Physics Grade 12

Physics Past Papers 12th Class Peshawar Board

Section A-(MCQs)

i)  An electron volt is a unit of __________.                                              (Mark 1)

A.  Electric power  
B.  Energy
C.  Potential difference
D.  Voltage
Answer:
B.  Energy

Section B-(Short Questions)

Q.2 i) What is the difference between the electric potential and electric potential energy at any point of an electric field?                                                     (Marks 4)

Answer:
Electric potential:
The amount of work needed to move a unit of positive charge from a reference point to a specific point inside the electric field is called electric potential.
Electric potential energy:
The electric potential energy is an energy that results from conservative Coulomb forces and is associated with the configuration of a particular set of point charges within a defined system.

Section C-(Long Questions)

Q.4 a) Show that the energy stored in the magnetic field of an inductor of self-inductance L and carrying current I is given by: Um = ½ LI2                                                                                                                                                                (Marks 9)

Answer:
Energy can be stored in the magnetic field of an inductor. Consider a coil connected to a battery and a switch in series. When the switch is turned on, voltage is applied across the ends of the coil and current through it rises from zero to its maximum value I. due to change of current, an emf is induced, which is opposite to that of the battery. Work is done by the battery to move charges against the induced emf.
Work done by the battery in moving a small charge ∆q is
                                                    W = ∆qεL -------------------------- (1)
Where εL is the magnitude of the induced emf, given by
                                                    εL = L.∆I/∆t
Putting the value of εL in equation (1), we have
                                                    W = (∆q. L)∆I/∆t = (∆q/∆t) L∆I ------------------- (2)
Total work done in establishing the current from 0 to I is found by inserting for ∆q/∆t, the average current, and the value of ∆I.
                            Average current = ∆q/∆t = (0+I)/2 = ½ I
Where change in current = ∆I = I-0 = I
                                Total work W = (½I)LI = ½ LI2
                                                   W = ½ LI2
This work is stored as potential energy in the inductor. Hence the energy stored in an inductor is:
                                                  Um = ½ LI2

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