3=\frac{\mathbf{s}}{3}\)

velocity, v₃ = 60 kmph

Average velocity = \(\frac{\text { total distance }}{\text { total time }}\)

Time, \(t=t_1+t_2+t_3=\frac{s}{3 v_1}+\frac{s}{3 v_2}+\frac{s}{3 v_3}\)

(because time = \(\frac{\text { distance }}{\text { velocity }}\)

∴ \(\mathrm{t}=\frac{\mathrm{s}}{3}\left(\frac{\mathrm{v}_2 \mathrm{v}_3+\mathrm{v}_3 \mathrm{v}_1+\mathrm{v}_1 \mathrm{v}_2}{\mathrm{v}_1 \mathrm{v}_2 \mathrm{v}_3}\right)\)

= \(\frac{\mathrm{s}}{3}\left[\frac{(60 \times 20)+(60 \times 10)+(20 \times 10)}{60 \times 20 \times 10}\right]\)

= \(\frac{\mathrm{s}}{3} \frac{(1200+600+200)}{12000}\)

= \(\frac{\mathrm{s}}{3} \frac{2000}{12000}=\frac{2000 \mathrm{~s}}{3 \times 12000}\) (Or)

∴ \(v_{\text {mean }}==\frac{3 v_1 v_2 v_3}{v_1 v_2+v_2 v_3+v_3 v_1}=18 \mathrm{kmph}\)

Average velocity  = \(\frac{\mathrm{s}}{\mathrm{t}}\)

= \(\frac{3 \times 12000}{2000 \mathrm{~s}}\) = 18 kmph

Question 4. A bullet moving with a speed of 150 m s^-1 strikes a tree and penetrates 3.5 cm before stopping. What is the magnitude of its retardation in the tree and the time taken for it to stop after striking the tree?
Solution:

Velocity of bullet, u = 150 m/s;

Final velocity, v = 0

Distance travelled, s = 3.5 cm = 3.5 x 10^-2 m,

Acceleration, \(a=\frac{v^2-u^2}{2 s}\)

= \(\frac{0^2-150^2}{2 \times 3.50 \times 10^{-2}}\)

= \(\frac{-22500}{7 \times 10^{-2}}\)

= \(-3.214 \times 10^5 \mathrm{~m} / \mathrm{sec}^2\) (-ve sign for retardation)

Time taken to stop, \(\mathrm{t}=\frac{\mathrm{v}-\mathrm{u}}{\mathrm{a}}\)

= \(\frac{-150}{-3.214 \times 10^5}\)

= \(4.67 \times 10^{-4} \mathrm{sec}\)

Motion In A Straight Line Problems in KSEEB Physics

Question 5. A motorist drives north for 30 min at 85 km/h and then stops for 15 min. He continues travelling north and covers 130 km in 2 hours. What is his total displacement and average velocity?
Solution:

In First Part:

Velocity, v₁ = 85 kmph

Time, t₁ = 30 min

Distance travelled, s₁ = v₁ t₁

= 85 x \(\frac{30}{60}\) =42.5 km

In Second Part:

Distance travelled, s₂ = 0;

Time, t₂ = 15.0 min.

In Third Part:

Distance travelled, s₃ = 130 km;

Time, t₃ = 120 min = 2 hours

1. Total distance of the motorist, s = s₁+ s₂ + s₃ = 42.5 + 0 + 130 = 172.5 km

2. Total time travelled, t = t₁ + t₂ + t₃ = 30 + 15 + 120

= 165 minutes

= 2 hrs 45 minutes

= 2\(\frac{3}{4}\) hrs. = \(\frac{11}{4}\) hrs.

∴ Average velocity, \(\mathrm{v}_{\mathrm{avg}}=\frac{\text { total displacement }}{\text { total time }}\)

= \(\frac{172.5}{(11 / 4)}=62.7 \mathrm{kmph}\)

Question 6. A ball A is dropped from the top of a building and at the same time an identical ball B is thrown vertically upward from the ground. When the balls collide the speed of A is twice that of B. At what fraction of the height of the building did the collision occur?
Solution:

Given at time of collision velocity of A = V_A

= 2 x V_B (velocity of B)

Let the body be dropped from a height ‘h’.

Let the two stones collide at x from the ground.

For the body dropped, \(\mathrm{s}=\mathrm{h}-\mathrm{x}=\frac{1}{2} \mathrm{gt}^2\)……(1)

For the body thrown up, x = ut – \(\frac{1}{2}\) gt²…….(2)

For the body dropped, v = u + at ⇒ V_A = gt…….(3)

For the body thrown up, v = u – gt ⇒ V_B = u – gt……..(4)

Given V_A = 2V_B

⇒ gt = 2 (u – gt) or u = \(\frac{3gt}{2}\)…… (5)

Divide equation (1) with equation (2)

⇒ \(\frac{h-x}{x}=\frac{\frac{1}{2} g t^2}{u t-\frac{1}{2} g t^2} \Rightarrow \frac{h}{x}-1=\frac{\frac{1}{2} g t^2}{u t-\frac{1}{2} g t^2}\)

⇒ \(\frac{h}{x}=\frac{\frac{1}{2} g t^2+u t-\frac{1}{2} g t^2}{u t-\frac{1}{2} g t^2}\)

∴ \(\frac{h}{x}=\frac{u t}{u t-\frac{1}{2} g t^2}\)

Put u = \(\frac{3 g t}{2}\) then \(\frac{h}{x}=\frac{3 g t^2 / 2}{3 g t^2 / 2-\frac{1}{2} g t^2}\)

∴ \(\frac{h}{x}=\frac{3 g t^2}{2 g t^2}=\frac{3}{2}\)

∴ x = \(\frac{2}{3} h\)

