What is friction in physics – Friction, Physics, class 11 – eSaral

Class 9-10, JEE & NEET

Friction is the opposing force that is set up between the surfaces of contact. If you want to learn what is friction in physics then keep reading.

## Friction

Friction is the opposing force that is set up between the surfaces of contact when one body slides or rolls or tends to do so on the surface of another body.

For example, if a book on a table slides from left to right along the surface of a table, a frictional force directed from the right to left acts on the book at the surface of contact i.e. In between the two surfaces and parallel to the surface.

Frictional forces may also exist between surfaces when there is no relative motion. Frictional forces arise due to molecular interactions.

1. If lubricants are used, then the force of friction generally reduces.
2. In some cases, friction acts as a supporting force, and in some cases, it acts as an opposing force.
(a) Supporting:

Frictional forces are essential in many walks of life. For example, the walking process can only take place because there is friction between the shoes and the ground. In the process of walking, in order to step forward, you must press your foot backward on the ground. A friction force tends to oppose this movement of the foot, the ground pushes you forward which allows you to walk. When there is no friction, as on ice, or polished granite, or oily surfaces, the walking is difficult.

(b) Opposing:

When a block slides over a surface the force of friction work as an opposing force in the opposite direction of the motion.

(c) Both Supporting and Opposing:

(i) Pedaling:

When a cyclist pedals the friction force on the rear wheel acts as a supporting force and on front-wheel as an opposing force.

(ii) Non-Pedaling:

When a cyclist does not pedal the friction force on the rear wheel & front wheel act as an opposing force.

So, that’s all from this Topic. I hope you get the idea about what is friction in physics. If you found this explanation helpful then please share it with your friends.

If you have any confusion related to this Topic then feel free to ask in the comments section down below.

What is Static Friction

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What is dynamic friction in physics – Friction, Class 11 – eSaral

Class 9-10, JEE & NEET

Hey, do you want to know What is dynamic friction in physics? If yes. Then keep reading.

## Dynamic Friction

When the force acting on the body is greater than the limiting friction, then the body comes into motion. The friction now acting between the surfaces of contact is dynamic friction.

The dynamic friction is always less than the limiting static friction.

Dynamic friction is of two types:
1. Sliding Friction
2. Rolling Friction

### Sliding Friction:

If one body begins to slides over the other, then the friction between the surfaces is called sliding friction.

If the limiting kinetic friction force is $\mathrm{F}_{\mathrm{k}}$ and the normal reaction is N.

Then the coefficient of kinetic friction is given by $\mu_{\mathrm{k}}=\frac{\mathrm{F}_{\mathrm{k}}}{\mathrm{N}}$

If on the moving body an external force F acts, then in the presence of kinetic friction, the acceleration produced in the body is given by

$a=\frac{F-F_{k}}{m}$

where m is the mass of the body, and $F_{k}=\mu_{k} N$

Where the proportionality constant $\mu_{k}$ is a dimensionless number and is called the coefficient of kinetic friction. $\mu_{k}$ connects only the magnitudes of $F_{k}$ and N. These forces have perpendicular directions and $F_{k}$ is opposite v.

Note:
1. $\mu_{k}$ depends on the nature and condition of the two surfaces, and $\mu_{k}$ usually falls in the range from about 0.1 to about 1.5.
2. $\mu_{k}$ is nearly independent of speed for low relative speeds of the surfaces, decreasing slightly as the speed increases. We shall use the approximation that $F_{k}$ is independent of the speed.
3. $F_{k}$ (or $\mu_{k}$) is nearly independent of the area of contact for a wide range of areas.The near independence of $\mu_{k}$ on the area of contact can be demonstrated by sliding a block that has sides with different areas. The surface of each side should consist of the same type of material and should be in the same condition. When the applied force required to slide the block at a given speed on different sides is measured, it is found to be nearly the same. Since N is the same in each case, we conclude that $\mu_{k}$ is approximately independent of the area.

Similar to $\mu_{k}$, coefficient $\mu_{s}$ depends on the condition and nature of the two surfaces and is nearly independent of the area of contact.

Table lists $\mu_{k}$ and $\mu_{S}$ for a few representative pairs of surfaces. Normally, for a given pair of surfaces, $\mu_{S}$ is noticeably larger than $\mu_{\mathrm{k}}$.

When two copper plates are highly polished and placed in contact with each other, then instead of decreasing, the force of friction increases. This arises due to the fact that for two highly polished surfaces in constant, the number of molecules coming in contact increases and as a result the adhesive forces increase. This in turn, increases the force of friction.

### Rolling Friction:

When a body (say wheel) rolls on a surface the resistance offered by the surface is called rolling friction.

The velocity of the point of contact with respect to the surface remains zero.

The rolling friction is negligible in comparison to static or kinetic friction which may be present simultaneously i.e., $\mu_{\mathrm{R}}<\mu_{\mathrm{K}}<\mu_{\mathrm{S}}$

#### Angle of Friction

The angle of friction is the angle which the resultant of limiting friction $\mathrm{F}_{\mathrm{S}}$ and normal reaction N makes with the normal reaction. It is represented by $\lambda$, Thus from the figure.

$\tan \lambda=\frac{\mathrm{F}_{\mathrm{S}}}{\mathrm{N}}$

$\left(\because \mathrm{F}_{\mathrm{s}}=\mu \mathrm{N}\right)$

Or

$\tan \lambda=\mu$

For smooth surfaces,

$\lambda=0$ (zero)

#### Angle of Repose ($\theta$)

If a body is placed on an inclined plane and if its angle of inclination is gradually increased, then at some angle of inclination $\theta$ the body will just on the point to slide down.

The angle is called angle of repose ($\theta$).

$\mathrm{F}_{\mathrm{S}}$ = mg $\sin \theta$ and N = mg $\cos \theta$

So

$\frac{F_{s}}{N}=\tan \theta$

Or

$\mu=\tan \theta$

Relation between angle of friction ($\lambda$) and angle of repose ($\theta$)

We know that tan $\lambda=\mu$ and $\mu=\tan \theta$

hence $\tan \lambda=\tan \theta$ or $\theta=\lambda$

Thus, angle of repose = angle of friction

So, that’s all from this blog. I hope you get the idea of What is dynamic friction in physics. If you liked this explanation then please share it with your friends.

What is Static Friction

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What is Static Friction – Friction, Physics, Class 11 – eSaral

Class 9-10, JEE & NEET

Hey, do you want to know what is Static Friction? If yes. Then keep reading.

## Static friction

The frictional force between two surfaces before the relative motion actually starts is called static frictional force or static friction. If we slowly increase the force with which we are pulling the box, the graph shows that the friction force increases with our force up to a certain critical value, $\mathrm{f}_{\mathrm{L}}$, the box suddenly begins to move, and as soon as it starts moving, a smaller force is required to maintain its motion as in motion friction is reduced. The friction value from 0 to $\mathrm{f}_{\mathrm{L}}$ is known as static friction, which balances the external force on the body and prevents it from sliding. The value $\mathrm{f}_{\mathrm{L}}$ is the maximum limit up to which the static friction acts is known as limiting friction, after which the body starts sliding and friction reduces to kinetic friction.

## Laws of Static Friction

1. The direction of the force of friction is always in a direction opposite to which the body moves or tends to move.
2. The force of friction is a self-adjusting force and increases with the applied force, so as to be equal and opposite to it until the motion is just about to start.
3. When one body just tends to move, the force of friction between the surfaces in contact is maximum. The force of friction is called limiting friction.
4. The limiting friction depends on the nature of surfaces in contact but is independent of the area of the surface in contact.
5. The limiting friction between the surface of two bodies is directly proportional to the normal reaction N i.e. $F_{s} \propto N$.
6. The frictional force is independent of the relative velocities of the two surfaces.

## Coefficient of Static Friction $\left(\mu_{s}\right)$

The limiting friction $\mathrm{F}_{\mathrm{S}}$ is proportional to the normal reaction N thus

$\mathrm{F}_{\mathrm{s}}=\mu_{\mathrm{S}} \mathrm{N}$

Or

$\mu_{S}=\frac{F_{s}}{N}$

$=\frac{\text { limiting friction }}{\text { normal reaction }}$

so, that’s all from this blog. I hope you get the idea about What is Static Friction/ If you liked this Explanation then please share it with your friends.

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Motion in a lift

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Motion in a lift – Laws of Motion, Physics, Class 11 – eSaral

Class 9-10, JEE & NEET

Hey, do you want to learn about the Motion in a lift? If yes. Then keep reading.

## Motion in a Lift

The weight of a body is simply the force exerted by the earth on the body. If the body is on an accelerated platform, the body experiences a fictitious force, so the weight of the body appears changed and this new weight is called apparent weight. Let a man of weight W = Mg be standing in a lift.

We consider the following cases:

Case (a)

If the lift moving with constant velocity v upwards or downwards.

In this case, there is no accelerated motion hence no pseudo force experienced by observer $0^{\prime}$ inside the lift.

So apparent weight $\mathrm{W}^{\prime}$ = Actual weight W.

