SolveIt is under development
SolveItNCERT · CBSE · NEET

NCERT Exemplar · Class 10 Science Electricity

35 questions · 35 still being checked

Multiple Choice Questions 1–10 (part 1 of 4)

  1. Exercise 1

    A cell, a resistor, a key and ammeter are arranged as shown in the circuit diagrams of Figure 12.1. The current recorded in the ammeter will be
    (a)
    maximum in (i)
    (b)
    maximum in (ii)
    (c)
    maximum in (iii)
    (d)
    the same in all the cases
    NCERT_Question_Class10_Science_Exemplar_Ch12_Q1

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    (d)
    (d) The same in all the cases.Each is one series loop: same cell, same \(\displaystyle R\), ammeter connected correctly. \[I = \frac{\varepsilon}{R} \] Only the order of the parts differs, so \(\displaystyle I\) is equal.

  2. Exercise 2

    In the following circuits (Figure 12.2\displaystyle 12.2), heat produced in the resistor or combination of resistors connected to a 12\displaystyle 12 V battery will be
    (a)
    same in all the cases
    (b)
    minimum in case (i)
    (c)
    maximum in case (ii)
    (d)
    maximum in case (iii)
    NCERT_Question_Class10_Science_Exemplar_Ch12_Q2

    Check this one against your book

    NCERT’s printed answer for this exercise does not match its own question. This working follows the question as printed.

    (d) Maximum in case (iii).\[P = \frac{V^2}{R} \] \[R_{(i)} = 2\,\Omega \] \[R_{(ii)} = 2\,\Omega + 2\,\Omega = 4\,\Omega \] \[\frac{1}{R_{(iii)}} = \frac{1}{2\,\Omega} + \frac{1}{2\,\Omega} = 1\,\Omega^{-1} \Rightarrow R_{(iii)} = 1\,\Omega \] \[P_{(i)} = \frac{(12\,\mathrm{V})^2}{2\,\Omega} = 72\,\mathrm{W} \] \[P_{(ii)} = \frac{(12\,\mathrm{V})^2}{4\,\Omega} = 36\,\mathrm{W} \] \[P_{(iii)} = \frac{(12\,\mathrm{V})^2}{1\,\Omega} = 144\,\mathrm{W} \] Lowest resistance draws the most power at fixed \(\displaystyle V\).

  3. Exercise 3

    Electrical resistivity of a given metallic wire depends upon
    (a)
    its length
    (b)
    its thickness
    (c)
    its shape
    (d)
    nature of the material

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    (d)
    (d) Nature of the material.\[R = \rho\,\frac{L}{A} \] Here \(\displaystyle \rho\) is the proportionality constant fixed by the material itself; length \(\displaystyle L\) and area \(\displaystyle A\) change \(\displaystyle R\), not \(\displaystyle \rho\).

  4. Exercise 4

    A current of 1\displaystyle 1 A is drawn by a filament of an electric bulb. Number of electrons passing through a cross section of the filament in 16\displaystyle 16 seconds would be roughly
    (a)
    1020\displaystyle 10^{20}
    (b)
    1016\displaystyle 10^{16}
    (c)
    1018\displaystyle 10^{18}
    (d)
    1023\displaystyle 10^{23}

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    (a)
    (a) \(\displaystyle 10^{20}\).\[Q = It \] \[Q = (1\,\mathrm{A})(16\,\mathrm{s}) = 16\,\mathrm{C} \] \[n = \frac{Q}{e} = \frac{16\,\mathrm{C}}{1.6\times10^{-19}\,\mathrm{C}} \] \[n = 1\times10^{20} \] Roughly \(\displaystyle 10^{20}\) electrons cross the section.

  5. Exercise 5

    Identify the circuit (Figure 12.3\displaystyle 12.3) in which the electrical components have been properly connected.
    (a)(i)
    (b)(ii)
    (c)(iii)
    (d)(iv)
    NCERT_Question_Class10_Science_Exemplar_Ch12_Q5

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    (b)
    (b) (ii)
    Ammeter A in series, current entering at its \(\displaystyle +\); voltmeter V across \(\displaystyle R\), \(\displaystyle +\) at the high-potential end.
    (iv)
    has A reversed; (i) puts V in series; (iii) swaps A and V.

