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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)
How was the CBSE Physics exam?
When the question was asked to teachers and students the response was a mixed one.
Not as expected
Time was not enough
Question paper was lengthy
1 and 2 marks questions were easy
Five marks questions were confusing
Etc … were some of the reactions
How do you feel?
How was the paper?
Respond now. Your response counts. It will also help the teachers and Principal to take up the matter to CBSE authorities.
Very Easy but Very Lengthy
“I appeared yesterday for the respective exam. It was quite surprising for us to see the very easy Physics Exam! I myself had prepared hard for it but the exam was much lengthy especially 3 marks questions. I myself couldn’t attempt 10 marks which i had already realized were quite easy questions. But I have to accept this disappointment now” – Shubham Dhingra
How to find the focal length of convex mirror using convex lens?
We cannot use an optical method to determine the focal length of a convex mirror without using a convex (converging) lens since a convex mirror does not form a real image of an object in front of it or even parallel rays coming from infinity.
Therefore we have to use a convex mirror to determine the focal length of the convex mirror by an indirect method.
STEP 1 DETERMINE THE ROUGH FOCAL LENGTH OF THE CONVEX LENS
First of all we take a convex lens and determine its rough focal length by forming a real image of a distant object on a screen. The distance between the convex lens and the screen gives the rough focal length, since when the object is at infinity, the image formed by the convex lens is at its focus. This method is called the distant object method to estimate the focal length of a convex lens. The focal length of a concave mirror may also be estimated using this method.
Now, as we know the approximate focal length of the convex lens, keep an object- an optical pin or a lit candle in front of the convex lens at a distance around 2 times the rough focal length determined and place a screen on the other side so as to form a real image of the object on it. This adjustment is done so as to make the distances involved quite manageable.
When you get a clear and sharp image on the screen, mark the positions of the convex lens and the screen on the table.
Place the convex mirror in between the screen and the lens without disturbing the position of the candle and the lens.
Place the screen close to the candle. Carefully adjust the position of the convex mirror alone so that you get a sharp image on the screen which is now kept at the position of and along with the candle. Mark the position of the convex mirror now.
The distance between the position of the convex mirror and the old position of the screen gives the radius of curvature of the convex mirror. (Since r = 2f; we can determine the focal length by dividing the distance with 2)
Why the distance is r (2f)?
We know that when a ray of light is incident normally on the surface of a mirror, it retraces its path.
We also know that a line drawn from the centre of curvature of the mirror to the surface of a mirror is normal to the surface.
Therefore, when we are getting the image at the same position as the candle, it is formed by retracing of the rays, which means that the old position of the screen is at the position of the centre of curvature of the convex lens.
Read the post
See the Videos below to understand how the focal length of a convex mirror is determined using a convex lens in an optic bench.
Explain the formation of depletion region and potential barrier in a pn junction diode.
Asked Sunder Bisht
The depletion region (also called depletion layer, depletion zone, junction region, space charge region or space charge layer) is an insulating region within a conductive, doped semiconductor material where the mobile charge carriers have been diffused away, or have been forced away by an electric field. The only elements left in the depletion region are ionized donor or acceptor impurities.
When the N-type semiconductor and P-type semiconductor materials are first joined together, a very large density gradient exists between both sides of the PN junction. The result is that some of the free electrons from the donor impurity atoms begin to migrate across this newly formed junction to fill up the holes in the P-type material producing negative ions.
A depletion region forms instantaneously across a p–n junction. It is most easily described when the junction is in thermal equilibrium or in a steady state: in both of these cases the properties of the system do not vary in time; they have been called dynamic equilibrium.
Electrons and holes diffuse into regions with lower concentrations of electrons and holes, much as ink diffuses into water until it is uniformly distributed. By definition, N-type semiconductor has an excess of free electrons compared to the P-type region, and P-type has an excess of holes compared to the N-type region. Therefore, when N-doped and P-doped pieces of semiconductor are placed together to form a junction, electrons migrate into the P-side and holes migrate into the N-side. Departure of an electron from the N-side to the P-side leaves a positive donor ion behind on the N-side, and likewise the hole leaves a negative acceptor ion on the P-side.
Following transfer, the diffused electrons come into contact with holes on the P-side and are eliminated by recombination. Likewise for the diffused holes on the N-side. The net result is the diffused electrons and holes are gone, leaving behind the charged ions adjacent to the interface in a region with no mobile carriers (called the depletion region). The uncompensated ions are positive on the N side and negative on the P side. This creates an electric field that provides a force opposing the continued exchange of charge carriers. When the electric field is sufficient to arrest further transfer of holes and electrons, the depletion region has reached its equilibrium dimensions. Integrating the electric field across the depletion region determines what is called the built-in voltage (also called the junction voltage or barrier voltage or contact potential).
Does air pressure in a capped bottle is different as that of open bottle??
If not and the atmospheric pressure in a closed container is same as that of the surroundings let 1bar at sea level, if i consider a
tube both end open and dip one end in water (like pipette in chemistry lab) and close the other by thumb, water remain
hanged in the tube.. if we say it is because the atmosphere that pushes up on the water in the tube is same as that of remaining air in tube pushing down on the water..won’t the water fall out due to its own weight as the upward and downward pressure is balanced…
Please explain the whole process and compare weight of water with up and down pressure by atmosphere..
Again would liquid ‘ll flow out of a container through a hole in vacuum??
Asked Shashank Patra
The Air Pressure in an open bottle is equal to the atmospheric pressure. The pressure inside a closed bottle can be different.
In the experiment described, when the tube is partially filled and the upper end is closed, the water tries to fall down creating a lower pressure above it inside the tube. This creates a pressure difference, the outside pressure greater than the pressure inside and the water can fall only upto the level where the weight of water column is balanced by the force due to difference in pressure created.
See one live demonstrations here
“In Pascal’s law we find that pressure does not increase with area. But when we study about pressure, we learn that pressure is inversely proportional to area. How is this possible? Please explain.”
Fahad Imtiaz asked via Speak Pipe
Pressure is defined as the thrust (the total force acting normal to a surface) per unit area.
Pascal’s law deals with fluid pressure and the statement goes –
“The pressure exerted anywhere in an enclosed incompressible and non-viscous fluid is transmitted equally and undiminished in all directions through out the fluid, provided the effect of gravity is neglected”
Read the statement carefully.
Here we are not changing the definition or meaning of pressure.
the fact to note that, in an enclosed fluid, the pressure is transmitted equally throughout the fluid. Therefore, if we apply some pressure somewhere in the fluid, the same pressure will be felt at any other place on the enclosed fluid. This gives us an opportunity to multiply the force. Since the pressure is equal everywhere, if we increase the area the force (thrust) is increased.
P = F/A or F = PA
So, pressure remaining constant, greater the area, greater is the force.
Hope you understand the matter now.