∴ Fraction of height of collision = \(\frac{2}{3}\)

KSEEB Class 11 Physics Motion in a Straight Line Key Concepts

Question 7. Drops of water fall at regular intervals from the roof of a building of height 16 m. The first drop strikes the ground at the same moment as the fifth drop leaves the roof. Find the distances between successive drops.
Solution:

Height of building, h = 16 m

∴ Time taken to reach the ground, t = \(\sqrt{\frac{2 h}{g}}\)

∴ Time taken to fall, t =\(\sqrt{\frac{2 \times 16}{9.8}}\)

= \(\sqrt{\frac{32}{9.8}}=\sqrt{3.26}=1.80\)

Number of drops, n = 5

∴ Number of intervals = n — 1 = 5 — 1 = 4

Time interval between drops = \(\frac{1.8}{4}\) = 0.45 sec

Time of travel of 1st drop, t₁ = 4 x 0.45 = 1.81

∴ Distance travelled by 1st drop, \(s_1=\frac{1}{2} g t_1^2=\frac{1}{2} \times 9.8 \times 1.8 \times 1.8=16 \mathrm{~m}\)

For 2nd drop, t₂ = 3 x0.45 = 1.35 sec., \(\mathrm{S}_2=\frac{1}{2} \mathrm{gt}_2^2\);

= \(4.9 \times 1.822 \simeq 1.822=9 \mathrm{~m}\)

For 3rd drop, \(\mathrm{t}_3=2 \times 0.45=0.9 \mathrm{sec}\)

Distance \(\mathrm{S}_3=\frac{1}{2} \mathrm{gt}_3^2=\frac{1}{2} \times 9.8 \times 0.9^2 =3.97\) = 4m

For 4th drop, \(t_4=1 \times 0.45=0.45 \mathrm{sec}\)

Distance travelled, \(\mathrm{S}_4=\frac{1}{2} \mathrm{gt}_4^2=\frac{1}{2} \times 9.8\) \(\times(0.45)^2 \simeq 1\)

For 5 th drop, t₅ = 0 ⇒ S₅ = 0

Distance between 1st and 2nd drop S_1,2 = S₁ – S₂ = 16-9 = 7 m

Distance between 2nd and 3rd drop S_2,3 = S₂ – S₃ = 9 – 4 = 5m

Distance between 3rd and 4th drop S_3,4 = S₃ – S₄ = 4 – 1= 3 m

Distance between 4th and 5th drop S_4,5 = S₄ – S₅ = 1-0 = 1 m

∴ Distances between successive drops are 7m, 5m, 3m and lm.

Question 8. Rain is falling vertically with a speed of 35 ms^-1. A woman rides a bicycle with a speed of 12 ms^-1 in east to west direction. What is the direction in which she should hold her umbrella?
Answer:

Velocity of rain V_R = 35 m/s (vertically)

Velocity of women V_w = 12 m/s (towards east)

Resultant angle = \(\theta=\tan ^{-1} \frac{\mathrm{V}_{\mathrm{W}}}{\mathrm{V}_{\mathrm{R}}}=\frac{12}{35}\)

∴ \(\theta=\tan ^{-1}\left[\frac{12}{35}\right]=0.343 \text {. or } \theta=19^{\circ}\) (Nearly)

She should hold an umbrella at an angle of 19° with the east.

Question 9. A hunter aims a gun at a monkey hanging from a tree some distance away. The monkey drops from the branch at the moment he fires the gun hoping to avoid the builet. Explain why the monkey made a wrong move.
Answer:

Let the bullet be fired with an angle α and distance from hunter’s rifle to monkey = x

The vertical component of velocity v_xy = v sin α

when exactly aimed at monkey S_y = v sinα t = h

But acceleration due to gravity \(h_1=u \sin \alpha \mathrm{t}-\frac{1}{2} g t^2=h-\frac{1}{2} g t^2\)….(1)

So the bullet passes through a height of \(\frac{1}{2}\) gt² below the monkey.

But when the monkey is falling freely height  of fall during time t = \(\frac{1}{2}\)

So new height is h – \(\frac{1}{2}\) gt² …… (2)

From equations (1) and (2) h₁ is the same i.e., if the monkey is dropped from the branch bullet will hit it exactly.

KSEEB Physics Motion In A Straight Line Questions And Answers

Question 10. A food packet is dropped from an aeroplane, moving with a speed of 360 kmph in a horizontal direction, from a height of 500m. Find

1. It is a time of descent
2. The horizontal distance between the point at which the food packet reaches the ground and the point above which it was dropped.

Solution:

The velocity of plane, V = 360 kmph = 360 x \(\frac{5}{18}\) = 100 m/s

Height above ground, h = 500 m; g = 10 m/s²

1. Time of descent, \(\mathrm{t}=\sqrt{\frac{2 h}{g}}=\sqrt{\frac{2 \times 500}{10}}=\sqrt{100}=10 \mathrm{sec}\)

2. Horizontal distance between the point of dropping and the point where it reaches the ground = Range R.

∴ R = \(u \sqrt{\frac{2 h}{g}}=100 \sqrt{\frac{2 \times 500}{10}}\) = 100 x 10 = 1000 m.

Question 11. A ball is tossed from the window of a building with an initial velocity of 8 ms^-1 at an angle of 20° below the horizontal. It strikes the ground 3 s later. From what height was the ball thrown? How far from the base of the building does the ball strike the ground?
Solution:

Initial velocity, u = 8 m/s;

The angle of projection, θ = 20°

Time is taken to reach the ground, t = 3 sec.

