Viresh Singh has shared 'Online AIEEE ,26 May 2012...Solution.pdf' with you using Dropbox

Hi,

Here's a link to "Online AIEEE ,26 May 2012...Solution.pdf" in my Dropbox:

https://www.dropbox.com/s/q0fkxvl4fgfbtol/Online%20AIEEE%20%2C26%20May%202012...Solution.pdf

Viresh

Fwd: Jee mains test with answers..24 march

Ans : b c a c c.
c a d a a d
c c a a b d
d c b d d

Jee mains :paper 2013 ..23 April online

Following link to the paper
https://www.dropbox.com/s/6w0xkohd21hlh22/Aieee%20013%20-23%20APRIL.pdf

Following is link to the key for this paper

https://www.dropbox.com/s/cxrqlnjgjn5sdj4/fromiphone.pdf

JEE MAINs ...motion in magnetic field

JEE mains ..solved problems .20 march 14

JEE Mains : Test on electro and mag..for jee mains

test-12th-25july-13-E+M.pdf: https://www.idrive.com/idrive/sh/sh?k=z1p2m3v7r0 Regards, © Pro Softnet Corporation

Viresh

JEE Mains :Work power energy ..a revision(in Hinglish)

Click the link below

https://secure1.storegate.com/Shares/Home.aspx?ShareID=90d238c8-2325-4700-aacb-eee2b0e77328

JEE Mains ..Some point related to Gravitation

• Speed of a satellite circulating around the earth is independent of it's mass.
• Speed of satellite rotating in larger orbits is lesser.
• If the kinetic energy of a satellite is doubled , it will escape to infinity.
• Polar satellites are used for remote sensing of earth but geostationary satellites are used for the purpose of communication.
• Orbits of geostationary satellite and polar satellites are mutually perpendicular.
• Dimensionally, gravitational potential is same as latent heat.
• If variation of g is accounted for then gravitational potential energy of a particle of mass m raised to height h from earth surface, will gain potential energy by amount. m g h / ( Re +h).
• A satellite rotating around the earth will sweep area at rate L /2M where L is it's angular momentum and M is it's mass.
• Polar satellites have height around 500 to 1000 km above the earth.Their time period is around 2 hrs.
• Taking potential energy zero at infinity, for a satellite rotating around the earth, total energy is negative, potential energy is negative and magnitude of potential energy is twice of that of the magnitude of total energy. Magnitude of the kinetic energy is same as magnitude of the total energy.For example if the kinetic energy is 1000 joules, then total energy will be -1000 joules and potential energy is -2000 joules.
• Any energy of rotating satellite is inversely proportional to the radius of the orbit.
• If the radius of two satellites differ by one percent, then their time periods will differ by 1.5 percent.

JEE MAINS questions / solutions

JEE Mains :Resolving power

Due to the diffraction effects i.e. due to bending of light around the corners of any obstacle, it becomes difficult to get precise position of any image. This produces another problem of difficulty in resolving the two different objects while observing them at same time. It is agreed by the senior academicians of physics that two objects are said to be resolved if diffraction central maxima of one coincides with first minima of other's diffraction pattern.

JEE Mains : Diffraction of light

Diffraction of wave means bending of wave around any obstacle. Bending is significant if the wavelength is comparable to the size of obstacle. For red color, bending is more than for voilet as bending increases with increase in the wavelength.

In case of diffraction of parallel beam of light , around a long narrow slit, following conclusions are important to know for students :

JEE MAINS 14..solved example on KTG.16 March

Q.Let there be a container of fixed volume 4 V.It contains ideal gas at pressure P. A pump is used to remove the gas from this container. In each stroke , the pump takes volume V from the container. We have to find the pressure in the gas after 2 stokes assuming the gas temperature to be constant.