Case (b)

If the lift is accelerated (i.e. a = constant upward) : Then net forces acting on the man are
1. weight W = Mg downward
2. fictitious force $\mathrm{F}_{0}$ = Ma downward.

So apparent weight

$W^{\prime}=W+F_{0}$

Or

$\mathrm{W}^{\prime}=\mathrm{Mg}+\mathrm{Ma}$

$=M(g+a)$

Effective gravitational acceleration

$g^{\prime}=g+a$

Case (c)

If the lift is accelerated downward with acceleration $a<g$: Then fictitious force $\mathrm{F}_{0}$ = Ma acts upward while the weight of man W = Mg always acts downward, therefore

So apparent weight

$W^{\prime}=W+F_{0}$

Or

$\mathrm{W}^{\prime}=\mathrm{Mg}-\mathrm{Ma}$

$=M(g-a)$

Effective gravitational acceleration

$g^{\prime}=g-a$

Special Case:

If g = a then

$W^{\prime}=0$ cond $^{n}$. Of weightlessness.

Thus, in a freely falling lift the man will experience weightlessness.

Case (d)

If lift accelerates downward with acceleration a > g

Then as in Case c

Apparent weight $W^{\prime}=M(g-a)$ = $-M(a-g)$ is negative, i.e., the man will be accelerated upward and will stay at the ceiling of the lift.

Ex. A spring weighing machine inside a stationary lift reads 50 kg when a man stands on it. What would happen to the scale reading if the lift is moving upward with (i) constant velocity, and (ii) constant acceleration?

Sol. (i) In the case of a constant velocity of lift, there is no fictitious force; therefore, the apparent weight = actual weight. Hence the reading of the machine is 50 kg wt.

(ii) In this case the acceleration is upward, the fictitious force R = ma acts downward, therefore apparent weight is more than actual weight i.e. $\mathrm{W}^{\prime}$ = W + R = m (g + a).

Hence scale shows a reading = m (g + a) newton

$=\frac{m g\left(1+\frac{a}{g}\right)}{g} k g$

$=\left(50+\frac{50 a}{g}\right) k g.w t$

So, that’s all from this blog. I hope you get the idea about the Motion in a lift. If you found this article helpful then don’t forget to share it with your friends.

Frame of reference

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Frame of reference in physics – Inertial and Non-inertial – eSaral

Class 9-10, JEE & NEET

A non-accelerating frame of reference is called an inertial frame of reference. If you want to learn more about the frame of reference in physics. Then keep reading.

## Inertial and accelerated frames of reference

### (a) Inertial frames of reference:

A non-accelerating frame of reference is called an inertial frame of reference. A frame of reference moving with a constant velocity is an inertial frame of reference.
1. All the fundamental laws of physics have been formulated in respect of the inertial frame of reference.
2. All the fundamental laws of physics can be expressed as having the same mathematical form in all the inertial frames of reference.
3. The mechanical and optical experiments performed in an inertial frame in any direction will always yield the same results. It is called the isotropic property of the inertial frame of reference.

#### Examples of inertial frames of reference:

1. A frame of reference remaining fixed w.r.t. distance stars is an inertial frame of reference.
2. A spaceship moving in outer space without spinning and with its engine cut-off is also an inertial frame of reference.
3. For practical purposes, a frame of reference fixed to the earth can be considered as an inertial frame. Strictly speaking, such a frame of reference is not an inertial frame of reference, because the motion of the earth around the sun is accelerated motion due to its orbital and rotational motion. However, due to negligibly small effects of rotation and orbital motion, the motion of earth may be assumed to be uniform and hence a frame of reference fixed to it may be regarded as an inertial frame of reference.

### (b) Non-inertial frame of reference:

An accelerating frame of reference is called a non-inertial frame of reference.

Newton’s laws of motion are not directly applicable in such frames, before application

Note:

A rotating frame of reference is a non-inertial frame of reference because it is also an accelerating one due to its centripetal acceleration.

## Pseudo force

The force on a body due to the acceleration of non-inertial frame is called fictitious or apparent or pseudo

force and is given by $\overrightarrow{\mathrm{F}}=-\mathrm{m} \overrightarrow{\mathrm{a}_{0}}$, where $\vec{a}_{0}$ is acceleration of non-inertial frame with respect to an inertial frame and m is mass of the particle or body.

The direction of pseudo force must be opposite to the direction of acceleration of the non-inertial frame.

When we draw the free body diagram of a mass, with respect to an inertial frame of reference we apply only the real forces (forces that are actually acting on the mass).

But when the free body diagram is drawn from a non-inertial frame of reference a pseudo force

(in addition to all real forces) has to be applied to make the equation $\overrightarrow{\mathrm{F}}=\mathrm{m} \overrightarrow{\mathrm{a}}$ to be valid in this frame also.

So, that’s all from this article. I hope you get the idea about the frame of reference in physics. If you have any confusion related to this article then feel free to ask in the comments section down below.

Motion of bodies connected by strings

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Motion of bodies connected by strings – Physics Class 11 – eSaral

Class 9-10, JEE & NEET

Hey, do you want to learn about the motion of bodies connected by strings? If yes. Then you are at the right place.

## The motion of bodies connected by strings

Two bodies :

Let us consider the case of two bodies of masses $\mathrm{m}_{1}$ and $\mathrm{m}_{2}$ connected by a thread and placed on a smooth horizontal surface as shown in fig. A force F is applied on the body of mass $\mathrm{m}_{2}$ in forwarding direction as shown. Our aim is to consider the acceleration of the system and the tension T in the thread. The forces acting separately on two bodies are also shown in the figure:

From fig

$\mathrm{T}=\mathrm{m}_{1} \mathrm{a}$…..(1)

$F-T=m_{2} a$…..(2)

$\mathrm{F}=\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right) \mathrm{a}$

Or

$a=\frac{F}{m_{1}+m_{2}}$…..(3)

The acceleration of the system can be calculated from eq. (3) and tension in the thread by eq. (1).

Three bodies :

In the case of three bodies, the situation is shown in fig.

Acceleration $\mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}_{1}+\mathrm{m}_{2}+\mathrm{m}_{3}}$….(4)

$T_{1}=m_{1} a=\frac{m_{1} F}{m_{1}+m_{2}+m_{3}}$…..(5)

$\mathrm{F}-\mathrm{T}_{2}=\mathrm{m}_{3} \mathrm{a}$

Or

$\mathrm{T}_{2}=\mathrm{F}-\frac{\mathrm{m}_{3} \mathrm{~F}}{\mathrm{~m}_{1}+\mathrm{m}_{2}+\mathrm{m}_{3}}$

$=\frac{\left(m_{1}+m_{2}\right) F}{m_{1}+m_{2}+m_{3}}$ …..(6)

Ex. Three blocks, are connected by string as shown in fig. below, and are pulled by a force $\mathrm{T}_{3}$ = 120N. If $\mathrm{m}_{1}$ = 5 kg, $\mathrm{m}_{2}$ = 10 kg and $\mathrm{m}_{3}$ = 15 kg. Calculate the acceleration of the system and $\mathrm{T}_{1}$ and $\mathrm{T}_{2}$.

Sol. Acceleration of the system

$a=\frac{F}{m_{1}+m_{2}+m_{3}}$

$=\frac{120}{5+10+15}$

$=4 \mathrm{~m} / \mathrm{sec}^{2}$

Now

$T_{1}=m_{1} a$

$=5 \times 4=20 \mathrm{~N}$

$T_{2}=\left(m_{1}+m_{2}\right) a$

$=(5+10) 4$

$=60 \mathrm{~N}$

So, that’s all from this blog, I hope you get the idea about the motion of bodies connected by strings. If you liked this article then please share it with your friends. If you have any confusion related to this article then feel free to ask in the comments section down below.

Motion of bodies in contact

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Motion of bodies in contact – Physics Class 11 – eSaral

Class 9-10, JEE & NEET

Hey, do you want to learn about the motion of bodies in contact? If yes. Then you are at the right place.

## The motion of bodies in contact

(a) When two bodies of masses $\mathrm{m}_{1}$ and $\mathrm{m}_{2}$ are kept on the frictionless surface and a force F is applied on one body, then the force with which one body presses the other at the point of contact is called Force of Contact.

These two bodies will move with same acceleration = a.

When the force F acts on the body with mass $\mathrm{m}_{1}$

as shown in fig. (1) and (1a) $F=\left(m_{1}+m_{2}\right) a$.

