JEE level test -GRAVITY and ROTATION

Hi,

Here's a link to "JEE level test on gravity and rotation

https://www.dropbox.com/s/is0gcsd9mfcsoy3/JEE%20level%20test%20-GRAVITY%20%20and%20ROTATION.pdf

Solve the following question also

viresh..

What makes the light waves in laser light parallel? | Science Questions with Surprising Answers

http://sciencequestionswithchris.wordpress.com/2012/12/20/what-makes-the-light-waves-in-laser-light-parallel/



Viresh

For 11th class and jee aspirant

Solution :

Heat ...problems...10 dec 13

1-An ideal gas is filled in a cylinder fitted with a movable piston at one end. Pressure in the cylinder is P. Now the piston is suddenly pushed to compress the gas to one fourth of it's initial volume and held in same position for long time so that the gas attains a thermal equilibrium with surrounding . If the adiabatic constant for the gas is 1.5 then what is the final pressure ? Show the whole process once in p-V graph and then in T - V graph.

Answer : 4P; fig-a and fig -b

2-An ideal gas is filled in a cylinder fitted with a movable piston at one end. Pressure in the cylinder is P and the volume is V . Now the piston is suddenly pushed to compress the gas to one fourth of it's initial volume and held in same position for long time so that the gas attains a thermal equilibrium with surrounding . Now the gas is allowed to expand isothermally to initial volume V. If the adiabatic constant for the gas is 1.5 ,what is net amount of heat absorbed by the gas ?

Hint :
When the gas is compressed adaibatically, no heat is exchanged .During constant volume cooling heat is released. This amount of the heat released can be obtained using the concept of heat capacity at constant volume i.e. Q = n Cv (Tf-Ti). During isothermal expansion,heat is absorbed according to the relation PV ln (Vf/Vi).

Transverse magnification by lenses...different relations ..

Following are the different relations for transverse magnification produced by a thin lens with same medium on both side .Here f stands for second focal length which will be positive for converging lens and negative for the diverging lens. Here x1 is distance of the object from first focal point and x2 is distance of the image from the second focal point.

Compound ..microscope...a problem..09 dec 13

Following fig shows, a compound microscope , size of object and size of the intermediate image , position of the final image and the focal lengths of the lenses. Using these informations, answer following two questions:



Hint: Since transverse magnification by the objective is easily visible, one can get distance of the intermediate image from the objective using the relation m = 1- v/f. Also since the position of the final image Is at near point, so the distance of the intermediate image from the eye piece should be. f D / ( f + D).

From the fig transverse magnification by objective can be obtained and since the final image is at near point so the angular magnification by the eye piece can easily be known isung the relation 1+ D / f.

For JEE aspirants ...temperature variation due to heat conduction ...

For JEE. aspirants thermal conduction in spherical and conduction

It is really useful to remember that how thermal current depends on inner and outer radius of a spherical conductor and of a cylindrical conductor.

Problem on Rigid body...for Jee aspirants

This problem can be solved as following :

1- conserve the angular momentum about the pivoted end and obtain the angular speed of the rod just after the collision.
2- the kinetic energy of the rod should be equated to maximum increase of the potential energy of the rod.

Compound microscope...Normal adjustment

In normal adjustment, of the compound microscope,, final image is formed at infinity, and eye is most relaxed ,intermediate image i.e. the image formed by the objective coincides with the (first ) focal point of the eye piece. The separation between the objective and eye piece can be found by adding distance of intermediate image from the objective and focal length of the eye piece.

The object is placed close to the (first) focus of the objective but kept away slightly away from the focus.

From different  desired considerations,it comes out that the focal lengths of both objective and eyepiece ,should be kept smaller.

For 12 th Simple Microscope

In simple microscope, only one converging lens is used.This lens acts as objective as well as eyepiece. The objects can be placed closer than D if they are observed through naked eye where as with the use of converging lens , objects can be placed closer than D as the lens will form it's image much behind where distance of the image will be greater or equal to infinity. But the lens used should have focal length much lesser than D. Following figures give some memorable results.

A very interesting question arise that when an object of height h is placed at the focal point of a simple microscope lens, what will be the angle formed by the image which is formed at infinity?

R C Circuit Revision

Refer the fig which shows a RC circuit. Let the charge on the capacitor be initially zero on , After closing the circuit, current suddenly shoots up to E / R decreases exponentially and the charge increases with time exponentially. KVL is always satisfied. Potential across the capacitor increases with time where as the potential across the resistor decreases with time. Long time after closing the circuit, current and potential across the resistor becomes zero and whole emf of the cell is applied across the capacitor as if there is no resistor.


Q1-Verify that the growth rate d U /dt of the potential energy stored in the capacitor changes with time as given in the graph.
.
Q2-Verify that the logarithmic of the current ln(I) vary linearly with time according to the graph given here.
Q3-When the current in the circuit becomes E/3R, what is the rate of growth of the capacitor's energy and what will be the current when the rate of growth of the capacitor's energy becomes maximum ?

Viresh

For 11 th class a Revisional tests sheet ...up to elasticity ..(p)

Link to the sheet ..27 nov 13

https://www.dropbox.com/s/esfko3i4ifcfb08/pt-17%20march-fluid-grav.doc

Viresh

For 11th and Jee aspirants ...Solved examples in hinglish language ..mechanics

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u-v graph for mirror and lens

These graphs are used to find the focal lengths of a spherical concave mirror or a converging lens ! A line drawn at 45 degree from the origin, will  cut the graph at point whose x and y coordinates will be same , equal to twice of the focal length in the magnitudes. This is true for  the graphs for both concave mirror  and  convex lens

Problems on magnetism ..good one for jee aspirants

Hi,

Here's a link

https://www.dropbox.com/s/eui7c2php1gvfxh/12th%202problems%20on%20magnetism...pdf

Viresh

Answers :

1-(I) 0 ; E /3 (ii) E/ 2eR. 2- mg tan(theeta)/(NBl) where l is the length of the cylinder.

For 12 th ...and jee aspirants

To solve this problem, use refraction formula for spherical surface and then use refraction formula for the plane surfaces .

Viresh

For 12 th class and JEE aspirants

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For 11th class and Aspirants for Jee and Aipmt

To view viresh _iitk's file, click this link:
equilibrium & rolling.doc

Viresh

For 12th class ..Displacement method to find focal length of a convex lens

In the following figure, the two positions of the lens corresponds to the real image at the same distance from the object and the speciality is that in these two positions, u and v are interchanged. By measuring delta and D we can calculate f. Note that D should be greater than 4 f.

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