The horizontal component of initial velocity, u_x = u cos θ = 8 cos 20° = 8 x 0.94 = 7.52 m/s

Vertical component of initial velocity, v_y = u sin θ = 8 sin 20° = 8 x 0.342 = 2.736 m/s

1. From equation of motion, s = ut + \(\frac{1}{2}\) at²

we can write h = (u sinθ)t + \(\frac{1}{2}\) gt²

⇒ h = (2.736)3 + \(\frac{1}{2}\) x 9.8 x(3)² .

⇒ h = 8.208+ 4.9 x 9

⇒ h = 8.208 + 44.1 or h = 52.308 m

2. Horizontal distance travelled, s_x = v_x x t = 7.52 x 3 = 22.56 m

Question 12. Two balls are projected from the same point in directions 30° and 60° with respect to the horizontal. What is the ratio of their initial velocities if they

1. Attain the same height?
2. Have the same range?

Solution:

The angle of projection of the first ball, θ₁ =30°

The angle of projection of the second ball, θ₂ = 60°

Let u₁ and u₂ be the velocities of projections of the two balls.

1. Maximum height of the first ball, \(\left(\mathrm{H}_{\max }\right)_1=\frac{\mathrm{u}_1^2 \sin ^2 30^{\circ}}{2 \mathrm{~g}}\)

Maximum height of second ball, \(\left(\mathrm{H}_{\max }\right)_2=\frac{\mathrm{u}_2^2 \sin ^2 60^{\circ}}{2 \mathrm{~g}}\)

Given that \(\left(H_{\max }\right)_1=\left(H_{\text {max }}\right)_2\)

⇒ \(\frac{u_1^2 \sin ^2 30^{\circ}}{2 g}=\frac{u_2^2 \sin ^2 60^{\circ}}{2 g}\)

⇒ \(u_1^2 \sin ^2 30^{\circ}=u_2^2 \sin ^2 60^{\circ}\)

⇒ \(\frac{u_1^2}{u_2^2}=\frac{\sin ^2 60^{\circ}}{\sin ^2 30^{\circ}} \Rightarrow \frac{u_1^2}{u_2^2}=\frac{(\sqrt{3} / 2)^2}{(1 / 2)^2}\)

⇒ \(\frac{u_1}{u_2}=\frac{\sqrt{3}}{1}\) or \(u_1: u_2=\sqrt{3}: 1\)

2. If the balls have the same range, then \(R_1=R_2\)

i.e., \(\frac{u_1^2 \sin \left(2 \times 30^{\circ}\right)}{g}=\frac{u_2^2 \sin \left(2 \times 60^{\circ}\right)}{g}\)

⇒ \(u_1^2 \sin 60^{\circ}=u_2^2 \sin 120^{\circ}\)

⇒ \(\frac{u_1^2}{u_2^2}=\frac{\sin 120^{\circ}}{\sin 60^{\circ}} \Rightarrow \frac{u_1^2}{u_2^2}=\frac{\cos 30^{\circ}}{\sin 60^{\circ}}\)

⇒ \(\frac{u_1^2}{u_2^2}=\frac{\frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{2}} \text { or } u_1: u_2=1: 1\)

Question 13. A ball is thrown vertically upwards with a velocity of 20 ms^-1 from the top of a multistory building. The height of the point from where the hall Is thrown Is 25.0 in from the ground.

1. How high will the ball rise?
2. How long will it be before the ball hits the ground?
3. Take g = 10 ms^-2 [Actual value of ‘g’ is 9.8 ms^-2]

Solution:

Initial velocity V₀ = 20 m/s;

Height above ground h₀ = 2.50 m ; g = 10 m/s²

1. For a body thrown up vertically height of rise \(\left(y-y_0\right)=\frac{u^2}{2 g}=\frac{20 \times 20}{2 \times 10}=20 \mathrm{~m}\)

2. Time spent in the air (t) is \(y_1-y_0=V_0 t+\frac{1}{2} g t^2\)

Where y₁ = Total displacement of the body from the ground = 0

∴ 0 = \(y_0+V_0 t+\frac{1}{2} g t^2=25+20 t-\frac{1}{2} \cdot 10 . t^2\)

(because \(g=-10 \mathrm{~m} / \mathrm{s}^2\)) while going up

∴ \(0=-5 t^2+20 t+25\) (or) \(t^2-4 t-5=0\)

i.e., (t-5)(t+1)=0

t=5 (or) t=-1

But time is not – ve.

∴ Time spent in air t = 5 sec

KSEEB Class 11 Physics Chapter 3 Motion In A Straight Line 

Question 14. A parachutist flying in an aeroplane jumps when it is at a height of 3 km above ground. He opens his parachute when he is about 1 km above ground. Describe his motion.
Answer:

1. Height of fall before opening, h = 2 km = 2000 m

∴ Velocity at a height of 1 km

= \(\sqrt{2 \mathrm{gh}_1}=\sqrt{2 \times 10 \times 2000}\)

= \(\sqrt{40000}=200 \mathrm{~ms}^{-1}\)

Time of fall, t = \(\sqrt{\frac{2 h}{g}}=\sqrt{\frac{2 \times 2000}{10}}\)

= \(\sqrt{400}=20 \mathrm{sec}\)

2. After the parachute Is opened It touches the ground with almost zero velocity.

∴ u = 200 m/sec, v = 0, S = h = 1000 m

From v² – u² = 2as

Retardation, \(a=\frac{0-200^2}{1000}=-40 \mathrm{~m} / \mathrm{sec}^2\)

Time taken to reach ground, t = \(\frac{u}{a}\) = \(\frac{200}{40}\) = 5 sec

The month is as shown.

Question 15. A car moving along a straight highway with a speed of 126 km/h is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform) and how long does it take for the car to stop?
Solution:

Initial speed = V₀ = 126 kmph

= 126 x \(\frac{5}{18}\) = 35 m/sec

Ending speed = V = 0

Distance taken to stop = X = 200 m.

