NEET 2025 Physics: Mastering Kepler’s Third Law of Planetary Motion

Are you preparing for NEET 2025 or IIT-JEE? In this video, we solve a high-yield question from the Gravitation chapter focusing on planetary orbits and time periods.

What You Will Learn:

In this tutorial, we calculate the orbital period of Mercury based on its relationship with Mars using Kepler’s Third Law [00:39].

  • The Law Explained: Understand the Law of Periods, which states that the square of the time period is directly proportional to the cube of the orbital radius  [00:47].

  • Step-by-Step Calculation: We show you how to set up the ratio  to solve for unknown values [00:59].

  • Practical Application: Learn how to handle "four times the radius" problems and simplify cubic/square root calculations effectively [01:24].

  • Exam Strategy: We discuss how to identify the "best possible answer" from multiple-choice options when your calculation results in an approximation [02:31].

Key Highlights:

  • Problem Statement: Comparing Mars' orbit (687 days) with Mercury’s orbit when the radius of Mars is 4x that of Mercury [00:08].

  • Simplified Math: Taking the square root of 64 to find the time ratio [01:50].

  • Final Answer: Why 88 days is the most accurate choice for Mercury's orbital period in competitive exams [02:25].




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Master NEET 2025 Physics: Thermal Properties & Ideal Gas Equation Explained!

Are you preparing for NEET 2025? In this video, we dive deep into a crucial physics problem that combines the Thermal Properties of Matter with the Ideal Gas Equation. Understanding how these concepts interact is key to scoring high in the physics section!



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Master NEET Physics 2025: Thermodynamics First Law Solved Problem

Are you preparing for NEET 2025 and looking to master Thermodynamics? In this video, we break down a high-yield question based on the First Law of Thermodynamics, specifically focusing on the relationship between heat, internal energy, and work done in cylindrical systems.

What You Will Learn:

In this tutorial, we solve a specific problem involving two gases (A and B) in separate cylinders with movable pistons. You will learn how to:

  • Apply the First Law of Thermodynamics formula: dQ = dU + dW [01:11].

  • Understand how constant pressure and equal heat supply affect work done [01:26].

  • Relate volume change to cylinder radius and displacement [02:01].

  • Step-by-step calculation to find the ratio of radii when given displacement values [02:44].

Problem Breakdown:

The problem eXplores two systems where the heat energy supplied  and the change in internal energy  are identical. By isolating the work done , we demonstrate that the product of the cross-sectional area and the displacement remains constant. This leads us to the final ratio of for the radii given the displacements of and [02:58].



 

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NEET 2025 Electro Magnetism Question on Magnetic moment of current carrying ciruclar coil

 

This video provides a step-by-step solution to a physics problem regarding the magnetic moment of current-carrying circular coils. Below is an SEO-friendly description written in simple English without special characters or symbols to help boost views for your video and blog.

NEET 2025 Physics Solution Magnetic Moment of a Circular Coil

In this video we solve a specific physics question from the NEET 2025 exam. The lesson focuses on the chapter of electromagnetism. We explain how to calculate the ratio of magnetic moments for two different circular copper coils when they carry the same amount of current.

What You Will Learn in This Video

This tutorial covers the fundamental formula for the magnetic moment of a circular conductor. You will see how the magnetic moment relates to the number of turns the current and the area of the coil. We break down the math simply so you can understand how changing the radius of a coil affects its magnetic properties.

Key Topics Covered

  • Understanding the magnetic field in a current carrying coil.

  • The formula for magnetic moment using current and area.

  • How to find the ratio of magnetic moments when radii are different.

  • Step by step calculation for a common entrance exam question.

Visit Our Blog for More Physics Tips

If you want more practice problems and detailed notes on electromagnetism please visit our blog. We have a full guide that explains these concepts in detail to help you prepare for your exams.

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If this explanation helped you please like the video and share it with other students. Subscribe to our channel for more NEET and JEE physics solutions. Click the bell icon to get updates on new lessons.

Leave a Comment

Do you have questions about electromagnetism or other physics topics? Let us know in the comments below and we will be happy to help.

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Physics JEE and NEET Complete Class 11 and Class 12 Chapters Video Lessons

Here in this place at Venkats Academy, you will be able to find detailed class room notes and video lessons for class 11 and class 12 right from basics to advanced topics in detail with example problems and solutions. You can also find all previous NEET and JEE Physics questions solved so that you can understand the trend in which the questions are being asked.

CLASS 11 Physics Chapters Video Lessons 
CLASS 12 Physics Chapters Video Lessons
Class 11 Physics Chapters NEET Previous Exam Questions
Class 12 Physics Chapters NEET Previous Exam Questions
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Electric Intensity and Torque due to Electric Dipole Video Lesson

Electric dipole is the combination of two charges of equal magnitude but opposite nature separated by small distance.Electric intensity is the force experienced by a unit positive charge placed in the electric field of dipole. Torque is the turning experience got by a electric dipole when placed in an external electric field.

