Q.Two moving coil meters, $M_1$ and $M_2$ have the following particulars:
$R_1 = 10\ \Omega$, $N_1 = 30$, $A_1 = 3.6 \times 10^{-3}\ \text{m}^2$, $B_1 = 0.25\ \text{T}$
$R_2 = 14\ \Omega$, $N_2 = 42$, $A_2 = 1.8 \times 10^{-3}\ \text{m}^2$, $B_2 = 0.50\ \text{T}$
(The spring constants are identical for the two meters). Determine the ratio of

Showing the 12 most recent of 15 on this concept.

1. KCET 2026Set C21 markMCQ
   Q.Which of the following circuits is correct for verification of Ohm's law? (A) Figure

   (1) (B) Figure

   (2) (C) Figure

   (3) (D) Figure (4)

   ›Reveal solutionSolution

   Verifying Ohm's law ($V=IR$) needs the ammeter in series with R (so it reads the actual current through R) and the voltmeter connected directly across R alone (so it reads the true potential difference across R, not across R plus the ammeter or any other extra component).

   Step 1 — Requirement on the ammeter's placement

   The ammeter must be placed in series with the resistor R itself (not in series with the parallel combination of R and the voltmeter), so that the current it measures is exactly the current flowing through R, with none diverted through the voltmeter branch first.

   Step 2 — Requirement on the voltmeter's placement …

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2. KCET 2025Set D-41 markMCQ
   Q.Two similar galvanometers are covered into an ammeter and a milliammeter. The shunt resistance of ammeter as compared to the shunt resistance of milliammeter will be (A) Zero (B) More (C) Less (D) Equal

   ›Reveal solutionSolution

   $S = I_g G/(I - I_g)$ — the shunt is inversely related to the range, so the higher-range instrument (the ammeter) needs the smaller shunt.

   Step 1 — How a galvanometer becomes an ammeter.

   A galvanometer of resistance $G$ gives full-scale deflection at a tiny current $I_g$. To read a much larger current $I$, a low resistance $S$ (the shunt) is connected in parallel, so that only $I_g$ passes through the coil and the rest, $(I-I_g)$, bypasses it.

   Step 2 — Derive the shunt formula.

   Because $G$ and $S$ are in parallel, the potential difference across them is the same:

   $I_g G = (I - I_g) S$

   $\boxed{S = \frac{I_g\,G}{I - I_g}}$

   Step 3 — Compare the two instruments.

   The galvanometers are similar, so $G$ and $I_g$ are the same for both. Then $S$ depends only on the range $I$:

   $S \propto \frac{1}{I - I_g}$

   A larger range $I$ means more current must be diverted around the coil, which demands an easier bypass path — i.e. a smaller $S$.

   Step 4 — Apply it.

   An ammeter measures currents of the order of amperes; a milliammeter only milliamperes. So $I_{\text{ammeter}} \gg I_{\text{milliammeter}}$, and therefore …

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3. KCET 2025Set D-41 markMCQ
   Q.The range of electrical conductivity $(\sigma)$ and resistivity $(\rho)$ for metals, among the following, is (A) $\rho \rightarrow 10^{-5} - 10^{-6}\Omega \mathrm{m}$ \ (B) $\rho \rightarrow 10^{11} - 10^{19}\Omega \mathrm{m}$ \ (C) $\rho \rightarrow 10^2 - 10^8\Omega \mathrm{m}$ \ (D) $\rho \rightarrow 10^{-2} - 10^{-8}\Omega \mathrm{m}$ \

   ›Reveal solutionSolution

   Metals are the low-resistivity class: $\rho \approx 10^{-2}$–$10^{-8}\ \Omega\,\mathrm{m}$ (equivalently $\sigma \approx 10^{2}$–$10^{8}\ \mathrm{S\,m^{-1}}$).

   Step 1 — The classification by resistivity/conductivity (NCERT, Semiconductor Electronics).

   Class	Resistivity $\rho\ (\Omega\,\mathrm{m})$	Conductivity $\sigma\ (\mathrm{S\,m^{-1}})$
   Metals (conductors)	$10^{-2}$ – $10^{-8}$	$10^{2}$ – $10^{8}$
   Semiconductors	$10^{-5}$ – $10^{6}$	$10^{5}$ – $10^{-6}$
   Insulators	$10^{11}$ – $10^{19}$	$10^{-11}$ – $10^{-19}$

   Recall $\sigma = \dfrac{1}{\rho}$, so the two columns are reciprocals of each other.

   Step 2 — Test the options.

   • (A) $\rho \to 10^{-5}$–$10^{-6}$ — far too narrow a window; it is not the quoted metallic range.
   • (B) $\rho \to 10^{11}$–$10^{19}$ — this is the insulator range. …

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4. KCET 2024Set D-21 markMCQ
   Q.In an experiment to determine the temperature coefficient of resistance of a conductor, a coil of wire X is immersed in a liquid. It is heated by an external agent. A meter bridge set up is used to determine resistance of the coil X at different temperatures. The balancing points measured at temperatures $t_1 = 0^\circ \mathrm{C}$ and $t_2 = 100^\circ \mathrm{C}$ are $50\,\mathrm{cm}$ and $60\,\mathrm{cm}$ respectively. If the standard resistance taken out is $S = 4\,\Omega$, in both trials, the temperature coefficient of the coil is (A) $0.05^\circ \mathrm{C}^{-1}$ (B) $0.02^\circ \mathrm{C}^{-1}$ (C) $0.005^\circ \mathrm{C}^{-1}$ (D) $2.0^\circ \mathrm{C}^{-1}$

   ›Reveal solutionSolution

   Use the meter bridge balance condition $X = S\cdot\dfrac{l}{100-l}$ to find the coil's resistance at $0^\circ$C and $100^\circ$C, then apply $\alpha = \dfrac{R_2-R_1}{R_1(t_2-t_1)}$. This gives $\alpha = 0.005\ ^\circ\text{C}^{-1}$, matching option (C).