From V² – V₀² = 2ax .

a = \(\frac{-35}{2 \times 200}\) = \(\frac{-35 \times 35}{2 \times 200}\) = \(\frac{-49}{16}\)

Time taken to stop t =?

From V = V₀ + at

0 = \(35+\frac{49}{16} \times t \Rightarrow t=\frac{35 \times 6}{49}\) = \(\frac{80}{7}=11.43 \mathrm{sec}\)

Units and Measurements Solutions KSEEB Class 11 Physics Chapter 2 Very Short Answer Questions

Question 1. Distinguish between accuracy and precision.
Answer:

Accuracy:

1. It is defined as the closeness of the measured value to the true value.
2. It depends on the minimization of errors.

Precision:

1. It is defined as to what resolution the quantity is measured.
2. It depends on the least count of the measuring instrument.

Read and Learn More KSEEB Class 11 Physics Solutions

Question 2. What are the different types of errors that can occur in a measurement?
Answer:

Types Of Errors

1. Systematic errors and
2. Random errors.

Systematic Errors Are Again Divided Into

1. Imperfection errors
2. Environmental errors and
3. Personal errors.

KSEEB Class 11 Physics Chapter 2

Question 3. How can systematic errors be minimized or eliminated?
Answer:

Systematic Errors Can Be Minimised

1. By improving experimental techniques,
2. By selecting better instruments.

Question 4. Illustrate how the result of a measurement is to be reported indicating the error involved.
Answer:

The result of the measurement is in the form of a number which indicates the precision of measurement along with the unit of the same physical quantity.

Question 5. What are significant figures and what do they represent when reporting the result
of a measurement?
Answer:

Significant figures represent all practically measured digits plus one uncertain digit at the end.

When a result Is reported in this way we can know up to what extent the value Is reliable and also the amount of uncertainty In that reported value.

Question 6. Distinguish between fundamental units| and derived units.
Answer:

The units of fundamental physical| quantities are called fundamental units.

Example: Kg, m, sec, etc.,

The units of derived physical quantities are called derived units.

Example: m/sec, J, m/sec² etc.,

Question 7. Why do we have different units for the same physical quantity?
Answer:

To measure the same physical quantity we have different units by keeping the magnitude of the quantity to be measured.

Example:

1. The measure astronomical distances we will use light year.
   • 1 light year = 9.468 x 10¹⁵ m.
2. To measure atomic distances we will use Angstrom Å (or) Fermi.

Question 8. What is dimensional analysis?
Answer:

Dimensional analysis is a tool to check the relations among physical quantities by using their dimensions.

Dimensional analysis is generally used to check the correctness of derived equations.

KSEEB Class 11 Physics Units And Measurements Notes And Solutions

Question 9. How many orders of magnitude greater is the radius of the atom than that of the nucleus?
Answer:

Size of atom = 10^-10 m,

Size of atomic nucleus = 10^-14 m.

Size of the atom + size of the nucleus is \(\frac{10^{-10}}{10^{-14}}=10^4\)

∴ The size of an atom is 10⁴ times greater than the size of the nucleus.

Question 10. Express unified atomic mass unit in kg.
Answer:

By definition,

1 a.m.u. = \(\frac{1}{12} \times \text { mass of an atom of }{ }_6^{12} \mathrm{C}\)

= \(\frac{1}{12} \times \frac{12 \mathrm{~g}}{6.023 \times 10^{23}}\)

= \(1.67 \times 10^{-24} \mathrm{~g}=1.67 \times 10^{-27} \mathrm{~kg}\)

Chapter 2 Units And Measurements Short Answer Questions

Question 1. The vernier scale of an instrument has 50 divisions which coincide with 49 main scale divisions. If each main scale division is 0.5 mm, then using this instrument what would be the minimum inaccuracy in the measurement of distance?
Answer:

Least count of Vernier Callipers = 1 MSD – 1 VSD

∴ L.C. = 1 MSD – \(\frac{49}{50}\) MSD = \(\frac{1}{50}\) MSD

= \(\frac{1}{50}\) x 0.5 = 0.01 m.m

KSEEB Physics Class 11 Units and Measurements Key Concepts

Question 2. In a system of units, the unit of force is 100N, the unit of length is 10m and the unit of time is 100s. What is the unit of mass in this system?
Answer:

Here, F = MLT^-2 = 100 N …….(1) ;

L = 10 m ; T = 100s

∴ From equation (1)

M X (10) X (100)^-2 = 100

⇒ M X 10^-3 = 100

⇒ M = \(\frac{100}{10_3}\)

⇒ M = 10⁵ kg

Question 3. The distance of a galaxy from Earth is of the. order of 10²⁵ m. Calculate the order of magnitude of the time taken by light to reach us from the galaxy.
Answer:

Size of galaxy = 10²⁵ m,

Velocity of light, c = 3 x 10⁸ ms^-1

Time taken by light to reach Earth,

t = \(\frac{d}{c}=\frac{10^{25}}{3 \times 10^8}=\frac{1}{3} \times 10^{17}=0.3333 \times 10^{17}\)

While calculating the order of magnitude we will consider the power of Ten only.

So the order of magnitude of time taken by light to reach Earth from the galaxy is 10¹⁷ seconds.

Question 4. The Earth-Moon distance is about 60 Earth radius. What will be the approximate diameter of the Earth as seen from the Moon?
Answer:

Earth-moon distance, D = 60r.

