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# NCERT Exemplar Solutions for CBSE Class 12 Physics, Chapter 3 (Part II)

Class 12 Physics NCERT Exemplar solutions for Chapter 3 (Current Electricity) are available here. In this article, students will find solutions from question number 3.7 to question number 3.11. These questions are multiple choice questions with multiple correct answers. The solutions of this chapter are helpful for the preparation of CBSE Class 12 Physics board exam 2017 and other competitive exams. NCERT Exemplar solutions for Class 12 Physics - Chapter 3 (Current Electricity) are available here. In this article, students will find solutions from question number 3.7 to question number 3.11. These questions are basically multiple choice questions with multiple correct answers. The solutions of this chapter are helpful for the preparation of CBSE Class 12 Physics board exam 2017.

NCERT Exemplar Solutions for CBSE Class 12 Physics, Chapter 3 (from question number 3.7 to 3.11) are given below:

Question 3.7: Kirchhoff ’s junction rule is a reflection of

(a) conservation of current density vector.

(b) conservation of charge.

(c) the fact that the momentum with which a charged particle approaches a junction is unchanged (as a vector) as the charged particle leaves the junction.

(d) the fact that there is no accumulation of charges at a junction.

Solution 3.7:

Kirchhoffs junction rule or Kirchhoffs current law is a reflection of the fact that there is no accumulation of charges at a junction and conservation of charge.

NCERT Solutions for Class 12 Physics

Question 3.8: Consider a simple circuit shown in Fig 3.2 stands for a variable resistance R ′. R ′ can vary from R0 to infinity. r is internal resistance of the battery (r << R << R0).

(a) Potential drop across AB is nearly constant as R′ is varied.

(b) Current through R′ is nearly a constant as R′ is varied.

(c) Current I depends sensitively on R′.

(d) I ≥ V/(r + R) always.

Solution 3.8: (a, d)

The current in the circuit is given by, I = V / [r + R R’/(R + R’)]

As, R’ >> R, so, R + R’ ≥ R

Therefore, I ≥ V/(r + R) and due to this reason option (d) is correct.

Now, potential difference across A and B, VAB = [(V × RR’)]/[r (R + R’) + RR’]

For very large R’, we can write R + R’ = R

VAB = VR/(r + R)

So, VAB is consent if R’ is varied. So, option (a) is correct.

Current through R’, I’ = VAB/R’ = VR/(rR’ + RR’)

Clearly I’ vary with R’ also the current I does not depend on sensitivity of R’.

Question 3.9: Temperature dependence of resistivity ρ (T) of semiconductors, insulators and metals is significantly based on the following factors:

(a) number of charge carriers can change with temperature T.

(b) time interval between two successive collisions can depend on T.

(c) length of material can be a function of T.

(d) mass of carriers is a function of T.

Solution 3.9:

The resistivity of a metallic conductor is given by, ρ = m/(ne2 τ), where n is number of charge carriers per unit volume which can change with temperature T and τ is time interval between two successive collisions which decreases with the increase of temperature.

From the relation, we can observe that options (a), (b) are correct.

Question 3.10: The measurement of an unknown resistance R is to be carried out using Wheatstones bridge (see Fig. 3.25 of NCERT Book). Two students perform an experiment in two ways. The first students takes R2 = 10 Ω and R1 = 5 Ω. The other student takes R2 = 1000 Ω and R1 = 500 Ω. In the standard arm, both take R3 = 5 Ω.

Both find R= (R2/R1) R3 = 10 Ω within errors.

(a) The errors of measurement of the two students are the same.

(b) Errors of measurement do depend on the accuracy with which R2 and R1 can be measured.

(c) If the student uses large values of R2 and R1, the currents through the arms will be feeble. This will make determination of null point accurately more difficult.

(d) Wheatstone bridge is a very accurate instrument and has no errors of measurement.

Solution 3.10: (b), (c)

According to the question, for first student, R2 = 10Ω, R1 = 5Ω, R3 = 5

For second student, R1 = 500Ω, R3 = 5Ω

For balanced wheatstone bridge,

R2/R = R1/R3 ⇒ R = R3 × (R2/R1)

Putting values in above relation, we will find that value of R for both the students comes out to be 10 Ω.

The errors of measurement of the two students depend on the accuracy and sensitivity of the bridge, which in turn depends on the accuracy with which R2 and R1 can be measured.

Question 3.11 In a meter bridge the point D is a neutral point (Fig 3.3). (a) The meter bridge can have no other neutral point for this set of resistances.

(b) When the jockey contacts a point on meter wire left of D, current flows to B from the wire.

(c) When the jockey contacts a point on the meter wire to the right of D, current flows from B to the wire through galvanometer.

(d) When R is increased, the neutral point shifts to left.

Solution 3.11: (a), (b), (c)

Here, the meter bridge can’t have other neutral points for this set of resistances provided we do not interchange the resistances. Also, VB = VD.

NCERT Solutions for CBSE Class 12 Physics ‒ Chapter 3: Current Electricity