 Electric Field intensity on the axial line of dipole

A line passing through two charges is called axial line. We would like to measure the electric field intensity at any point on the axial line of electric dipole. Let us consider a point on the axial line at a distance r from the center of the dipole. We shall imagine a unit positive charge at the considered point. It experience force both due to positive and negative charge. Due to Positive charge force is repulsive and due to negative charge it is attractive. Using Coulomb's inverse square law, we need to write equations  for the force experienced by unit positive charge at the given location.We need to measure the effective force as the difference between the two charges and it can be further simplified as shown in the video lesson below. Electric dipole moment is the product of any one charge of the dipole to the distance between the two charges of dipole. It is a vector quantity and its direction is from negative charge towards positive charge. Intensity is expressed in terms of dipole moment as shown below.



Electric field intensity on equatorial line of Dipole

Equatorial line is a line passing through the center of dipole and perpendicular to the axial line. Let us consider a point on that line that is at a finite distance  from center of dipole. We shall imagine a unit positive charge at that point and it experience force due to both positive and negative charge of the dipole. Its magnitude can be determined using inverse square law and its value is shown in the video below. The force due to positive charge is repulsive on unit positive charge and force due to negative charge is attractive. Their directions were identified and the resultant is determined using the vector laws of addition as  shown in the video lesson below. It can be noticed that the electric intensity on the equatorial line is half that of intensity on the axial line. A detailed proof is given in the video lesson below. Its direction is also shown.



Electric Intensity at any point on the dipole

Let us consider a point around the dipole that is neither on the axial line or equatorial line and the point is at a distance and is making some angle with the horizontal line. To find the electric field intensity at that point, we can consider the dipole as the combination of two dipoles that are perpendicular to each other as shown in the video lesson. For one dipole the considered point is on the axial line and for the other imagined dipole the point is on the equatorial line. As we have derived the equations for the intensity on axial line and equatorial line, we can use that equations and they two are perpendicular to each other. By simplifying them further as shown in the video lesson, we can get the resultant equation as shown. This is a generic equation and in that equation, if the angle is zero, the point will be on axial line and if the angle is ninety degree, the point goes to equatorial line. 


Torque experienced by dipole when placed in a uniform electric field

Let us consider a dipole of two charges separated by a small distance and let us apply a electric field of known intensity on it. Each charge experience a force and and the two forces are equal in magnitude but opposite in direction. But they don't cancel each other as the two forces are acting on different points of the electric dipole. Thus it experience a turning effect in anti clock wise direction and we can measure the torque as shown in the video lesson below. Torque is defined as the product of any one force and the perpendicular distance between the two forces acting on the dipole. It is a vector and we can find the direction using the right hand thumb rule or cork screw rule as shown in the video lesson below.



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Neutral Point and Equilibrium of Electric charge in a line Video Lesson

Electrically  neutral point is a location at which the resultant electric force is zero. It means the charge kept at that point is experiencing equal and opposite force due to the other two charges kept in that line. To find the neutral point, we can equate the two forces and when the two charges are of similar nature, we can get null point between the charges existing in one line and beyond them, we are not going to find the neutral point. At a location beyond the neutral point, the force acting on the third charge are in the same direction and hence there will  not be any neutral point. It is explained in the following video lesson and the location of neutral point is also derived.



Neutral Point between opposite charges

When the charges are of opposite nature, the resultant force on a third charge in between them is not zero and it is experiencing the two force in the same directions. If we consider a location out side the two charges and consider a third charge, it experience forces due to two charges in opposite direction. If they are equal in magnitude, we can get null point as shown in the video lesson. The expression for the location of null point is also derived here.


Equilibrium of system of three charges

Now we are considering a scenario of three charges kept on a straight line. If the two charges at the ends of the line, we can get null point between them weather we keep positive or negative charges in between them. But to keep any of the positive charge at the end of the system, we shall only keep a negative charge but not positive charge in between them. Thus to keep system of three charges in equilibrium, we need to keep two positive charges at the ends and a negative charge in between. We need to write the conditions  for zero resultant force location in at least on two charges so that we can get the magnitude,location and nature of the charge in between to get the answer as shown in the video lesson below.
 

Finding separation between two charges suspended from rigid support

Let us consider two identical ball of same mass and similar charge kept suspended from a rigid support using a light wire and because of the repulsion there will be some angular as well as physical separation between them. We need to find that angular separation in terms of the given data and the video solution is as shown below.

Motion of Charged particle in Electric Field

Let us consider a charged particle is moving horizontally with some constant velocity and an electric field is applied perpendicular to the motion. Gravitational force is small and we can ignore it in this case. Thus there is no force in the horizontal direction and hence the charged particle will have constant velocity over the horizontal direction. 

There is no initial velocity along the vertical direction but there is electric force due to intensity along the Y axis. Thus the particle experience acceleration along Y axis and hence the velocity of the particle increases on that axis. We can find the final velocity along each direction and its displacement along both the axes. It is further shown in the video lesson that the path of the charged particle is parabola as shown below.


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Electric force and Coulomb's Inverse square law Video Lesson

Electric force of attraction or repulsion exists between every two charges and it follows Coulomb's inverse square law. According to the law, the force between two charges is directly proportional to the product of charges and is inversely proportional to the square of the distance of separation.