Diameter of earth, b = 2r

As seen from the moon

⇒ \(\theta=\frac{\text{Arc~} A B}{\text{Rad}}=\frac{2 \mathrm{r}}{60 \mathrm{r}}=\frac{1}{30} \mathrm{Rad}\)

∴ The size of the earth as seen from the moon is \(\frac{1}{30}\)

radians, or \(\frac{57^{\circ} 30^{\prime}}{30} \cong 1.9^{\circ} \text { (or) } 2^{\circ}\)

Units And Measurements Questions And Answers KSEEB Physics

Question 5. Three measurements of the time for 20 oscillations of a pendulum give t₁ = 39.6 s, t₂ = 39.9 s, and t₃ = 39.5 s. What is the precision of the measurements? What is the accuracy of the measurements?
Answer:

Precision is the least measurable value with that instrument in our case precision is = 0.1 sec.

Calculation Of Accuracy:

Average value of measurements = \(\frac{39.6+39.9+39.5}{3}=\frac{119}{3}=39.67\)

Error in each measurement

⇒ \(\Delta a_1=39.6-39.6=0.07\);

⇒ \(\Delta a_2=39.9-39.67=0.23\);

⇒ \(\Delta a_3=39.67-39.5=0.17\)

∴ \(\Delta a_{\text {mesn }}=\frac{0.07+0.23+0.17}{3}=0.156\)

Precision ± 1 sec. In these measurements only two significant figures are believable. 3rd one is uncertain.

Adjustment of Δa_mean up to the given significant figure = 0.156 adjusted to 0.2.

So our value is accurate up to ±0.2

So our result is 39.67 ± 0.2, when significant figures are taken into account it is 39.7 ± 0.2 sec.

Question 6. 1 calorie = 4.2 J where 1 J = 1 kg m²s^-2. Suppose we employ a system of units in which the unit of mass is α kg, the unit of length is β m, and the unit of time γ s, showing that a calorie has a magnitude 4.2 α^-1 β^-2 γ² in the new system.
Answer:

Here, 1 calorie = 4.2 J = 4.2 kg m²/s² ……(1)

As a new unit of mass = α kg

∴ 1 kg = \(\frac{1}{\alpha}\) new unit of mass

⇒ α^-1 new unit of mass

Similarly 1 m = β^-1 new unit of length and 1 s = γ^-1 new unit of time

Putting these values in (1) we get

1 calorie = 4.2 (α^-1 new unit of mass) (β^-1 new unit of length)² (γ^-1  new unit of time)^-2

= 4.2 α^-1β^-2γ² new unit of energy, which was proved.

Question 7. A new unit of length is chosen so that the speed of light in vacuum is 1 ms^-2. If light takes 8 min and 20s to cover this distance, what is the distance between the Sun and Earth in terms of the new unit?
Answer:

Given the velocity of light in a vacuum,

c = 1 new unit of length s^-1

Time taken by the light of the Sun to reach Earth,

t = 8 min 20s = 8 x 60 + 20 = 500 s

∴ Distance between the Sun and Earth, x = c x t

= 1 new unit of length s^-1 x 500s

= 500 new units of length.

KSEEB Class 11 Physics Chapter 2 Units and Measurements Solutions

Question 8. A student measures the thickness of a human hair using a magnification 100 microscope. He makes 20 observations and finds that the average thickness (as viewed in the microscope) is 3.5 mm. What is the estimate of the thickness of hair?
Answer:

Magnification, \(\mathrm{m}=\frac{\text { observed width }(\mathrm{y})}{\text { real width }(\mathrm{x})}\)

⇒ x = \(\frac{\mathrm{y}}{\mathrm{m}}=\frac{3.5 \mathrm{~mm}}{100}=0.035 \mathrm{~mm}\)

∴ The thickness of the hair is 0.035 mm

Question 9. A physical quantity X is related to four mea¬surable quantities a, b, c, and d as follows?

X = a²b³C^5/2d^-2

The percentage error in the measurement of a, b, c, and d are 1%, 2%, 3% and 4% respectively. What is the percentage error in X?

Answer:

Here, X = a²b³C^5/2d^-2

⇒ \(\frac{\Delta X}{X}= \pm\left[2\left(\frac{\Delta a}{a}\right)+3\left(\frac{\Delta b}{b}\right)+\frac{5}{2}\left(\frac{\Delta c}{c}\right)+2\left(\frac{\Delta d}{d}\right)\right] \)

= \(\pm\left[2(1 \%)+3(2 \%)+\frac{5}{2}(3 \%)+2(4 \%)\right]\)

= ± 423.5 %

The percentage error in X is ± 23.5 %

Question 10. The velocity of a body is given by v = At² + Bt + C. If v and t are expressed in SI what are the units of A, B, and C?
Answer:

From the principle of Homogeneity, the terms At², Bt, and C must have the same dimensional formula of velocity V.

v = \(\text { Velocity }=\mathrm{LT}^{-1} \Rightarrow \mathrm{LT}^{-1}=\mathrm{A}\left[\mathrm{T}^2\right]\)

∴ A = \(\frac{\mathrm{LT}^{-1}}{\mathrm{~T}^2}=\mathrm{LT}^{-3}\) So unit of A is m/\(\mathrm{sec}^3\)

∴ \(\mathrm{LT}^{-1}=\mathrm{BT} \Rightarrow \mathrm{B}=\mathrm{LT}^{-2}\) So unit of B is \(\mathrm{m} / \mathrm{sec}^2\)

∴ \(\mathrm{LT}^{-1}=\mathrm{C}\)

So the unit of C is \(\mathrm{m} / \mathrm{sec}\).