Ans: 16 P / 25

Solution:

Q. Let there be a vessel of volume V cubic meter ,containing an ideal gas at pressure P0. There is a small hole in the vessel from where the gas is leaking at rate. kV cubic meter per unit time. The temperature is maintained to be constant T and molar mass of the gas is M. Find the pressure after time t.
Solution:

JEE MAINS 2014....problems on rolling..15. March

JEE Mains :Convex lens and plane mirror

JEE Mains Thin prism

JEE Mains :SNELL's Law and wave

The Snell's law is applicable for any wave i.e. sound and light. Using this law it can be concluded that the ray corresponding to the wave will be at greater angle from the normal if the if the wave velocity is more in a medium and vice versa .

JEE MAINS 2013 APRIL 22

Link to the paper

Following is keyLink

JEE main 2013 -9april2013online

Here's a link to "Aieee 013 -09 APRIL.pdf" held last year , solve it.
https://www.dropbox.com/s/j7ttb8fovpvltjf/Aieee%20013%20-09%20APRIL.pdf

Here is the link to key .
Link to the key

For JEE Mains solved examples 6 Feb 2014

Hint: 1-All medium particles perform SHM with same frequency.
2-All medium particles do not have same energy.
3-Amplitude at distance x from a node is sinusoidal function of position x.
4- When mass m oscillates , it's oscillation energy is equal to half multiplied by mass, square of the angular frequency , square of the amplitude.
4- Consider a small length say dx ,get it's mass and it's oscillation energy and then integrate with respect to x with limits 0 and half wavelength which is distance between nearby nodes.

JEE MAINS PHYSICS-solved examples-2 march-14

• In the first problem, one can simply conserve the momentum in two directions to reach on the answer. Note that had the collision been head on then there could have been 100% transfer of the energy.
• In the second problem only notable point is that don't use the relation E= hc/ l for electron or any matter to find the wavelength associated with it.
• In third problem the concept here used that beat frequency is the inverse of the time interval of the intensity function.
  
  Good Luck
  

Dual Nature and Photoelectric effect

• Photoelectric effect could be explained satisfactorily after it is assumed  that the light consists of particles which travel with the velocity of light and these particles have energy and momentum , where energy is proportional to the frequency of the light and momentum is inversely proportional to the wavelength.
• The behaviour of light  as particle is not a universal  i.e. in every situation the light does not behave as particle but in most situations it behaves as wave.
• So the  light behaving as wave and also as particle, is of dual nature.
• Just like the light has duality in nature, de Broglei proposed that all matter must possesses wave nature also and therefore each matter should be associated with  a wave known as  de Broglei wave whose wavelength  is related to the momentum of the matter in the same manner  for the light where  l = h / p . 
• 
  
• For photons energy    E = h c/ l    and effective mass  = E/ c²                                                       
  

   therefore momentum  p = effective mass  x  c 
  = (E/ c² ) x c

  \ p = E/c                                                                                                                        
  But    E = hc/ l   for photons , so      p  = (hc/ l )/c                                                                    
                                    or    p  = h /  l
   
• When we calculate  wavelength for the matter, we must  not    use   the relation                               E = hc/ l  as this relation is strictly valid only for light.  For matters, first calucate momentum and then using the relation    l = h / p  we can find the deBroglei wavelength.
• For example if we have to find deBroglei wavelength. of an electron  of kinetic energy 4 ev , first we calculate  its momentum  using the relation  P= ( 2 m E )^1/2   and then find its wavelength. 
   


• Specially for the electron, there is a relation  12.27 / ( E )^1/2  angstrom  where E is the kinetic energy of the electron in the electron volt . So  using this relation, we get wavelength of the electron as  12.27/ 2= 6.13  A^{0 .} 
• Suppose  there is a light consisting of the photons of energy  4 ev and we have to find its wavelength. here we will use the relation  
  E = hc/ l  which on simplification comes out to be a simple fomula as   l = 12400 / E  A⁰  and so  wavelength will be 3100 A⁰  which  is clearly different than wavelength of the electron of same kinetic energy.
• Note that the wavelength associated with the electron is much  shorter than  that of the photon of same energy. 
• Lets compare the de Broglei wavelength of a proton and that of a alpha particle having been accelerated to same electric potential :