If the force exerted by $\mathrm{m}_{2}$ on $\mathrm{m}_{1}$ is $\mathrm{f}_{1}$ (force of contact)

then for body $\mathrm{m}_{1}$

$\left(F-f_{1}\right)=m_{1} a$

Or

$f_{1}=F-m_{1} a$

$=\left(m_{1}+m_{2}\right) a-m_{1} a=m_{2} a$

or action on $\mathrm{m}_{2}$

$\mathrm{f}_{1}=\mathrm{m}_{2} \cdot \frac{\mathrm{F}}{\mathrm{m}_{1}+\mathrm{m}_{2}}$

When the force F acts on the body with mass $\mathrm{m}_{2}$ as shown in fig. 2 and 2(a)

$\mathrm{F}=\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right) \mathrm{a}$

$\mathrm{F}=\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right) \mathrm{a}$

for body with mass $\mathrm{m}_{2}$

$\mathrm{F}-\mathrm{f}_{2}=\mathrm{m}_{2} \mathrm{a}$

Or

$\mathrm{f}_{2}=\mathrm{F}-\mathrm{m}_{2} \mathrm{a}$

$=\left(m_{1}+m_{2}\right) a-m_{2} a=m_{1} a$

or action on $\mathrm{m}_{1}$

$\mathrm{f}_{2}=\frac{\mathrm{m}_{1} \mathrm{~F}}{\mathrm{~m}_{1}+\mathrm{m}_{2}}$

Ex. Two blocks of mass m = 2 kg and M = 5 kg are in contact on a frictionless table. A horizontal force F (= 35 N) is applied to m. Find the force of contact between the block, will the force of contact remain same if F is applied to M?

Sol. As the blocks are rigid under the action of a force F, both will move with same acceleration

$a=\frac{F}{m+M}=\frac{35}{2+5}=5 \mathrm{~m} / \mathrm{sec}^{2}$

Now as the mass of larger block is M and its acc a,

so, force of contact = action on it

$f_{M}=M a=5 \times 5=25 \mathrm{~N}$

If the force is applied to M then its action on m will be

$\mathrm{f}_{\mathrm{m}}=\mathrm{ma}=2 \times 5=10 \mathrm{~N}$

From this problem, it is clear that acceleration does not depend on the fact that whether the force is applied to m or M, but the force of contact does.

(b) Let us now consider three bodies of masses $m_{1}$, $\mathrm{m}_{2}$ and $\mathrm{m}_{3}$ placed one after another and in contact with each other. Suppose a force F is applied horizontally on mass $\mathrm{m}_{1}$

$\mathrm{F}-\mathrm{f}_{1}=\mathrm{m}_{1} \mathrm{a}$

$f_{1}-f_{2}=m_{2} a$

$\mathrm{f}_{2}=\mathrm{m}_{3} \mathrm{a}$

Here $\mathrm{F}=\left(\mathrm{m}_{1}+\mathrm{m}_{2}+\mathrm{m}_{3}\right)$ a

$a=\frac{F}{\left(m_{1}+m_{2}+m_{3}\right)}$

$\mathrm{f}_{1}=\frac{\left(\mathrm{m}_{2}+\mathrm{m}_{3}\right) \mathrm{F}}{\left(\mathrm{m}_{1}+\mathrm{m}_{2}+\mathrm{m}_{3}\right)}$ (action on both $\mathrm{m}_{2}$ and $\left.\mathrm{m}_{3}\right)$

And

$f_{2}=\frac{m_{3} F}{\left(m_{1}+m_{2}+m_{3}\right)}$

(action on $\mathrm{m}_{3}$ alone)

when the force F is applied on $m_{3}$, then

$\mathrm{f}_{1}=\frac{\mathrm{m}_{1} \mathrm{~F}}{\left(\mathrm{~m}_{1}+\mathrm{m}_{2}+\mathrm{m}_{3}\right)}$ (action on $\mathrm{m}_{1}$ alone)

and $f_{2}=\frac{\left(m_{1}+m_{2}\right) F}{\left(m_{1}+m_{2}+m_{3}\right)}$

(action on $\mathrm{m}_{1}$ and $\mathrm{m}_{2}$ )

So, that’s all from this blog. I hope you get the idea about the Motion of bodies in contact. If you enjoyed this explanation then please share it with your friends.

Application of impulse

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Applications of impulse – Impulse, Physics – eSaral

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Hey do you want to learn about the applications of impulse? If so, Then you are at the right place

## Impulse

It is not convenient to measure the varying force of impact. Suppose that $\overrightarrow{\mathrm{F}}$ av is the average force acting during impact and $t$ is the small-time for which the impact lasts.

In such situations, it is found that the quantity Fav. It is very easy to measure.

The quantity $\overrightarrow{\mathrm{F}}_{\text {av }} . \mathrm{t}$ is called impulse.

If a variable force acts on a body or particle for a interval of time $t_{1} \rightarrow t_{2}$

Impules is

$\overrightarrow{\mathrm{I}}=\int_{t_{1}}^{t_{2}} \overrightarrow{\mathrm{F}} \mathrm{dt}=\overrightarrow{\mathrm{p}_{2}}-\overrightarrow{\mathrm{p}_{1}}$

Impules momentum theorem

The integral $\int_{0}^{t} \overrightarrow{\mathrm{F}} \mathrm{dt}$ is measure of the impulse, when the force of impact acts on the body and from equation, we find that it is equal to total change in momentum of the body.

Dimensional formula $\&$ unit. The dimensional formula of impulse is the same as that of momentum i.e. [ML $T^{-1}$ ]. In SI, the unit of impulse is N-s or kg ms $^{-1}$ and in cgs system it is dyne-s or $\mathrm{g} \mathrm{cm} \mathrm{s}^{-1}$

## Applications of Concept of Impulse :

(a) Bogies of a train are provided with buffers: The buffers increase the time interval of jerks during shunting and hence reduce the force with which the bogies pull each other. $(\because \mathrm{F} \Delta \mathrm{t}=\mathrm{constant})$

(b) A cricket player draws his hands back while catching a ball: The drawing back of hands increases time interval and hence reduces force with which the ball hits the hand.

(c) A person jumping on a hard cement floor receives more injuries than a person jumping on a muddy or sandy road: Because on the hard cemented floor the feet of the man immediately comes to rest and the time is small, therefore, the force experienced by the man is large while on the sandy soil the feet embedes into the soil, thus the time taken for same change in momentum is comparatively increased. Thus the force is decreased.

(d) Cars, buses, trucks, bogies of the train, etc are provided with a spring system to avoid severe jerks. Due to the spring system, the time interval of the jerks increases. As the rate of change of momentum will be smaller, comparatively lesser force acts on the passengers during the jerks.

(e) China wares are wrapped in straw or paper before packing. China wares are wrapped in straw or paper so that when they receive jerks during transportation, the time of impact may be more. As such, only a small force can act on the china wares during jerks and there is no fear of damage to them.

(f) If a graph is plotted between $\overrightarrow{\mathrm{F}}$ and $\mathrm{t}$ with $\overrightarrow{\mathrm{F}}$ as ordinate (Y-axis)

and $t$ as $(X$ -axis)

Impulse = Area enclosed by F-t curve and time axis for specified duration.

For example impulse between time $t_{1}$ and $t_{2}=$ area abdc.

Newton’s Laws of Motion

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Ex. A hammer weighing $2.5 \mathrm{kg}$ moving with a speed of $1 \mathrm{m} / \mathrm{s}$ strikes the head of a nail driving it $10 \mathrm{cm}$ into the wall. What is the acceleration during impact and the impulse imparted to wall.

Sol. The initial velocity of nail is same as that of hammer i.e. $\mathrm{u}=1 \mathrm{m} / \mathrm{s}, \mathrm{v}=0, \mathrm{s}=10 \mathrm{cm}=0.10 \mathrm{m}$

$$\begin{array}{l}\therefore \quad v^{2}=u^{2}+2 a s \text { gives } \\0=1^{2}+2 a \times 0.1\end{array}$$

This gives $a=-\frac{1}{0.2}=-5 \mathrm{m} / \mathrm{s}^{2}$

i.e. Retardation $\mathrm{a}=5 \mathrm{m} / \mathrm{s}^{2}$

Impulse imparted to wall = – change in momentum of hammer

$$=-\{\mathrm{M}(\mathrm{v}-\mathrm{u})\}=-\{0.5(0-1)\}=+2.5 \mathrm{N}-\mathrm{s}$$

Ex. Magnitude of the force in newton acting on a body varies with time $t$ (in mili-second) as shown in fig. below. Calculate magnitude of total impulse of the force.

Sol. The impulse

\begin{aligned} \mathrm{I}=\int \mathrm{F} \mathrm{dt} &=\frac{1}{2}(0.014 \mathrm{s}-0.003 \mathrm{s})(10 \mathrm{N}) \\ &=0.055 \mathrm{N} \cdot \mathrm{s}=0.055 \mathrm{kg} \cdot \mathrm{m} / \mathrm{s} \end{aligned}

Area of the graph gives the magnitude of the Impulse.

Ex. $\quad$ A force exert an impulse I on a body changing its speed from u to v. The force and object’s motion are along the same line. Show that the work done by the force is $\frac{\mathrm{I}(\mathrm{u}+\mathrm{v})}{2}$.