Dimensional Formulae Of Physical Quantities

Units And Measurements Problems In KSEEB Physics Chapter 2

Question 1. In the expression, P = El² m^-5 G^-2 the quantities E, l, m in and G denote energy, angular momentum, mass, and gravitational constant respectively. Show that P is a dimensionless, quantity.
Solution:

Here, P = El² m^-5 G^-2

Here, I = energy, l = angular momentum

m = mass, G = gravitational constant

= \(\left[M \mathrm{~L}^2 \mathrm{~T}^{-2}\right]\left[\mathrm{M} \mathrm{L}^2 \mathrm{~T}^{-1}\right]^2[\mathrm{M}]^{-5}\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]^{-2}\)

= \(M^{1+2-5+2} \mathrm{~L}^{2+1-6} \mathrm{~T}^{-2-2+4}\)

P = \(\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0\right]\)

Hence, P is a dimensionless quantity.

Question 2. If the velocity of light c, Planck’s constant, h, and the gravitational constant G are taken as fundamental quantities; then express mass, length, and time in terms of dimensions of these quantities.
Solution:

Here, \(c=\left[L \cdot T^{-1}\right] ; h=\left[M L^2 \mathrm{~T}^{-1}\right]\);

G = \(\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]\)

∴ E = \(\mathrm{h} v, \mathrm{~h}=\frac{\mathrm{E}}{v} ; \mathrm{G}=\left(\frac{\mathrm{Fd}^2}{\mathrm{~m}_1 \mathrm{~m}_2}\right)\)

Let m = \(c^x h^y G^z \rightarrow(1)\)

⇒ \(\left[M^1 L^0 T^0\right]=\left(L^{-1}\right)^x\left(M L^2 T^{-1}\right)^y\left(M^{-1} L^3 T^{-2}\right)^2\)

⇒ \(\left[M^1 L^0 T^0\right]=M^{y-z} L^{x+2 y+3 z} T^{-x-y-2 z}\)

Applying the principle of homogeneity of dimensions, we get

y – z = 1 → (2); x + 2y. + 3z = 0 → (3);

– x – y – 2z = 0 → (4)

Adding equation (2), equation (3) and equation (4),

2y = 1

⇒ y = \(\frac{1}{2}\)

∴ From eq. (2) z = y – 1 = \(\frac{1}{2}\) – 1 = \(\frac{-1}{2}\)

From eq. (4) x = – y – 2z = \(\frac{-1}{2}\) + 1 = \(\frac{1}{2}\)

Substituting the values of x, y, and z in eq. (1), we get

m = \(\mathrm{c}^{1 / 2} \mathrm{~h}^{1 / 2} \mathrm{G}^{-\sqrt{1 / 2}} \Rightarrow \mathrm{m}=\sqrt{\frac{\mathrm{ch}}{\mathrm{G}}}\)

Proceeding as above we can show that

L = \(\sqrt{\frac{\mathrm{hG}}{c^3}}\) and \(\mathrm{T}=\sqrt{\frac{\mathrm{hG}}{\mathrm{c}^5}}\)

Question 3. An artificial satellite revolves around a planet of mass M and radius R, in a circular orbit of radius r. Using dimensional analysis shows that the period of the satellite.

T = \(=\frac{k}{R} \sqrt{\frac{r^3}{g}}\)

where k is a dimensionless constant and g is the acceleration due to gravity.

Solution:

Given that T² ∝ r³ or T ∝ r^3/2

Also, T is a function of g and R

Let T ∝ r^3/2 g^a R^b where a, b are the dimensions of g and R.

(or) T = k r^3/2 g^a R^b → (1)

where k is a dimensionless constant of proportionality

From equation (1) \(\left[M^0 L^0 T^1\right]=L^{3 / 2}\left(L T^{-2}\right)^a(L)^b=M^0 L^{a+b+\frac{3}{2}} T^{-2 a}\)

Applying the principle of homogeneity of dimensions, we get

a + b + \(\frac{3}{2}=0\)…..(2)

∴ -2 a=1

⇒ a = \(\frac{-1}{2}\)

From eq (1), \(\frac{-1}{2}+b+\frac{3}{2}=0 \Rightarrow b=-1\)

Substituting the values of ‘a’ and ‘b’ in eq. (1), we get

T = \(k r^{3 / 2} \cdot g^{-1 / 2} R^{-1}\)

T = \(\frac{k}{R} \sqrt{\frac{r^3}{g}}\)

This is the required relation.

Units And Measurements KSEEB Class 11 Physics

Question 4. State the number of significant figures in the following

1. 6729
2. 0.024
3. 0.08240
4. 6.032
5. 4.57 x 10⁸

Solution:

1. In 6729 all are significant figures.
   • A number of significant figures are Four.
2. In 0.024 the zeroes to the left of 1st non-zero digit of a number less than one are not significant.
   • A number of significant figures Two.
3. 0.08240 – Significant figures Four.
   • In 6.032 the zero between two non-zero digits is significant.
   • So, a number of significant figures in 6.023 are 4.
4. 4.57 x 10⁸ Significant figures Three.

[In the representation of powers of Ten our rule is only significant figures must be given].

Question 5. One stick has a length of 12.132 cm and another has a length of 12.4 cm. If the two sticks are placed end what is the total length? If the two sticks are placed side by side, what Is the difference in their lengths?
Solution:

1. When placed end to end total length is

l = l₁ + l₂

l₁ = 12.132 cm and l₂ = 12.4 cm.

∴ l₁ + l₂ = 12.132 + 12.4 = 24.532 cm.

In addition, the final answer must have the least number of significant numbers in that addition, i.e., one after the decimal point.

So our answer is 24.5 cm.

2. For difference use l₁ – l₂

i.e., 12.4- 12.132 = 0.268 cm.

In subtraction final answer must be adjusted to the least number of significance figures in that operation.

Here least number is one digit after I decimal. By applying the round-off procedure] our answer is 0.3 cm.

Question 6. Each side of a cube is measured to be 7.203 m. What is

1. The total surface area and
2. The volume of the cube, to appropriate significant figures?

Solution:

Side of the cube, a = 7.203 m.

