How do you figure out the size of a nucleus? Especially, a Helium-4 nucleus.
Do you have any good charts online I could use?
I would appreciate it if you would write back.
An estimate of the size of nucleus (nuclear radius) can be made from the empirical formula where R0 is an empirical constant whose value is 1.25 x 10-15m, A is the mass number.
The following document may be helpful to you for further information
Did you get very low marks in Physics and finding it too difficult to get pass marks?
Here is a systematic and easy to follow plan to succeed. Just believe me, you can get pass mark by following the instructions religiously.
1. Don’t leave a single question.
During the cool off time go through all the questions and decide the questions known well to you and mark them to attempt in the first round. Then have a look at the questions which may take some time to solve or of which you have some idea but the answer is not completely known. The rest of questions can be kept for third round where you will write an answer with the concepts known to you based on the hints which you may get from the question itself.
Attempting all questions is very important, because if you write an answer there is a chance to get some marks but if you do not write there is no chance of getting any mark for the question.
2. Practice the frequently asked portions
Please go through “MUST PRACTICE PORTIONS FOR CLASS XII” and practise the portions thoroughly.
3. Practice the Diagrams
You can find a list of important diagrams here shortly.
- Electric field lines due to a points charge, a dipole, two negative charges, two positive charges etc.
- Electric field due to a dipole (axial position and equatorial position) diagram for derivation.
- Torque on a dipole in a uniform electric field
- Equipotential surfaces due to a point source, a line of charge and an electric dipole.
- Combination of resistors (in series and in parallel)
- Combination of cells. (diagram for derivation)
- Motional emf
- Mutual inductance of two solenoids
- AC generator
- AC circuit with R only
- AC circuit with L only
- AC circuit with C only
- LCR AC series circuit
- Impedance diagram
- LC oscillations
- Resonance in LCR series circuit
- Displacement current
- Ray diagram for image formation by Simple Microscope
- Ray diagram for image formation by Compound Microscope
- Ray diagram for image formation by Telescope in normal adjustment
- Ray diagram for image formation by Telescope when final image is at the least distance of distinct vision
- Prism formula
- Laws of reflection using Huygen’s wave theory
- Laws of reflection using Huygen’s wave theory
- Experimental arrangement for Young’s double slit experiment
- Expression for fringe width in Young’s double slit experiment
- Diffraction at a single slit
- Binding Energy per nucleon vs Mass number graph
- Distinction between conductors, insulators and semiconductors on the basis of energy band diagram.
- PN junction diode characteristics (circuit diagram and graph for forward bias and reverse bias)
- Half wave rectifier
- full wave rectifier
- zener diode as voltage regulator
- Transistor action
- Transistor characteristics in CE configuration (input characteristics and output characteristics)
- Transistor amplifier in CE configuration
All Block diagrams from Communication (posted below)
My Physics Class 11 question says that a climber who is 65kg climbs 600m. What is the work done by the climber?
I thought I would have to add both the change in KE and PE. But the answer says that you should simply use mgh (PE).
This is confusing because there should be some work done through movement. Why is KE omitted from calculating the work done that in this case?
Many problems in Physics are solved by making certain approximations and assumptions to avoid complications. If we consider all factors contributing towards the expenditure of energy there may still different factors, but of less importance for the problem under consideration.
In the question we assume that there is no change in Kinetic Energy involved. Work is not to be done to maintain the KE but to change the KE. Though there may be some changes in speed during the process, we disregard the changes and concentrate only on the change in potential energy, which is actually the work done against gravity.
Even if there is a change in Kinetic Energy, then also the work done against gravity will only be the change in potential energy (mgh). He may be doing work against friction and also the change his speed during the climbing process, but the work done against gravity would remain the same (mgh)
If the question had mentioned this (work done against gravitation) then confusion could have been avoided.
Hope that the idea is clear now. If you need further clarifications please respond via the comment form.
Download the following question bank on optics based on CBSE board question papers
A patient lying in bed is completely covered by a 10mm thick blanket. Above the blanket the air is still and the convective heat transfer coefficient in still air is approximately 3 Wm^-2K^-1. The patient’s skin temperature is stable at 36 degrees Celsius and the air in the hospital is 24 degrees Celsius. What is the rate at which the patient loses heat due to convection and conduction over 1 square metre (in W)?
[The thermal conductivity of the blanket is 0.03 Wm^-1K-1
Asked Anurag Ganugapati
If I take a medical syringe, position the plunger in the middle of the length of the tube and seal off the tip, I will encounter substantial resistance when I try to move the plunger.
But, if I take the syringe into outer space – no atmosphere – and do the SAME procedure will I notice NO resistance when moving the plunger? Since there is no air to compress or decompress I suspect there will be no resistance. Is this correct?
Asked George Creegan
Yes, the only thing keeping you from pulling the plunger on a syringe with a sealed end is the surrounding air pressure. If there is perfect vacuum inside and outside the syringe!
The question was posted by Subhdeep Sarkar.
The Hall effect is the production of a voltage difference (the Hall voltage) across an electrical conductor, transverse to an electric current in the conductor and a magnetic field perpendicular to the current. It was discovered by Edwin Hall in 1879
Refer to the following links for more information