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

Jul 24, 2017 12:00 IST

    NCERT Exemplar Solutions for CBSE Class 12 Physics, Chapter 3 (Solutions of Very Short Answer Type Questions or VSA)

    Get NCERT Exemplar Solutions for CBSE Class 12 Physics – Chapter 3 (Current Electricity). Here, you will find solutions of very short answer type questions (i.e. question number 3.12 to question number 3.21). Solutions of questions from 3.1 to 3.11 are available in Part I and Part II. Solution from question number 3.22 onwards will be available in further parts. These questions can be asked in CBSE Class 12th Physics board exam and various engineering & medical entrance exams like JEE Main, NEET, WBJEE, UPSEE etc.

    NCERT Exemplar Solutions for CBSE Class 12 Physics, Chapter 3 (from question number 3.12 to 3.21) are given below:

    Question 3.12: Is the motion of a charge across junction momentum conserving? Why or why not?

    Solution 3.12:

    Drift velocity is given by, vd = (eE τ)/m

    When an electron approaches a junction, in addition to the uniform E that it normally faces (which keep the drift velocity vd fixed), there are accumulation of charges on the surface of wires at the junction. These produce electric field. These fields alter direction of momentum.

    NCERT Solutions for Class 12 Physics

    Question 3.13: The relaxation time τ is nearly independent of applied E field whereas it changes significantly with temperature T. First fact is (in part) responsible for Ohm’s law whereas the second fact leads to variation of ρ with temperature. Elaborate why?

    Solution 3.13:

    Relaxation time is bound to depend on velocities of electrons and ions.

    Applied electric field affects the velocities of electrons by speeds at the order of 1 mm/s, an insignificant effect.

    Change in T, on the other hand, affects velocities at the order of 102 m/s. This can affect τ significantly.

    [ρ = ρ (E, T) in which E dependence is ignorable for ordinary applied voltages.]

    Question 3.14: What are the advantages of the null-point method in a Wheatstone bridge? What additional measurements would be required to calculate Runknown by any other method?

    Solution 3.14:

    The advantage of null point method in a Wheatstone bridge is that the resistance of galvanometer does not affect the balance point and there is no need to determine current in resistances and galvanometer and the internal resistance of a galvanometer.

    Runknown can be calculated applying Kirchhoff’s rules to the circuit.

    We would need additional accurate measurement of all the currents in resistances and galvanometer and internal resistance of the galvanometer.

    Question 3.15: What is the advantage of using thick metallic strips to join wires in a potentiometer?

    Solution 3.15:

    The metal strips have low resistance and need not be counted in the potentiometer length l1 of the null point. One measures only their lengths along the straight segments (of lengths 1 meter each). This is easily done with the help of centimeter rulings or meter ruler and leads to accurate measurements.

    Question 3.16: For wiring in the home, one uses Cu wires or Al wires. What considerations are involved in this?

    Solution 3.16:

    Two considerations are required: (i) cost of metal, & (ii) good conductivity of metal.

    Cost factor inhibits silver.

    Cu and Al are the next best conductors.

    Question 3.17: Why are alloys used for making standard resistance coils?

    Solution 3.17:

    Alloys have low value of temperature co-efficient (less temperature sensitivity) of resistance and high resistivity.

    Question 3.18: Power P is to be delivered to a device via transmission cables having resistance RC. If V is the voltage across R and I the current through it, find the power wasted and how can it be reduced.

    Solution 3.18:

    Power wasted PC = I2RC where, RC is the resistance of the connecting wires.

    PC = (P2/V2) RC [as, P = VI]

    In order to reduce PC, power should be transmitted at high voltage.

    Question 3.19: AB is a potentiometer wire (figure given below). If the value of R is increased, in which direction will the balance point J shift?

    Solution 3.19:

    If R is increased, the current through the wire will decrease and hence the potential gradient will also decrease, which will result in increase in balance length. So J will shift towards B.

    Question 3.20: While doing an experiment with potentiometer (Fig 3.5) it was found that the deflection is one sided and (i) the deflection decreased while moving from one end A of the wire to the end B; (ii) the deflection increased. while the jockey was moved towards the end B.

    (i) Which terminal +or –ve of the cell E1, is connected at X in case (i) and how is E1 related to E ?

    (ii) Which terminal of the cell E1 is connected at X in case (ii)?

    Solution 3.20:

    (i)

    Deflection in galvanometer is one sided and the deflection decreases while moving from end A to end B, this is possible only when positive terminal of the cell E1, is connected at X and E1>E.

     (ii)

    The deflection in galvanometer is one sided and the deflection increased, while moving from one end A of the wire to the end S, this is possible only when negative terminal of the cell Ei is connected  at  X.

    Question 3.21: A cell of emf E and internal resistance r is connected across an external resistance R. Plot a graph showing the variation of P.D. across R, verses R.

    Solution 3.21:

    When the cell of emf E and internal resistance r is connected across an external resistance R, the relationship between the voltage across R is given by

    V = E / [1 + (r/R)]

    From the above relation,

    Now, as R = 0 ⇒ V = 0 & R = ∞ ⇒ V = E. So, this variation is shown in the figure given below.

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

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