So a number of significant figures is Four.

1. Surface area of cube = 6a²

= 6x 7.203x 7.203 = 311.2991

But our final answer must be rounded to the least number of significant figures is four digits.

So the surface area of the cube = 311.3 m³

2. Volume of cube, V = a³ = (7.203)³ = 373.1471

But the answer must be limited to Four! significant figures.

∴ The volume of the sphere, V = 373.1 m³.

KSEEB Class 11 Physics solutions for Chapter 2 Units and Measurements

Question 7. The measured mass and volume of a body are 2.42 g and 4.7 cm³ respectively with possible errors 0.01 g and 0.1 cm³. Find the maximum error in density.
Solution:

Mass, m = 2.42 g ; Error, Δm = 0.01 g. Volume, V = 4.7 cm³, Error, ΔV = 0.1 cc.

% error in mass = \(\frac{\Delta \mathrm{m}}{\mathrm{m}} \times 100=\frac{0.01}{2.42} \times 100\)

= \(\frac{1}{2.42}\)

% error in volume = \(\frac{0.1}{4.7} \times 100\) = \(\frac{10}{4.7}\)

Density = \(\frac{\text { Mass }}{\text { Volume }}=\frac{M}{V}\)

Maximum % error in density = % error in mass + % error in volume

∴ Maximum percentage error in density = \(\frac{1}{2.42}+\frac{10}{4.7}=0.413+2.127=2.54 \%\)

Question 8. The error in the measurement of the radius of a sphere is 1%. What is the error in the measurement of volume?
Solution:

Percentages Errors in radius = 1% = \(\frac{\Delta r}{r} \times 100\)

The volume of sphere V ∝ r³

⇒ ΔV = 3r²Δr

⇒ \(\frac{\Delta V}{V}=3 \frac{r^2 \cdot \Delta r}{r}\)

Percentage error in volume \(\frac{\Delta \mathrm{V}}{\mathrm{V}}=3\left(\frac{\Delta \mathrm{r}}{\mathrm{r}} \times 100\right)=3 \times 1=3 \%\)

Question 9. The percentage error in the mass and speed are 2% and 3% respectively. What is the maximum error in kinetic energy calculated using these quantities?
Solution:

Percentage change in mass = \(\frac{\Delta m}{m} \times 100\) = 2%

Percentages chage in speed = \(\frac{\Delta v}{v} \times 100\) = 3%

But K.E = \(\frac{1}{2}\) = mv

Percentages change in K.E = 1(% change mass) + 2(% change in velocity)

∴ Percentage change in K.E

= \(1\left(\frac{\Delta \mathrm{m}}{\mathrm{m}} \times 100\right)+2\left(\frac{\Delta \mathrm{v}}{\mathrm{v}} \times 100\right)\)

= 1 x 2 + 2 x 3 = 8%

Question 10. One mole of an ideal gas at standard temperature and pressure occupies 22.4 L (molar volume). If the size of the hydrogen molecule is about 1Å, what is the ratio of molar volume to the atomic volume of a mole of hydrogen?
Solution:

Size of Hydrogen atom = 1Å =10^-10 m = 10^-8 cm

V₁ = Atomic volume = number of atoms x volume of atom.

One mole gas contains ‘n’ molecules.

Avogadro Number, n = 6.022 x 10²³

∴ \(V_1=\frac{4}{3} \pi r^3 \times n\)

= \(\frac{4}{3} \times \frac{22}{7} \times\left[10^{-8}\right]^3 \times 6.022 \times 10^{23}\)

= \(25.23 \times 10^{-1}\)or \(2.523 \mathrm{cc} \text {. }\)

∴ \(\mathrm{V}_2\) = Molar volume of 1 mol gas = 22.4 lit

= \(2.24 \times 10^4 \mathrm{c} . \mathrm{c}\)

∴ 1 lit =1000 c.c.

∴ The ratio of molar volume to atomic volume

= \(\mathrm{V}_2: \mathrm{V}_1\)

= \(2.24 \times 10^4: 2.523=10^4 \mathrm{~m}\)

KSEEB Class 11 Physics  Chapter 1 Physical World Synopsis

Physics: Physics is the study of nature and natural phenomena.

Fundamental Forces In Nature: In physics

1. Gravitational Force
2. Electromagnetic Force
3. Strong nuclear Force
4. Weak nuclear Force.

Gravitational Force: It is the force of attraction between any two objects by virtue of their masses.

• These are very weak forces. They are very long-distance forces. For heavy bodies like planets and stars etc., the magnitude of these forces is high.
• These forces are very important in planetary motion and in the formation of Stars and Galaxies.

Read and Learn More KSEEB Class 11 Physics Solutions

Electromagnetic Forces: It is the force between two charged particles.

• Between like charges they are “repulsive forces” and between unlike charges, they are “attractive forces”.
• These forces are very strong forces. These are long-distance forces.

Strong Nuclear Forces: Strong nuclear forces bind protons and neutrons in a nucleus.

These are very strong attractive forces. They are 100 times stronger than electromagnetic forces. They are short-range forces. Their effect is up to very few fermi.

Weak Nuclear Forces: Weak nuclear forces will appear only in certain nuclear processes such as β-the decay of the nucleus where the nucleus emits electrons and neutrino.

These are weak forces, their range is up to a few fermi.

KSEEB Class 11 Physics Chapter 1 Solved Examples

Conserved Quantities: In physics, any physical phenomenon is governed by certain forces. Several physical quantities will change with time blit some special physical quantities will remain constant with time. Such physical quantities are called conserved quantities of nature.

Conserved Quantities Example: For motion under an external conservative force such as a gravitational field the total mechanical energy (i.e., P.+K.E) is constant or energy is conserved.