Gravitational and electric forces are fundamental in nature. Gravitational force is due to mass of the particle and it exits between every two masses where as electric force exits between the the particles or bodies having only excess charges. Both of them obeys inverse square law but gravitational force is independent of medium but electric force depends on the medium between charges.

Gravitational force is the weakest of fundamental forces but electric force is much stronger than that. Gravitational force is further always attractive force but electric force is either attractive or repulsive. Further classification is explained in the video lesson below.


Coulomb's inverse square law

To identify the relation between the force and charges different experiments were conducted and we got to a conclusion that the electric force between the charges is directly proportional to the product of charges and is inversely proportional to the square of the distance of the separation.  The force also depends on the medium between the charges and its impact is studied basing on the permitivity of the medium. The ratio of permitivity of the medium to the permitivity of the free space is called dielectric constant or relative permitivity. Its value for the vacuum is one and for any non conducting medium it is greater than one. It is explained as shown in the video lesson below.



Effect of the medium on force between the charges

As we have explained earlier, we can understand that the force between the charges gets reduced when the medium is placed instead of the vacuum. It is further explained in the video lesson below. We can find out the effective distance in the presence of medium as shown below.



Impact of force between two charges when third charge is placed 

Electric force between the charges is not effected even if we keep a new charge in between them. What makes a difference is, more forces are acting on each  charge and hence resultant force on each charge is more. But the force between two charges is not affected because of the third charge. To find the resultant charge, we can use vector laws of addition. It is explained in the video lesson below.


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Properties of Electric Charges Video Lesson

Electric charge is the property of a particle of certain mass due to which it can either attract the opposite charge or repel the similar charge.Charges are classified as positive and negative and many bodies around  are neutral which means they have equal and opposite charges. 

When one body with excess charges is rubbed with the other, there is transfer of electrons from one body to other. The body lost electrons will have excess positive charges and the body that gained electrons will have excess electrons. Thus the body that gained electrons is negatively charged and the body that lost electrons is positively charged.

Thus body that is positively charged losses a small portion of mass due to loss of electrons and the body  that gained electrons is not only negatively charged but its mass is also  slightly increases.



Properties of Charges

Similar charges repel each other and opposite charges attract each other. This force of attraction obeys inverse square law. According to it the force of attraction or repulsion is inversely proportional to the square of distance of separation. It is a long distance range force. Further properties are explained in the video lesson below.


Number of electrons in one coulomb charge

We know that electron is treated like a fundamental charge and it is measured in a unit called coulomb. We also know that the charge is always available as integral multiples of charge of electron and this concept is called quantization of the charge. Taking that concept into consideration, we can find the number of electrons in one coulomb charge as shown in the video lesson below.



Methods of electric charging

We can charge the body by rubbing one body with the other. This is called charging a body by friction and during the rubbing, one body loose the electrons and the other body gains the electrons. We can also charge a body by conducting where we are passing the charge through the materials which has the ability to pass the current. We can also charge a body by keeping the body in contact. In that case excess charges from one body to other get transferred until there is some potential difference between them. It is explained in the following video lesson.

Charging a body by induction

Induction is the phenomenon of rearrangement of charges in a neutral body in the presence of a charged body. This separation of charges is called polarisation of charges and we can get a body charged using induction method with out getting in touch and contact with the body. It is explained in the following video lesson below.



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Free body diagram and Connected Bodies Video Lesson

Free body diagram is the representation of all the forces acting on a body at a given instant. We shall consider only the forces acting on a body but not the forces applied by the same body. When there are multiple bodies, all of them can be treated as one system if they have same acceleration. When we are solving a system of bodies and we are interested in finding the acceleration and may be tension and contact forces  on a given body. We shall draw free body diagram for each body and we shall further write the equation for the resultant force. We shall consider the force along the direction of motion as positive and vice versa. We are also dealing with contact force in the given video below. Contact force is t he force applied by one body on the other when they are in contact. There will be reaction force also according to Newton's third law of motion.A detailed video lesson is presented below regarding free body diagrams as shown below.



Connected bodies and free body diagram

Here we are discussing about the bodies of different masses connected with the help of string and a force is applied on the system. When force is applied, the string becomes tight and a tension is developed in the wire. We need to identify all the forces acting on each body. We need to be careful and identify only forces acting on the body but not the forces applied by the body. This can be done with the help of free body diagram for each body. Thus we can write equation for resultant force on each body  and we can solve that equations and get the tension in each wire and acceleration of the system as shown in the video lesson below.



Atwood's machine

Two bodies of different masses connected over a smooth pulley with the help of light weighted string could be called as Atwood's machine. We need to find the tension int the wire and the acceleration of the system. We can write free body diagram for each body and equations for resultant force as shown in the video lesson below.


Connected bodies one in vertical mode and other in horizontal position

Let us consider a system of two bodies where one body is hanging vertically from a table and the other body is on a smooth horizontal smooth surface and the two bodies are connected with a light weighted string over a smooth pulley. We need to find the tension in the string and the acceleration of the system. We can draw free body diagram to identify the forces acting on each body and we need to write the equations for resultant force as shown in the video lesson below. We can solve the equations as shown below.




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