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 ...