Sol. According to work-energy theorem we have

$$W=\Delta K=\frac{1}{2} m v^{2}-\frac{1}{2} m u^{2} \quad \text { or } \quad W=\frac{1}{2} m(v-u)(v+u)$$

But according to impulse-momentum theorem

$$\mathrm{I}=\mathrm{m}(\mathrm{v}-\mathrm{u})$$

So, eliminating $\mathrm{m}$ between $\mathrm{eq}^{\mathrm{n}} .(1)$ and (2)

$$W=\frac{1}{2} \frac{I}{(v-u)}(v-u)(v+u)=\frac{I(v+u)}{2} \quad \text { This is the required result. }$$

Ex. A cricket ball of mass $150 \mathrm{g}$ is moving with a velocity of $12 \mathrm{m} / \mathrm{s}$ and is hit by a bat so that the ball is turned back with a velocity of $20 \mathrm{m} / \mathrm{s}$. If the duration of contact between the ball and bat is $0.01 \mathrm{s},$ find ther impulse and the average force exerted on the ball by the bat.

Sol. According to given problem change in momentum of the ball

$$\Delta p=p_{f}-p_{i}=m(v-u)=150 \times 10^{-3}[20-(-12)]$$

So by impulse-momentum theorem $I=\Delta p=4.8 \mathrm{N}-\mathrm{s}$

And by time averaged definition of force in case of impulse $\mathrm{F}_{\mathrm{av}}=\frac{\mathrm{I}}{\Delta \mathrm{t}}=\frac{\Delta \mathrm{p}}{\Delta \mathrm{t}}=\frac{4.80}{0.01}=480 \mathrm{N}$
Horizontal Projectile motion – Motion in Plane – eSaral

Class 9-10, JEE & NEET

Hey, do you want to learn about Horizontal projectile motion? If yes. Then you are at the right place.

## Horizontal Projectile motion

Suppose a body is thrown horizontally from point O, with velocity u. Height of O from ground = H. Let X-axis be along horizontal and Y-axis be vertically downwards and origin O is at point of projection as shown in fig. Let the particle be at P at a time t. The co-ordinates of P are (x, y) Distance travelled along X-axis at time t with uniform velocity i.e. Velocity of projection and without acceleration.

The horizontal component of velocity $\mathrm{V}_{\mathrm{x}}$ = u

and horizontal displacement x = u . t….(1)

to calculate y, consider vertical motion of the projectile

initial velocity in vertical direction $\mathrm{u}_{\mathrm{y}}$ = 0.

acceleration along y direction $a_{y}$ = g (acc. due to gravity)

so

$v_{v}=a_{y} t$

(y comp. of velocity at time t)

Or

$v_{y}=g t$….(2)

(as body were dropped from a height)

Resultant velocity at time t is

$\overrightarrow{\mathrm{v}}=\mathrm{u} \hat{\mathrm{i}}+(\mathrm{gt}) \hat{\mathrm{j}}$

$v=\sqrt{u^{2}+(g t)^{2}}$

if $\beta$ is the angle of velocity with X-axis (horizontal) $\tan \beta=\frac{\mathrm{gt}}{\mathrm{u}}$

and

$y=\frac{1}{2} g t^{2}$……(3)

Or

$y=\frac{1}{2} g\left(\frac{x}{u}\right)^{2}$

[from equation (i) $\left.\mathrm{t}=\frac{\mathrm{x}}{\mathrm{u}}\right]$

Or

$y=\frac{g}{2 u^{2}} \cdot x^{2}$

Or

$y=k x^{2}$

here $\mathrm{k}=\frac{\mathrm{g}}{2 \mathrm{u}^{2}} \quad(\mathrm{k}$ is constant)

This is eqn. of a parabola.

A body thrown horizontally from a certain height above the ground follows a parabolic trajectory till it hits the ground.
1. Time of flight$\mathrm{T}=\sqrt{\frac{2 \mathrm{H}}{\mathrm{g}}}$ [as $\left.\mathrm{y}=\frac{1}{2} \mathrm{gt}^{2}, \mathrm{~T}=\sqrt{\frac{2 \mathrm{H}}{\mathrm{g}}}\right]$

2. Range horizontal distance covered = R.R = u × time of flight$\mathrm{R}=\mathrm{u} \cdot \sqrt{\frac{2 \mathrm{H}}{\mathrm{g}}}$

$\left[\because \mathrm{H}=\frac{\mathrm{g}}{2 \mathrm{u}^{2}} \mathrm{R}^{2}\right]$

3. Velocity when it hits the ground$\mathrm{v}_{\mathrm{g}}=\sqrt{\mathrm{u}^{2}+2 \mathrm{gH}}$

Oblique Projectile motion

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Hey, do you want to learn about the Oblique Projectile motion? If yes. Then keep reading.

## Oblique Projectile motion

Consider the motion of a bullet which is fired from a gun so that its initial velocity $\overrightarrow{\mathrm{u}}$ makes an angle $\theta$ with the horizontal direction. Let us take $X$ -axis along ground and Y-axis along vertical.

$\overrightarrow{\mathrm{u}}$ can be resolved as

$\mathrm{u}_{\mathrm{x}}=\mathrm{u} \cos \theta \quad$ (along horizontal)

$\& u_{y}=u \sin \theta$ (along vertical)

motion of bullet can be resolved into horizontal and vertical motion.

(i) In horizontal direction there is no acc. so it moves with constant velocity $v_{x}=u_{x}=u \cos \theta$

So distance traversed in time $t$ is $x=u_{x} t$ or  $t=\frac{x}{u \cos \theta}$

The motion in the vertical direction is the same as that of a ball thrown upward with an initial velocity $\mathrm{u}_{y}=\mathrm{u} \sin \theta$ and $\mathrm{acc}=-\mathrm{g}$ (downward).

So at time $t$ vertical component of velocity $v_{y}=u_{y}-g t=u \sin \theta-g t$

Displacement along y direction

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

Substituting the value of $t$ from eqn. (i) in eqn. (iii)

we get

$\quad y=(u \sin \theta)\left(\frac{x}{u \cos \theta}\right)-\frac{1}{2} g\left(\frac{x}{u \cos \theta}\right)^{2}$

or $\left.y=x \tan \theta-\frac{g}{2 u^{2} \cos ^{2} \theta} \cdot x^{2}\right]$ This is eqn. of parabola.

The trajectory of projectile is parabolic

The projectile will rise to maximum height $\mathrm{H}$ (where $\mathrm{v}_{x}=\mathrm{u} \cos \theta, \mathrm{v}_{y}=0$ ) and then move down again to reach the ground at a distance $\mathrm{R}$ from origin.

Setting $x=R$ and $y=0$ (since projectile reaches ground again)

$\mathrm{O}=\mathrm{R} \tan \theta-\frac{\mathrm{g}}{2 \mathrm{u}^{2} \cos ^{2} \theta} \cdot \mathrm{R}^{2}$

We get $R=\frac{2 u^{2} \cos ^{2} \theta}{g} \times \frac{\sin \theta}{\cos \theta}$

or $\mathrm{R}=\frac{2 \mathrm{u}^{2}}{\mathrm{~g}} \cdot \sin \theta \cos \theta$

or Range $R=\frac{u^{2} \sin 2 \theta}{g}$

If time for upward journey is t

at highest point $\quad v_{y}=0$ so $0=(u \sin \theta)-g t$

$\left(v_{y}=u_{y}-g t\right)$

or $\mathrm{t}=\frac{\mathrm{u} \sin \theta}{\mathrm{g}}$

$\mathrm{T}=2 \mathrm{t} \quad$ (it will take same time for downward journey)

$\therefore$

$T=\frac{2 u \sin \theta}{g}$ Time of flight

At the highest point $y=H$ and $v_{y}=0$ Maximum Height and Time of

Flight Depends on Vertical

Component of Initial Velocity So that $H=\frac{u_{y}^{2}}{2 g} \quad\left[v_{y}^{2}=u_{y}^{2}-2 g y\right]$

or $\mathrm{H}=\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}$ Maximum Height

we can also determine $R$ as follows

so $x=u_{x} t$

$R=u_{x} \cdot T$

$=(u \cos \theta)\left(\frac{2 u \sin \theta}{g}\right)$

or $\mathrm{R}=\frac{\mathrm{u}^{2} \sin 2 \theta}{\mathrm{g}}$

velocity at time t $\vec{v}_{t}=v_{x t} \hat{i}+v_{y t} \hat{j}$

$=(u \cos \theta) \hat{i}+(u \sin \theta-g t) \hat{j}$

$v=\sqrt{u^{2} \cos ^{2} \theta+(u \sin \theta-g t)^{2}}$

Note :

(i) Alternative $e q^{n}$, of trajectory $y=x \tan \theta\left(1-\frac{x}{R}\right)$ where $R=\frac{2 u^{2} \sin \theta \cos \theta}{g}$

(ii) Vertical component of velocity $v_{y}=0,$ when particle is at the highest point of trajectory.

(iii) Linear momentum at highest point = mu cos $\theta$ is in horizontal direction.

(iv) Vertical component of velocity is +ive when particle is moving up.

(v) Vertical component of velocity is -ive when particle is moving down.

(vi) Resultant velocity of particle at time $t \vee=\sqrt{v_{x}^{2}+v_{y}^{2}}$ at an angle $\phi=\tan ^{-1}\left(\frac{v_{y}}{v_{x}}\right) .$

(vii) Displacement from origin, $s=\sqrt{x^{2}+y^{2}}$

Special Points :

(1) The three basic equation of motion, i.e.