  6. Exercise 6

    What is the maximum resistance which can be made using five resistors each of 1/5Ω\displaystyle 1 / 5 \Omega ?
    (a)
    1/5Ω\displaystyle 1 / 5 \Omega
    (b)
    10Ω\displaystyle 10 \Omega
    (c)
    5Ω\displaystyle 5 \Omega
    (d)
    1Ω\displaystyle 1 \Omega

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    (d)
    (d) \(\displaystyle 1\,\Omega\).\[R_{\text{series}} = n\,r = 5 \times \frac{1}{5}\,\Omega \] \[R_{\text{series}} = 1\,\Omega \] Series addition gives the largest possible resistance; the same five in parallel would instead give the smallest, \(\displaystyle r/n = \frac{1/5}{5} = \frac{1}{25}\,\Omega\).

  7. Exercise 7

    What is the minimum resistance which can be made using five resistors each of 1/5Ω\displaystyle 1 / 5 \Omega ?
    (a)
    1/5Ω\displaystyle 1 / 5 \Omega
    (b)
    1/25Ω\displaystyle 1 / 25 \Omega
    (c)
    1/10Ω\displaystyle 1 / 10 \Omega
    (d)
    25Ω\displaystyle 25 \Omega

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    (b)
    (b) \(\displaystyle \dfrac{1}{25}\,\Omega\) — the minimum from equal resistors needs all five in parallel. \[\frac{1}{R_{eq}}=\frac{1}{R}+\frac{1}{R}+\frac{1}{R}+\frac{1}{R}+\frac{1}{R}=\frac{5}{R} \] \[R_{eq}=\frac{R}{5}=\frac{(1/5)\,\Omega}{5} \] \[R_{eq}=\frac{1}{25}\,\Omega \]

  8. Exercise 8

    The proper representation of series combination of cells (Figure 12.4\displaystyle 12.4) obtaining maximum potential is
    (a)(i)
    (b)(ii)
    (c)(iii)
    (d)(iv)
    NCERT_Question_Class10_Science_Exemplar_Ch12_Q8

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    (a)
    (a) (i) — four cells all face the same way, each short (\(\displaystyle -\)) plate wired to the next long (\(\displaystyle +\)) plate, so the EMFs add; (ii) and (iv) have no leads between the cells.\[V_{(i)}=\varepsilon+\varepsilon+\varepsilon+\varepsilon=4\varepsilon \] \[V_{(iii)}=\varepsilon-\varepsilon-\varepsilon+\varepsilon=0 \]

  9. Exercise 9

    Which of the following represents voltage?
    (a)
     Work done  Current × Time \displaystyle \frac{\text { Work done }}{\text { Current × Time }}
    (b)
    Work done × Charge
    (c)
     Work done × Time  Current \displaystyle \frac{\text { Work done × Time }}{\text { Current }}
    (d)
    Work done × Charge × Time

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    (a)
    (a) \(\displaystyle \dfrac{\text{Work done}}{\text{Current}\times\text{Time}}\) — voltage is work done per unit charge, and charge is current times time. \[V=\frac{W}{Q} \] \[Q=It \] \[V=\frac{W}{I\,t} \]

  10. Exercise 10

    A cylindrical conductor of length l\displaystyle l and uniform area of cross-section A\displaystyle A has resistance R\displaystyle R. Another conductor of length 2l\displaystyle 2 l and resistance R\displaystyle R of the same material has area of cross section
    (a)
    A/2\displaystyle A / 2
    (b)
    3A/2\displaystyle 3 A / 2
    (c)
    2A\displaystyle 2 A
    (d)
    3A\displaystyle 3 A

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    (c)
    (c) \(\displaystyle 2A\) — resistance depends on length and area together; doubling the length while holding R fixed needs the area doubled too. \[R=\rho\,\frac{l}{A} \] \[R=\rho\,\frac{2l}{A'} \] \[\rho\,\frac{l}{A}=\rho\,\frac{2l}{A'}\ \Rightarrow\ A'=2A \]