NEET Physics Current Electricity Questions Solved

BOARD

(a) (i) State the principle on which a potentiometer works. How can a given potentiometer be made more sensitive?

(ii) In the graph shown below for two potentiometers, state with reason which of the two potentiometers, A or B, is more sensitive.

(b) Two metallic wires, P1 and P2 of the same material and same length but different cross - sectional of emf. Find the ratio of the drift velocities of free electrons in the two wires when they are connected (i) in series, and (ii) in parallel.                                                      5 marks

(a) (i) The potential difference across any length of wire is directly proportional to the length provided current and area of cross section are constant i.e., E(l) = l where is the potential drop per unit length.                                                                                                       1mark

It can be made more sensitive by decreasing current in the main circuit / decreasing potential gradient / increasing resistance put in series with the potentiometer wire.                   12mark

(ii) Potentiometer B                                                                                                12mark

     Has smaller value of Vl  (slope / potential gradient).                                           12mark

(b) In series, the current remains the same.                                                              12mark

I =neA1Vd1=neA2Vd2                                                                                             12mark

      Vd1Vd2=A2A1                                                                                                     12mark

In parallel potential difference is same but currents are different.

V=I1R1=neA1Vd1ρlA1=neρVd1l                                                                            12mark

Similarly, V=I2R2=neρVd2l                                                                                    12mark

                             I1R1=I2R2   Vd1Vd2=1

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