$v=u+a t$

$\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2} \quad \mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{as}$

For projectile motion give:

$T=\frac{2 u \sin \theta}{g} \quad R=\frac{u^{2} \sin 2 \theta}{g}$

$H=\frac{u^{2} \sin ^{2} \theta}{2 g}$

(2) In case of projectile motion,

The horizontal component of velocity (u $\cos \theta$ ), acceleration (g), and mechanical energy remain constant.

Speed, velocity, vertical component of velocity (u $\sin \theta)$, momentum, kinetic energy and potential energy all change. Velocity and K.E. are maximum at the point of projection, while minimum (but not zero) at the highest point.

(3) If angle of projection is changed from

then range

$\theta-\theta=(90-\theta)$

$R=\frac{u^{2} \sin 2 \theta^{\prime}}{g}=\frac{u^{2} \sin 2(90-\theta)}{g}=\frac{u^{2} \sin 2 \theta}{g}=R$

So a projectile has same range for angles of projection $\theta$ and $(90-\theta)$

But has different time of flight (T), maximum height (H) \& trajectories

Range is also same for $\theta_{1}=45^{\circ}-\alpha$ and $\quad \theta_{2}=45^{\circ}+\alpha .\left[\right.$ equal $\left.\frac{u^{2} \cos 2 \alpha}{g}\right]$

(4) For maximum Range $\quad R=R_{\max } \Rightarrow 2 \theta=90^{\circ}$

$\text { for } \quad \theta=45^{\circ}$

$R_{\max }=\frac{u^{2}}{g} \quad\left[\right.$ For $\sin 2 \theta=1=\sin 90^{\circ}$ or $\left.\theta=45^{\circ}\right]$

When range is maximum $\Rightarrow$ Then maximum height reached

$\mathrm{H}=\frac{\mathrm{u}^{2} \sin ^{2} 45}{2 \mathrm{~g}}\left(\mathrm{When} \mathrm{R}_{\max }\right)$

or

$\mathrm{H}=\frac{\mathrm{u}^{2}}{4 \mathrm{~g}}$

hence maximum height reached (for $\left.R_{\max }\right) \quad H=\frac{R_{\max }}{4}$

(5) For height H to be maximum

$\mathrm{H}=\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}=\max \quad \text { i.e. } \sin ^{2} \theta=1(\max ) \text { or for } \theta=90^{\circ}$

So that $\mathrm{H}_{\max }=\frac{\mathrm{u}^{2}}{2 \mathrm{~g}} \quad$ When projected vertically (i.e. at $\left.\theta=90^{\circ}\right)$

in this case Range $R=\frac{u^{2} \sin \left(2 \times 90^{\circ}\right)}{9}=\frac{u^{2} \sin 180^{\circ}}{9}=0$

$\mathrm{H}_{\max }=\frac{\mathrm{u}^{2}}{2 \mathrm{~g}}$ (For vertical projection) and $\mathrm{R}_{\max }=\frac{\mathrm{u}^{2}}{\mathrm{~g}}$ (For oblique projection with same velocity)

so $\mathrm{H}_{\max }=\frac{R_{\max }}{2}$

If a person can throw a projectile to a maximum distance (with $\theta=45^{\circ}$ ) $R_{\max }=\frac{\mathbf{u}^{2}}{\mathrm{~g}}$.
.
The maximum height to which he can throw the projectile (with $\theta=90^{\circ}$ ) $\mathrm{H}_{\max }=\frac{R_{\text {max }}}{2}$

(6) At highest point

Potential energy will be max and equal to $(\mathrm{PE})_{4}=\mathrm{mgH}=\mathrm{mg} \cdot \frac{\mathrm{U}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}$ or $(\mathrm{PE})_{\mathrm{H}}=\frac{1}{2} \mathrm{mu}^{2} \sin ^{2} \theta$

While K.E. will be minimum (but not zero) and at the highest point as the vertical component of velocity is zero.

$(\mathrm{KE})_{\mathrm{H}}=\frac{1}{2} \mathrm{mv}_{\mathrm{H}}^{2}=\frac{1}{2} \mathrm{~m}(\mathrm{u} \cos \theta)^{2}$

$=\frac{1}{2} \mathrm{mu}^{2} \cos ^{2} \theta$

so $(\mathrm{PE})_{\mathrm{H}}+(\mathrm{KE})_{\mathrm{H}}$

$=\frac{1}{2} m u^{2} \sin ^{2} \theta+\frac{1}{2} m u^{2} \cos ^{2} \theta$

= $\frac{1}{2} \mathrm{mu}^{2}=$ Total M.E.

So in projectile motion mechanical energy is conserved.

$\left(\frac{P E}{K E}\right)_{H}=\frac{\frac{1}{2} m u^{2} \sin ^{2} \theta}{\frac{1}{2} m u^{2} \cos ^{2} \theta}=\tan ^{2} \theta$

(7) In case of projectile motion if range $R$ is $n$ times the maximum height $H,$ l.e. $R=n H$

then $\frac{u^{2} \sin 2 \theta}{g}=n \cdot \frac{u^{2} \sin ^{2} \theta}{2 g}$

or $2 \cos \theta=\frac{n \sin \theta}{2}$

or $\tan \theta=\frac{4}{n} \quad \Rightarrow \quad \theta=\tan ^{-1}\left(\frac{4}{n}\right)$

(8) Weight of a body in projectile motion is zero as it is a freely falling body.

So, that’s all from this article. I hope you get the idea about the Oblique projectile motion. If you liked this explanation then please share it with your friends.

Motion in Two Dimension

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## Motion in Two Dimensions

An object moving in a plane is said to have two-dimensional motion. The two-dimensional motion is equal to the vector sum of two one-dimensional motions along a mutually perpendicular direction.

Let the position of point P at a time t be given by position

vector $\overrightarrow{\mathrm{OP}}=\overrightarrow{\mathrm{r}}$

$\vec{r}=\hat{i} r \cos \theta+\hat{j} r \sin \theta$

$=\hat{i} x+\hat{j} y$

## Displacement

Let the position of point P at time $\mathrm{t}_{1}$ be described by position vector $\vec{r}_{1}=x_{1} \hat{i}+y_{1} \hat{j}$ and at time $\mathrm{t}_{2}$

position Q is given by position vector $\vec{r}_{2}=x_{2} \hat{i}+y_{2} \hat{j}$

from $\Delta \mathrm{OPQ} \Rightarrow \overrightarrow{\mathrm{OP}}+\overrightarrow{\mathrm{PQ}}=\overrightarrow{\mathrm{OQ}}$

or $\quad \overrightarrow{\mathrm{PQ}}=\overrightarrow{\mathrm{OQ}}-\overrightarrow{\mathrm{OP}}$

Displacement $\overrightarrow{P Q}=\delta \vec{r}=\vec{r}_{2}-\vec{r}_{1}$

in time interval $\delta \mathrm{t}=\left(\mathrm{t}_{2}-\mathrm{t}_{1}\right)$

or

$\vec{\delta}_{r}=\left(x_{2} \hat{i}+y_{2} \hat{j}\right)-\left(x_{1} \hat{i}+y_{1} \hat{j}\right)$

$=\left(x_{2}-x_{1}\right) \hat{i}+\left(y_{2}-y_{1}\right) \hat{j}$

$=\delta \mathrm{x} \hat{\mathrm{i}}+8 \mathrm{y} \hat{\mathrm{j}}$

displacement along X-axis $\delta x=x_{2}-x_{1}$

displacement along Y-axis $\delta y=y_{2}-y_{1}$

Thus, displacement in 2 dimensions is equal to the vector sum of two one dimensional displacements along mutually perpendicular directions.

Let particle move with uniform velocity at $\overrightarrow{\mathrm{V}}$ an angle $\theta$ with X-axis.

Then in component form $\overrightarrow{\mathrm{v}}=\mathrm{v}_{\mathrm{x}} \hat{\mathrm{i}}+\mathrm{v}_{\mathrm{y}} \hat{\mathrm{j}}$

here $v_{x}=v \cos \theta$

and $\quad v_{y}=v \sin \theta$

and $\quad \delta \mathrm{x}=\mathrm{v}_{\mathrm{x}} \delta \mathrm{t}$

$\delta \mathrm{y}=\mathrm{v}_{\mathrm{y}} \delta \mathrm{t}$

or $\delta x=(v \cos \theta) \delta t$

$\delta y=(v \sin \theta) \delta t$

so with $\mathrm{V}_{\mathrm{x}}$ we get displacement along X-axis only and $\mathrm{v}_{\mathrm{y}}$ gives displacement along Y-axis only.

