CBSE Class 12 Physics Solved Question Paper 2016

According to Gauss's law, flux through a closed surface is given by,

                                 

i) Mutual Induction: It is the phenomenon in which a change of current in one coil induces an emf in another coil places near it. The coil in which the current changes is called the primary coil and the coil in which the emf is induced is called the secondary coil. 

ii) 

EMF induced is, e = - M . 


So, the flux linked with the other coil is given by, 

a) 

Force acting on a charged particle q, which is moving with velocity v in a magnetic field B, is given by, 


                                          
The right-hand rule gives the direction of this force. The direction of the force is perpendicular to the plane containing velocity v and magnetic field B. 

b) 

A charged particle experiences a force when it enters the magnetic field. Due to the presence of magnetic field, the charged particle will move in a circular path. This is because the force is perpendicular to the velocity of the charged particle. 

Radius of the circular path in which the charged particle is moving is given by, 


 

Since the neutron has no charge, it will move along a straight line. 

The electron will follow a circular path which has a radius smaller than that of the alpha particle. This is because the mass to charge ratio of the alpha particle is more than that of the electron. 

Therefore, the electron will move in the clockwise direction and the electron will move in the anticlockwise direction as per the Right Hand Rule. 

Given, 

EMF, E₁ = 1.5 V ; E₂ = 2 V 

Internal resistance, r = 0.2 

Effective emf of two cells connected in parallel is, 

E_eff =   

This implies, 



The effective resistance can be calculated as:

R_eff = 

That is, 

R_eff = 

Quality factor relates the maximum or peak energy stored in the circuit (the reactance) to the energy dissipated (the resistance) during each cycle of oscillation. 

Quality factor is expressed as, 



Q-factor is a dimensionless quantity. 

i) 

Capacitance of a parallel plate capacitor is given by, 


 

ii) 

Potential difference between plates X and Y can be calculated as, 

            Q = CV 

 

iii) 

The ratio of electrostatic energies can be calculated as: 

Electric potential at a distance r from the point charge +Q is given by, 

  

Potential at point A is, 


 

Similarly, potential at point B is given by, 



Since r_A < r_B 

Consider a ring of radius 'a' which carries uniformly distributed positive total charge Q. 




To find: electric field due to a ring at a point P lying at a distance x from its centre along the central axis perpendicular to the plane of the ring.
 

                                  
The perpendicular component of electric field due to charge on the ring along the x-axis cancels each other out.

As there is same charge on both sides of the ring, the magnitude of the electric field at P due to the segment of charge dQ is given by, 

dE = k_e 
E_x =  

   

1. At the centre (X = 0) , electric field is zero. 

2. When x>> a, a can be neglected in the denominator. 

Therefore,