Physical World Solutions KSEEB Class 11 Physics Chapter 1 Very Short Answer Questions

Question 1. What is Physics?
Answer:

Physics is a branch of science that deals with the study of nature and natural phenomena.

Question 2. What is the discovery of C.V. Raman?
Answer:

C.V. Raman’s contribution to physics is the Raman effect. It deals with the scattering of light by molecules of a medium when they] are excited to vibrational energy levels.

Question 3. What are the fundamental forces in nature?
Answer:

There Are Four Fundamental Forces In Nature

1. Gravitational force
2. Electromagnetic force
3. Strong nuclear force
4. Weak nuclear force

Question 4. Which of the following has symmetry?

1. Acceleration due to gravity.
2. Law of gravitation.

Answer:

1. Acceleration due to gravity varies from place to place. So it has no symmetry.
2. The law of gravitation has symmetry because it does not depend on any physical quantity.

Question 5. What is S. Chandra Seklmi’s contribution to physics?
Answer:

S. ChandraSekhnrdlsoovorod Hit-slniclurn and evolutIon ol stars. I defined the “Chandra Sekhar limit” which Is used In the study of black holes.

Question 6. What Is beta (β) decay? Which force is a function of it?
Answer:

In β-decay, the nucleus emits an electron and an uncharged particle called neutrino.

β – decay is due to weak nuclear forces.

Some Physicists And Their Major Contributions

Fundamental Forces Of Nature

Fundamental Constants Of Physics

KSEEB Class 11 Physics Physical World Notes And Solutions

Conversion Factors:

• 1 metre 100 cm
• 1 millimeter 10^-3 m
• 1 inch 2.54 x 10^-2 m
• 1 micron (μ) 10^-4 cm
• 1 Angstrom (A°) 10^-8 cm
• 1 fermi (f) 10^-13 cm
• 1 kilometer 10³ m
• 1 light year 9.46 x 10¹⁵ m
• 1 litre 10³ cm³
• 1 kilogram 1000 gm
• 1 metric ton 1000 kg
• 1 pound 453.6 gm
• 1 atomic mass
• unit (a.m.u) 1.66 x 10^-27 kg
• 1 a.m.u 931 MeV
• 1 day 8.640 x 10⁴ seconds
• 1 km/hour \(\frac{5}{8} m/sec or 0.2778 meter/sec
• 1 Newton 10⁵ dynes
• 1 gm wt 980.7 dynes
• 1 kg. wt 9.807 Newton
• 1 Newton/meter² 1 Pascal
• 1 atmospheric pressure 1.0133 x 10⁵ pascal (N/m²) or, 76 cm of Hg
• 1 Pascal 10 dyne/cm²
• 1 Joule 10⁷ erg
• 1 kilo watt hour 3.6 x 10⁶ joule
• 1 electron volt (ev) 1.602 x 10^-19 joule
• 1 watt 1 joule/sec
• 1 horsepower (HP) 746 watt
• 1 degree (°) 60 minute (‘)
• 1 Radian 57.3 degree (° )
• 1 Poise 1 dyne. sec / cm²
• 1 Poiseuille 10 poise (Newton, sec/m2² (or) Pascal sec.)

Important Prefixes:

The Greek Alphabet:

Formulae Of Geometry:

1. Area of triangle = 1/2 x base x height
2. Area of parallelogram = base x height
3. Area of square = (length of one side)²
4. Area of rectangle = length x breadth
5. Area of circle = πr² (r = radius of circle)
6. Surface area of sphere = 4πr² (r = radius of sphere)
7. The volume of cube = (length of one side of the cube)³
8. The volume of parallelopiped = length x breadth x height
9. Volume of cylinder = πr²l
10. Volume of sphere = [latex]\frac{4}{3}\) πr³and Circumference of square = 4l
11. Volume of cone = \(\frac{1}{3}\) πr²h
12. Circumference of circle = 2πr

Formulae Of Algebra:

1. (a + b)² = a² + b² + 2ab
2. (a-b)² = a² + b² -2ab
3. (a² – b²) = (a + b) (a – b)
4. (a + b)³ = a³ + b³ + 3ab (a + b)
5. (a – b)³ = a³ – b³ – 3ab (a – b)
6. (a + b)² – (a – b)² = 4ab
7. (a + b)² + (a – b)² = 2(a² + b²)

Formulae Of Differentiation:

1. \(\frac{d}{d x}\) (constant) = 0 dx; Differentiation with respect to x = \(\frac{d}{d x}\)
2. \(\frac{d}{d x}\left(x^m\right)=n x^{n-1}\)
3. \(\frac{d}{d x}(\sin x)=\cos x\)
4. \(\frac{d}{d x}(\cos x)=-\sin x\)

KSEEB Class 11 Physics Physical World Key Concepts

Formulae of Integration: Integration with respect to x = dx

1. \(\int d x=x\)
2. \(\int x^n d x=\frac{x^{n+1}}{n+1}\)
3. \(\int \sin x d x=\cos x+c\)
4. \(\int \cos x d x=\sin x+c\)

Formulae Of Logarithm:

1. log mn = (log m + log n)
2. log(\(\frac{m}{n}\)) = (log m – log n)
3. log mⁿ = n log m

Value Of Trigonometric Functions:

Signs Of Trigonometrical Ratios:

sin (90° – θ) = cos θ; sin (180° – θ) = sin θ

cos (90° – θ) = sin θ ; cos (180° – θ) = – cos θ

tan (90° – θ) = cot θ ; tan (180° – θ) = _-tanθ

According To Binomial Theorem: (1 + x)ⁿ ≈(1 + nx) if x < < 1

Quadratic Equation: ax² + bx + c = 0

x = \(\left(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\right)\)