And if particle is moving with uniform acceleration $\overrightarrow{\mathrm{a}}$, then

$\overrightarrow{\mathrm{a}}=\mathrm{a}_{\mathrm{x}} \hat{\mathrm{i}}+\mathrm{a}_{\mathrm{y}} \hat{\mathrm{j}}$

If direction of $\vec{a}$ makes angle $\phi$ with X-axis then $a_{x}=a \cos \phi$ and $\mathrm{a}_{\mathrm{y}}=\mathrm{a} \sin \phi$

are components of $\overrightarrow{\mathrm{a}}$.

Due to $\mathrm{a}_{\mathrm{x}}$, there is a change in the X component of velocity only with no change in Y-component.

Similarly, $\mathrm{a}_{\mathrm{y}}$ will change only the Y component of velocity at time t

So $v_{x}=u_{x}+a_{x} t$

(here $\mathrm{u}_{\mathrm{x}}$ and $\mathrm{u}_{\mathrm{y}}$ are components of initial velocity)

And

$v_{y}=u_{y}+a_{y} t$

Hence

$v_{x} \hat{i}+v_{y} \hat{j}=\left(u_{x}+a_{x} t\right) \hat{i}+\left(u_{y}+a_{y} t\right) \hat{j}$

$=\left(u_{x} \hat{i}+u_{y} \hat{j}\right)+\left(a_{x} \hat{i}+a_{y} \hat{j}\right) t$

Or

$\overrightarrow{\mathrm{v}}=\overrightarrow{\mathrm{u}}+\overrightarrow{\mathrm{a}} \mathrm{t}$

and similarly, component of displacement are

$s_{x}=u_{x} t+\frac{1}{2} a_{x} t^{2}$

And

$s_{y}=u_{y} t+\frac{1}{2} a_{y} t^{2}$

Hence

$s_{x} \hat{i}+s_{y} \hat{j}=\left(u_{x} \hat{i}+u_{y} \hat{j}\right)+\frac{1}{2}\left(a_{x} \hat{i}+a_{y} \hat{j}\right) t^{2}$

Or

$\vec{s}=\vec{u} t+\frac{1}{2} \vec{a} t^{2}$

Newtons Laws of Motion

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Hey, students do you want to know how to state the law of radioactive decay? If yes. Then read this article till the end.

The rate of decay (number of disintegrations per second) is proportional to number of radioactive atoms (N) present at that time t

rate of decay $\frac{-\mathrm{d} \mathrm{N}}{\mathrm{dt}} \propto \mathrm{N}$

or $\frac{\mathrm{d} \mathrm{N}}{\mathrm{dt}}=-\lambda \mathrm{N}$

or $\quad N=N_{0} e^{-\lambda t}$…..(1)

where $\lambda$ is disintegration constant, $\mathrm{N}_{0}$ = number of active atoms at t = 0
1. Equation one is the radioactive decay law. It shows that the number of active nuclei decreases exponentially with time.
2. The fraction of active atoms remaining at time t is$\frac{\mathrm{N}}{\mathrm{N}_{0}}=\mathrm{e}^{-\lambda \mathrm{t}}$
3. The number of atoms that have decayed in time t is$N_{0}-N$

$=N_{0}\left(1-e^{-\lambda t}\right)$
4. The fraction of atoms that have decayed in time t is$\frac{N_{0}-N}{N_{0}}$

$=1-e^{-\lambda t}$

## Decay constant

1. Decay constant$\lambda=\frac{-\mathrm{d} \mathrm{N} / \mathrm{dt}}{\mathrm{N}}$

$=\frac{\text { rate of decay }}{\text { number of active atoms }}$

2. at $t=\frac{1}{\lambda}$$\mathrm{N}=\frac{\mathrm{N}_{0}}{\mathrm{e}}$

The decay constant of a radioactive element is equal to the reciprocal of the time after which the number of remaining active atoms reduces to $\frac{1}{\mathrm{e}}$ times of original value.

3. at $t=\frac{1}{\lambda}$fraction of active nuclei left

$\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{1}{\mathrm{e}}=0.37$

or $37 \%$

fraction of decayed nuclei

$1-\frac{N}{N_{0}}$

$=0.63=63 \%$
4. $\lambda=\frac{\mathrm{dN} / \mathrm{N}}{\mathrm{dt}}$ The decay constant is the probability of decay per active atom per unit time.

5. The decay constant depends on the nature of the radioactive substance and is independent of temperature, pressure, force, etc.

6. The decay constant for a stable substance is zero

7. Unit of decay constant is second $^{-1}$ and dimension is $\mathrm{T}^{-1}$

8. If there are more than one radioactive elements in a group then the resultant decay constant is equal to sum of individual decay constants

$\lambda=\lambda_{1}+\lambda_{2}+\lambda_{3}+$……..

or

$\frac{1}{\mathrm{~T}}=\frac{1}{\mathrm{~T}_{1}}+\frac{1}{\mathrm{~T}_{2}}+\ldots$

## Half life

The time in which the number of radioactive atoms reduce to half of its initial value is known as half-life i.e. at

$\mathrm{t}=\mathrm{T}$

$N=\frac{N_{0}}{2}$

$\frac{N_{0}}{2}=N_{0} e^{-\lambda T}$

or

$\mathrm{T}=\frac{0.693}{\lambda}$
1. The half-life depends on the nature of radioactive elements.
2. The half-life of an element indicates the rate of decay. When half-life is a large rate of decay is small.
3. After t = nT number of active atoms left$N=\frac{N_{0}}{2^{n}}$

$=\frac{1}{2^{t / T}} \cdot \mathrm{N}_{0}$

where T = half-life and n = number of half-lives.
4. Number of radioactive atoms decayed in n half-lives$N_{0}-\frac{N_{0}}{2^{n}}$
$=N_{0}\left(\frac{2^{n}-1}{2^{n}}\right)$
5. The half-life for a given radioactive substance is constant. It does not change with time. It is unaffected by pressure, temperature, etc.

so, that’s all from this blog, I hope you get the idea about how to state the law of radioactive decay. If you found this article informative then please share it with your friends.

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Hey, do you want to learn about the uses of radioactive isotopes? If yes. Then you are at the right place.

1. In Medicine

a. $\mathrm{Co}^{60}$ for treatment of cancerb. $\mathrm{Na}^{24}$ for circulation of blood

c. $\mathbf{I}^{131}$ for thyroid

d. $S r^{90}$ for treatment of skin & eye

e. $\mathrm{Fe}^{59}$ for location of brain tumor

f. Radiographs of castings and teeth

2. In Industries

a. For detecting leakage in water and oil pipe lines

b. For investigation of wear & tear, study of plastics & alloys, thickness measurement.

3. In Agriculture

a. $\mathrm{C}^{14}$ to study kinetics of plant photosynthesis

b. $\mathrm{P}^{32}$ to find nature of phosphate which is best for given soil & crop

c. $\mathrm{Co}^{60}$ for protecting potato crop from earth worm

d. sterilization of insects for pest control.

a. $\mathrm{K}^{40}$ to find age of meteorites

b. $S^{35}$ in factories
5. Carbon dating

a. It is used to find age of earth and fossils

b. The age of earth is found by Uranium disintegration and fossil age by disintegration of $\mathrm{C}^{14}$.

c. The estimated age of earth is about $5 \times 10^{9}$ years.

d. The half-life of $\mathrm{C}^{14}$ is 5700 years.

6. As Tracers

a. A very small quantity of radio isotope present in any specimen is called tracer.

b. This technique is used to study complex biochemical reactions, in detection of cracks, blockages etc, tracing sewage or silt in sea.

7. In Geology

a. For dating geological specimens like ancient rocks, lunar rocks using Uranium.

b. For dating archaeological specimens, biological specimens using $\mathrm{C}^{14}$.

Also, see
Nuclear Fission and Fusion

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Working of NPN and PNP transistor – Electronics, Physics – eSaral

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Hey, do you want to learn about the Working of NPN and PNP transistor? if yes. Then you are at the right place.

## P-N-P Transistor:

The holes of the P region (emitter) are rippled by the positive terminal of battery $\mathrm{V}_{\mathrm{EE}}$ towards the base. The potential barrier at the emitter junction is reduced as it is forward bias and hence the holes cross this junction and penetrate into the N region. This constitutes the emitter current. The width of the base region is very thin and it is lightly doped and hence only two to five percent of the holes recombine with the free electrons of the N region. This constitutes the base current $\mathrm{I}_{\mathrm{B}}$. which of course, is very small. The remaining holes (95% to 98%) are able to drift across the base and enter into the collector region. This constitutes the collector current.

## N-P-N Transistor:

Working:
The electron in the emitter region is rippled from the negative terminal of the battery towards the emitter junction. Since the potential barrier at the junction is reduced due to forward bias and the base region is very thin and lightly doped, electrons cross the p-type base region. A few electrons combine with the holes in P-region and are lost as the charge carriers. Now the electrons in N-region (collector region) readily swept up by the positive collector voltage $\mathrm{V}_{\mathrm{CC}}$.

The current conduction in the N-P-N transistor is carried out by electrons.

## Transistor configuration

Three types of transistor circuit configuration are

Common base (CB)

Common emitter (CE)

Common collector (CC)

The term ‘Common’ is used to denote the transistor lead which is common to the input and the output.

