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

Jul 21, 2017 16:00 IST

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

    Find Class 12 Physics NCERT Exemplar Solutions for CBSE Class 12 Physics – Chapter 3 (Current Electricity). Here solutions from question number 3.1 to 3.6 are available. These questions are basically multiple choice questions with single correct answers. These questions are important for CBSE Class 12 Physics board exam and other competitive exams like JEE Mains, NEET etc.

    NCERT Exemplar Solutions for CBSE Class 12 Physics, Chapter 3 (from question number 3.1 to 3.6) are given below:

    Question 3.1: Consider a current carrying wire (current I) in the shape of a circle. Note that as the current progresses along the wire, the direction of j (current density) changes in an exact manner, while the current I remain unaffected. The agent that is essentially responsible for is

    (a) source of emf.

    (b) electric field produced by charges accumulated on the surface of wire.

    (c) the charges just behind a given segment of wire which push them just the right way by repulsion.

    (d) the charges ahead

    Solution 3.1: (b)

    Current density at a point in a conductor is defined as the amount of current flowing per unit area of the conductor around that point provided the area is held in a direction normal to the current.

    It is a vector quantity and its direction is along the direction of motion of positive charge.

    As its direction is decided by the flow of charge which in turn s decided by the charges accumulated on the surface of the wire.

    NCERT Solutions for Class 12 Physics

    Question 3.2: Two batteries of emf ε1 and ε22 > ε1) and internal resistances r1 and r2 respectively are connected in parallel as shown in figure given below

    NCERT Exemplar Solutions for CBSE Class 12 Physics - Chapter 3: Figure of question 3.2

    (a) The equivalent emf εeq of the two cells is between ε1 and ε2, i.e. ε1 < εeq < ε2.

    (b) The equivalent emf εeq is smaller than ε1.

    (c) The εeq is given by εeq = ε1 + ε2 always.

    (d) εeq is independent of internal resistances r1 and r2.

    Solution 3.2:

    In the above combination, the equivalent emf is given by

    eeq = (e2r1 + e1 r2)/(r1 + r2)

    The above equation shows that for any value of r1 and r2, eeq will always lie between e1 and e2

    In the given question, ε2 > ε1, therefore, e1 < eeq < e2.

    Question 3.3: A resistance R is to be measured using a meter bridge. Student chooses the standard resistance S to be 100Ω. He finds the null point at l1 = 2.9 cm. He is told to attempt to improve the accuracy.

    Which of the following is a useful way?

     (a) He should measure l1 more accurately.

    (b) He should change S to 1000Ω and repeat the experiment.

    (c) He should change S to 3Ω and repeat the experiment.

    (d) He should give up hope of a more accurate measurement with a meter bridge.

    Solution 3.3:

    The error in measuring R can be minimized by choosing suitable S and by adjusting the balance point near the middle of the bridge

    R/S = [R l1]/[R (100 ‒ l1)] or R/S = l1/(100 ‒ l1)

    ⇒R/100 = 2.9/(100 – 2.9)

    ⇒ R is nearly equal to 3 ohm.

    Question 3.4: Two cells of emf’s approximately 5V and 10V are to be accurately compared using a potentiometer of length 400cm.

    (a) The battery that runs the potentiometer should have voltage of 8V.

    (b) The battery of potentiometer can have a voltage of 15V and R adjusted so that the potential drop across the wire slightly exceeds 10V.

    (c) The first portion of 50 cm of wire itself should have a potential drop of 10V.

    (d) Potentiometer is usually used for comparing resistances and not voltages.

    Solution 3.4:

    In the potentiometer experiment, the emf of a cell can be measured, if the potential drop along the potentiometer wire is more than the emf of the cells to be determined.

    Here, values of emfs of two cells are given as 5V and 10V, therefore, the potential drop along the potentiometer wire must be more than 10V.

    Question 3.5: A metal rod of length 10 cm and a rectangular cross-section of 1cm × (&frac12;) cm is connected to a battery across opposite faces. The resistance will be

    (a) maximum when the battery is connected across 1 cm × (&frac12;) cm faces.

    (b) maximum when the battery is connected across 10 cm × 1 cm faces.

    (c) maximum when the battery is connected across 10 cm × (&frac12;) cm faces.

    (d) same irrespective of the three faces.

    Solution 3.5: (a)

    As, R = ρ (l/A)

    If we calculate R with the help of above relation then,

    From option (a) we will get R = 20 ρ Ω

    From option (b) we will get R = (ρ/10) Ω

    From option (c) we will get R = (ρ/5) Ω

    Clearly, the resistance will be maximum when the battery is connected across 1 cm × (&frac12;) cm faces.

    Question 3.6: Which of the following characteristics of electrons determines the current in a conductor?

    (a) Drift velocity alone.

    (b) Thermal velocity alone.

    (c) Both drift velocity and thermal velocity.

    (d) Neither drift nor thermal velocity.

    Solution 3.6:

    The relationship between current (I) and drift speed (vd) is given by

    I = ne A vd

    I µ vd

    Thus, drift velocity determines the current in a conductor.

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

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