CBSE Class 12 Physics Solved Question Paper 2013

Since the cells are in opposition,

E _net = E1 – E₂ = (200-10) = 190 V

So, current in the circuit is given by, I = 

Given, a convex lens placed coaxially in contact with a concave lens. 

Power of convex lens, P₁ = 
Power of concave lens, P₂ = 

So, Power of combination, P = P₁ + P<sub>2 

                                                       =</sub>  
                                                      
Since, power has negative magnitude; system of lenses is diverging in nature.

The equipotential surface for an isolated charge is concentric circles. The distance between the shell decreases with increase in electric field.

The charge on an unknown capacitor is given by,

Q = CV

On reducing the potential by 120 V, charge on the capacitor is reduced which is given by,

Q’ = C(V-120)

C(V-120) = 120 C                          … (2)

On solving equation (1) and (2), we have

 

Unknown capacitance from equation (1),

Q = CV

 

ii) Charge on the capacitor, if voltage is increased by 120V

Q = C (V+120)

   = 2 (180+120)

   = 600 C

                                                          OR




i) Electric flux through a surface, 
Flux through the left surface, 

Since x = 1m,

   

ii) Using Gauss Theorem, we can calculate the charge inside the cylinder.

When capacitor is charged by a dc source, 

Conduction current is equal to the displacement current.

i.e.,                           
Because,  is maximum when, capacitor is fully charged.

The polarity of the induced emf at the open ends of a closed loop is such that it tends to produce a current which opposes the change in magnetic flux that produced it.



Emf will be induced at the ends. Magnetic flux changes in a metallic rod falling freely under gravity and hence, emf is induced.

Substances that can retain their magnetism for a long duration of time at room temperature are called permanent magnets. They have high retentivity and coercivity.

Eg: Steel, cobalt 

Given, Resistance, R = 0.80

Maximum current, I_max = 1 A

So, Voltage across ammeter, V = IR = 1.0 0.80 = 0.8 V

i) If a shunt is connected in parallel,

Current I flowing through S is (I – I₁)

For parallel combination of resistors, 

I₁. R_A = ((I – I₁) S

 

This is the required value of shunt resistance. 

ii) Combined resistance of the ammeter and the shunt, 

V= E – I r;

The emf of a cell is greater than terminal voltage because there is some potential drop across the cell due to its small internal resistance.