### Common Base Configuration

In this configuration the input signal is applied between emitter and base and the output is taken from collector and base.

In common base as an amplifier phase difference between input and output is zero.

## Characteristics:

### Input characteristic:

The curve between emitter current $\mathrm{I}_{\mathrm{E}}$ and emitter base voltage $V_{\mathrm{EB}}$ at constant collector base voltage $\mathrm{V}_{\mathrm{CB}}$ represents the input characteristics. Collector base voltage $\mathrm{V}_{\mathrm{CB}}$ is kept fixed.
• There exist a cut in, offset or threshold voltage $V_{\mathrm{EB}}$ below which the emitter current is very small.
• The emitter current $\mathrm{I}_{\mathrm{E}}$ increases rapidly with small increase in emitter-base voltage $V_{\mathrm{EB}}$. This shows that the input resistance is very small.

### Output characteristic:

Curve between collector current $\mathrm{I}_{\mathrm{C}}$ and collector base voltage $V_{\mathrm{CB}}$ at constant emitter $\mathrm{I}_{\mathrm{E}}$ represents the output characteristic.

## Common emitter configuration

In this configuration the input signal is applied between base and emitter and the output is taken from collector and emitter.

In common emitter as an amplifier phase difference between input and output is $\pi$.

## Characteristic

### Input characteristics:

The curve between base current $\mathrm{I}_{\mathrm{B}}$ and base emitter voltage $\mathrm{V}_{\mathrm{BE}}$ at constant collector-emitter voltage $\mathrm{V}_{\mathrm{CE}}$ represents the input characteristic.

In this case, $\mathrm{I}_{\mathrm{B}}$ increases less rapidly with $V_{B E}$ as compared to common-base configuration. This shows input resistance of the common-emitter circuit is higher than that of the common-base circuit.

### Output characteristic:

The curve between collector current $\mathrm{I}_{\mathrm{C}}$ and collector emitter voltage $V_{C E}$ at constant base current $\mathrm{I}_{\mathrm{B}}$ represents the output characteristic.

Relationship between $\alpha, \beta$ and $\gamma$:

Emitter current $I_{E}=I_{B}+I_{C}$……(1)

divide by $\mathrm{I}_{\mathrm{C}}$ gives, $\frac{\mathrm{I}_{\mathrm{E}}}{\mathrm{I}_{\mathrm{C}}}=\frac{\mathrm{I}_{\mathrm{B}}}{\mathrm{I}_{\mathrm{C}}}+1$

or

$\frac{1}{\alpha}=\frac{1}{\beta}+1$

$\beta=\frac{\alpha}{1-\alpha}$

## Comparison table between CB, CE and CC configuration

So, that’s all from this blog, I hope you get the idea about the Working of NPN and PNP transistor. If you enjoyed this explanation then don’t forget to share it with your friends.

What is Transistor

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A Transistor is an electronic device formed by P and N types of semiconductors. If you want to know about what is transistor and its types then read this article till the end.

## Transistor

• Transistor is used in place of triode valve, it is an electronic device formed by P and N types of semiconductors. It was discovered by American scientist J Barden, W.H. Bratain, W. Shockley.
• The transistor is three-terminal and two P-N Jn. device.

## There are two types of transistors

1. N-P-N transistor
2. P-N-P transistor

A transistor (P-N-P or N-P-N) has the following sections:

Emitter: It emits the charge carriers and it is heavily doped.

Base: The middle section of the transistor is known as the base. This is very lightly doped and very thin ($10^{-6}$ m).

Collector: This is moderately doped. It collects the charge carriers

Important points
• The cross-sectional area of the base is very large as compared to the emitter.
• The cross-sectional area of the collector is less than the base but greater than the emitter.
• The base is much thinner than the emitter while the collector is wider than both because the emitter emits charge carriers and the collector has to receive a maximum of them. Since the base is very thin, so recombination in the base region is very less maximum 5%.
• The emitter is heavily doped so that it can provide a large number of charge carriers (electrons or holes) into the base. The base is lightly doped and is very thin, hence the majority of charge carriers move on to the collector. This lower doping decreases the conductivity (increases the resistance) of the base material by limiting the number of charge carriers to the collector. The collector is moderately doped.
• The emitter is always connected in forward biased and the collector is always connected in reverse biased.
• The resistance of the emitter-base junction (forward-biased) is very small as compared to a collector-base junction (reverse biased). Therefore, the forward bias applied to the emitter base is generally very small whereas reverse bias on the collector base is much higher.
• Arrowhead always shows the emitter’s current direction.
• Current conduction within the P-N-P transistor takes place by hole conduction from emitter to collector.
• Conduction in the external circuit is carried out by electrons.

So, that’s all from this blog. I hope you get the idea about what is transistors and its types. If you have any confusion related to this topic then feel free to ask in the comments section down below.

Different types of PN-junction diodes

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Diode is a PN junction device. If you want to know about the different types of PN-junction diodes then keep reading.

## Different types of PN-junction diodes

### 1. Light Emitting Diode (LED)

Very much used in a dancing light display of music systems and information boards on railway stations.
In a forward-biased diode, the energy produced by recombination of electrons and holes at the junction can be emitted. If the energy is in the visible region, such a diode is called a light-emitting diode or LED.

The color of emitted light depends on the type of material used

Ga-P Red or green light

Ga-As-P Red or yellow light

A circuit for LED is shown in fig. The brightness can be controlled by $\mathrm{R}_{\mathrm{L}}$.

### 2. Photodiode

A PN junction diode made of photosensitive semiconductor is called a photodiode.

In photodiode, provision is made for allowing the light to fall at the junction.

Its function is controlled by the light allowed to fall on it.

In a semiconductor, the electrons jump from the valence band to the conduction band by absorbing energy from some external source of energy.

Energy of light photon

$E=\frac{h c}{\lambda}$

If this energy is sufficient to break a valence bond, when such light falls on the junction, new hole-electron pairs are created.

The number of charge carriers increases and hence the conductivity of the junction increases.

Fig. shows a circuit in which the photodiode is reverse-biased.

The applied voltage is less than the breakdown voltage. When visible light of suitable energy (hn > forbidden energy gap) is made incident on the photodiode, current begins to flow due to shifting of electrons from valence band to conduction band.

This current increases with the increase in the intensity of incident light. If the intensity of light increases to a value, the current becomes maximum. This maximum current is called saturation current.

### 3. Solar Cell

A pn-junction diode in which one of the P-or N regions is made very thin (so that the light energy is not greatly absorbed before reaching the junction) is used in converting light energy to electrical energy. Such diodes are called solar cells. In the solar cell, the thin region is called the emitter and the other base. When a light incident on the emitter, a current in the resistance $R_{\mathrm{L}}$ (Fig.). The magnitude of the current depends on the intensity of light.

Unlike a photodiode, a solar cell is not given any biasing.

It supplies emf like an ordinary cell.

The solar cell is based on the photovoltaic effect. When the light of suitable frequency is made incident on an open-circuited solar cell, an emf is produced across its terminals. This emf is called photo-voltaic emf, the effect is called the photovoltaic effect.

Uses
1. We can use a set of solar cells to charge storage batteries in the daytime. These batteries can be used for power during the night.
2. Solar cells are extensively used in calculators, wristwatches, and light meters (in photography).
3. The power source in artificial satellites is a solar panel which is an array of solar cells.

### 4. Zener Diode

Zener diode is a PN junction diode.

By careful adjustment of the concentration of acceptor and do not impurity atoms near the junction, the characters beyond the turn over-voltage become almost a vertical line.

Thus, in this region of its characteristic curve, the reverse voltage across the diode remains almost constant for a large change of the reverse current.

Therefore, a Zener diode is used as a voltage reference device for stabilizing a voltage at a predetermined value.

Zener diodes have been designed to operate from 1 to several hundred volts. In the diodes which are operated below 6V the breakdown of the junction is due to the Zener effect.

In those operated between 5 and 8V, the breakdown is due to both the Zener effect and the avalanche multiplication.

In general, all diodes which are operated in the breakdown region of their reverse characteristic are known as Zener diodes.

Zener diode is a reverse-biased heavily doped P-N junction diode. Which is operated in the breakdown region.

So, that’s all from this article. I hope you get the idea about the Different types of PN-junction diodes. If you found this Explanation helpful then please share it with your friends and social media followers.

Bridge rectifier circuit Diagram

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Bridge Rectifier is also a full-wave rectifier. If you want to learn about the Bridge rectifier and Bridge rectifier circuit diagram then keep reading.

## Bridge Rectifier

It is also a full-wave rectifier.

Use of bridge rectifier

The bridge rectifier is used in the rectifier type voltmeter. The circuit arrangement is shown in fig. The rectifier elements are p-n junction diodes with a sensitive DC. ammeter as a load. This circuit can be used for the measurement of AC as well as DC voltage and currents. The DC ammeter reads the average value of currents. This may be calibrated to give r.m.s. values.

The bridge rectifier has the following advantages
• No center tap is required in the transformer secondary, hence both half cycles are similar.
• PIV across each diode is $\mathrm{E}_{0}$ (which is 2 $\mathrm{E}_{0}$ in a full wave rectifier).
• For a given power, output power transformer of small size can be used as the current in both the primary and secondary of the plate supply transformer flows for the entire cycle.
The main disadvantages of bridge rectifier are
• The circuit requires two extra diodes
• It has poor voltage regulation.

## The efficiency of rectifier:

The efficiency rectifier is defined as the ratio of DC output power to the AC input power

$\eta=\frac{\text { dc power delivered to the load }}{\text { ac input power from transformer sec ondary }}$

$=\frac{P_{d c}}{P_{a c}}$

$=\frac{\mathrm{I}^{2} \mathrm{dc} \mathrm{R}_{\mathrm{L}}}{\mathrm{I}_{\mathrm{rms}}^{2}\left(\mathrm{R}_{\mathrm{F}}+\mathrm{R}_{\mathrm{L}}\right)}$

For bridge rectifier

$\eta=\frac{0.812 R_{L}}{2 R_{f}+R_{L}}$

for ideal diode

$R_{f}=0$

$\eta=81.2 \%$

Full Wave Rectifier circuit diagram

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A device that rectifies both halves of the ac input is called a full-wave rectifier. If you want to learn about the full-wave rectifier and full-wave rectifier circuit diagram then keep reading.

## Full-wave rectifier:

A rectifier that rectifies both halves of the ac input is called a full-wave rectifier.

During the first half of the input cycle, the upper end of S coil is at positive potential and the lower end is at the negative potential the junction diode $\mathrm{D}_{1}$ will get forward biased, while the diode $\mathrm{D}_{2}$ reverse biased. The conventional current due to the diode $\mathrm{D}_{1}$ will flow.

When the second half of the input cycle comes, the situation will be exactly reverse. Now, the junction diode $\mathrm{D}_{2}$ will conduct and the current will flow.

## The efficiency of rectifier:

The efficiency rectifier is defined as the ratio of dc output power to the ac input power

$\eta=\frac{\text { dc power delivered to the load }}{\text { ac input power from transformer sec ondary }}$

$=\frac{P_{d c}}{P_{a c}}$

$=\frac{\mathrm{I}^{2} \mathrm{dc} \mathrm{R}_{\mathrm{L}}}{\mathrm{I}_{\mathrm{rms}}^{2}\left(\mathrm{R}_{\mathrm{F}}+\mathrm{R}_{\mathrm{L}}\right)}$

for Full wave rectifier

$\eta=\frac{.812}{1+\frac{R_{\mathrm{F}}}{R_{\mathrm{L}}}}$

if $\frac{R_{F}}{R_{L}}<<1$

$\eta=81.2 \%$

## Ripple and Ripple factor:

AC components are present in rectifier output these are known as ripple and they are measured in a factor which is known as Ripple Factor

Total Current Output

$\mathrm{I}_{\mathrm{rms}}=\sqrt{\mathrm{I}_{\mathrm{ac}}^{2}+\mathrm{I}^{2} \mathrm{dc}}$

Ripple Factor $=r=\frac{I_{\mathrm{ac}}}{\mathrm{I}_{\mathrm{dc}}}$

$r=\sqrt{\left(\frac{I_{r m s}}{I_{d c}}\right)^{2}-1}$

For full-wave or bridge rectifier

$\mathrm{I}_{\mathrm{rms}}=\frac{\mathrm{I}_{\mathrm{m}}}{\sqrt{2}}$

$I_{\mathrm{dc}}=\frac{2 \mathrm{I}_{\mathrm{m}}}{\pi}$

$r=0.48$

So, that’s all from this article. I hope you get the idea about the full-wave rectifier and full-wave rectifier circuit diagram. If you have any confusion related to this blog then feel free to ask in the comments section down below.

Half Wave Rectifier Circuit Diagram

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A device that converts alternating current into Direct current is called a rectifier. If you want to learn about the half-wave rectifier and half-wave rectifier circuit diagram then you are at the right place.

## Application of diode as a rectifier:

An electronic device that converts alternating current into Direct current is called a rectifier.

### Half wave rectifier:

A rectifier, which rectifies only one half of each ac supply cycle is called a half-wave rectifier.

During the first half of the input cycle, the junction diode gets forward bias. The conventional current will flow. The upper end of $R_{L}$ will be positive potential with respect to the lower end during the second half cycle junction diode will get reverse biased and hence no output will be obtained across $R_{L}$.

Input voltage

$\mathrm{V}_{\mathrm{i}}=\mathrm{V}_{\mathrm{m}} \sin \omega \mathrm{t}$

$\mathrm{i}=\mathrm{I}_{\mathrm{m}} \sin \omega \mathrm{t}$

for $0 \leq \omega t \leq \pi$

$\mathrm{i}=0$

for $\pi<\omega t<2 \pi$

$I_{m}=\frac{V_{m}}{R_{f}+R_{L}}$

here $\mathrm{R}_{\mathrm{f}}=$ forward resistance of diode

$R_{L}$ = load resistance

a. dc output current :

$\mathrm{I}_{\mathrm{dc}}=\frac{1}{2 \pi} \int_{0}^{2 \pi} \mathrm{idt}$

$=\frac{1}{2 \pi}\left[\int_{0}^{\pi} \mathrm{I}_{\mathrm{m}} \sin t \mathrm{dt}+\int_{\pi}^{2 \pi} 0 \mathrm{dt}\right]$

$\mathrm{I}_{\mathrm{dc}}=\frac{\mathrm{I}_{\mathrm{m}}}{\pi}=0.318 \mathrm{I}_{\mathrm{m}}$

b. dc output voltage:

$V_{d c}=I_{d c} \times R_{L}$

$=\frac{I_{m}}{\pi} \times R_{L}$

$=\frac{\mathrm{V}_{\mathrm{m}}}{\pi\left[1+\left(\mathrm{R}_{\mathrm{f}} / \mathrm{R}_{\mathrm{L}}\right)\right]}$

$V_{d c}=\frac{V_{m}}{\pi}=0.318 \mathrm{~V}_{\mathrm{m}}$

c. (Root mean square) RMS current:

$I_{r m s}=\left[\frac{1}{2 \pi} \int_{0}^{2 \pi} i^{2} d(t)\right]^{1 / 2}$

$=\frac{I_{m}}{2}$

same

$V_{r m s}=\frac{V_{m}}{2}$

So, that’s all from this blog. I hope you enjoyed this explanation of the half-wave rectifier and half-wave rectifier circuit diagram. If you liked this article then please share it with your friends.

What is Diode in electronics

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A diode is a PN junction device. If you want to know what is diode in electronics. Then keep reading.

### Diode

A diode is a PN junction device

## Ideal diode

• Conducts with zero resistance when forward biased.
• Offer an infinite resistance when reverse biased.

## Practical Circuit For diode

### The practical characteristic curve for a diode:

Important terms related to diode:
1. Knee VoltageKnee voltage is defined as the forward voltage at which the current through the junction starts increasing rapidly.For silicon = 0.7 volt for germanium = 0.3 volt

2. Forward resistance or ac resistance:It is defined as the reciprocal of the slope of the forward characteristic curve.forward resistance

$r_{f}=\frac{1}{\text { slope of forwardcharacteristic }}$

$=\frac{1}{\Delta \mathrm{I}_{\mathrm{f}} / \Delta \mathrm{V}_{\mathrm{f}}}$

$=\frac{\Delta \mathrm{V}_{\mathrm{f}}}{\Delta \mathrm{l}_{\mathrm{f}}}$

3. Junction breakdown :When the reverse voltage is increased a point is reached when the junction breaks down with sudden rise in reverse current. This value of the voltage is known as the breakdown voltage. Two types of breakdown occur:a. Zener breakdown:

a. Zener breakdown:
Takes place in junction which are heavily doped so having narrow depletion layers. A very strong electric field appears across the narrow depletion layer which breaks the bond.

b. Avalanche breakdown :

Occur in junctions which are lightly doped. (having wide depletion layer) so at a high electric field, the minority charge carriers, while crossing the junction acquire very high velocities. A chain reaction is established, giving rise to the high current.

4. Diffusion Current :Some electrons and holes have more kinetic energy $\left[\frac{1}{2} \mathrm{mv}^{2}>\mathrm{eV}\right]$ So $\mathrm{e}^{-}$ diffuse from n to p side and hole diffuse from p to n side due to diffusion of the charge carriers a current will flow known as diffusion current.

a. Because of the concentration difference diffusion occurs.

b. Diffusion results in an electric current from p side to the n side

c. When P-N. Jn is in no Bias = diffusion current = drift current

Net charge flow = 0

Net current = 0

5. Drift Current: Due to thermal collisions, the covalent bond is broken. If an electron-hole pair is created in the depletion region, there is a regular flow of electrons towards the n side and of holes towards the p side. Current flow n side to p side called drift current.Drift current and the diffusion current are in the opposite direction

So, that’s it from this blog. I hope you get the idea about what is a diode in electronics. If you found this Explanation helpful then share it with your friends and followers.