   The concept: meter bridge

   A meter bridge is a practical application of the Wheatstone bridge. A 1 m resistance wire is stretched along a scale; the unknown resistance $X$ (here, the coil) is connected in one gap and a known standard resistance $S$ in the other. A jockey slides along the wire until the galvanometer shows no deflection — the balancing length $l$ (measured from the end connected to $X$). At balance,

   $\frac{X}{S} = \frac{l}{100-l} \quad\Rightarrow\quad X = S\cdot\frac{l}{100-l}.$

   By repeating this at different temperatures, we can track how the coil's resistance changes with temperature — which is exactly what's needed to find its temperature coefficient of resistance.

   Step 1 — Resistance of the coil at $t_1 = 0^\circ$C

   Given $l_1 = 50$ cm and $S = 4\ \Omega$:

   $X_1 = S\cdot\frac{l_1}{100-l_1} = 4 \times \frac{50}{100-50} = 4 \times \frac{50}{50} = 4\ \Omega$

   Step 2 — Resistance of the coil at $t_2 = 100^\circ$C

   Given $l_2 = 60$ cm and the same $S = 4\ \Omega$:

   $X_2 = S\cdot\frac{l_2}{100-l_2} = 4 \times \frac{60}{100-60} = 4 \times \frac{60}{40} = 6\ \Omega$

   Step 3 — Apply the temperature coefficient formula

   The temperature coefficient of resistance is defined by

   $R_2 = R_1\left[1 + \alpha(t_2 - t_1)\right] \quad\Rightarrow\quad \alpha = \frac{R_2 - R_1}{R_1(t_2-t_1)}.$

   Substituting $R_1 = X_1 = 4\ \Omega$, $R_2 = X_2 = 6\ \Omega$, and $t_2-t_1 = 100^\circ$C: …

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5. KCET 2024Set D-21 markMCQ
   Q.Magnetic susceptibility of Mg at 300 K is $1.2 \times 10^{-5}$. What is its susceptibility at 200 K? (A) $18 \times 10^{-5}$ (B) $180 \times 10^{-5}$ (C) $1.8 \times 10^{-5}$ (D) $0.18 \times 10^{-5}$

   ›Reveal solutionSolution

   Magnesium is paramagnetic, so its susceptibility obeys Curie's law $\chi \propto 1/T$. Cooling from 300 K to 200 K increases the susceptibility by the factor $300/200 = 1.5$, giving $\chi_{200} = 1.8\times10^{-5}$ — option (C).

   The concept: Curie's law

   For a paramagnetic material, the individual atomic magnetic moments are only weakly aligned by an external field because thermal agitation constantly randomizes their orientation. Curie's law states that the magnetic susceptibility of such a material varies inversely with the absolute temperature:

   $\chi = \frac{C}{T}$

   where $C$ is the material-specific Curie constant. Lower temperature means less thermal disruption, so the moments align more easily with the field, and the susceptibility increases as $T$ decreases. Magnesium is a paramagnetic metal, so this law applies directly.

   Step 1 — Set up the ratio between two temperatures

   Since $\chi T = C$ is constant for a given sample,

   $\chi_1 T_1 = \chi_2 T_2 \quad\Rightarrow\quad \chi_2 = \chi_1\cdot\frac{T_1}{T_2}.$

   Step 2 — Substitute the given values

   Here $T_1 = 300$ K, $\chi_1 = 1.2\times10^{-5}$, and $T_2 = 200$ K:

   $\chi_2 = 1.2\times10^{-5} \times \frac{300}{200} = 1.2\times10^{-5} \times 1.5$

   Step 3 — Compute the final value

   $\chi_2 = 1.8\times10^{-5}$ …

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6. COMEDK 2024Set 2024-M1 markMCQ
   Q.The resistance of the galvanometer and shunt of an ammeter are $90 \mathrm{~ohm}$ and $10 \mathrm{~ohm}$ respectively, then the fraction of the main current passing through the galvanometer and the shut respectively are: (A) $\frac{1}{90}$ and $\frac{1}{10}$ (B) $\frac{1}{10}$ and $\frac{1}{90}$ (C) $\frac{1}{10}$ and $\frac{9}{10}$ (D) $\frac{9}{10}$ and $\frac{1}{10}$

   ›Reveal solutionSolution

   The ammeter uses a shunt to divert most of the current; the fraction through the galvanometer is the shunt resistance divided by the total resistance, giving 1/10 through the galvanometer and 9/10 through the shunt — so the correct option is (C).

   Concept & Intuition

   An ammeter is built by connecting a small resistance (shunt) in parallel with a sensitive galvanometer. Because the shunt has much lower resistance, most of the main current flows through it, protecting the delicate galvanometer. The current divides in inverse proportion to the resistances — the smaller resistor carries the larger fraction. Here, the galvanometer is 90 Ω and the shunt is 10 Ω, so the shunt should carry most of the current.

   Step-by-step reasoning

   1. Identify the circuit configuration

      The galvanometer (resistance $R_g = 90\ \Omega$) and the shunt (resistance $R_s = 10\ \Omega$) are connected in parallel. The main current $I$ splits into $I_g$ through the galvanometer and $I_s$ through the shunt.

   2. Apply the current divider rule

      For two resistors in parallel, the current through one branch is the total current multiplied by the opposite resistance divided by the sum of the resistances:

   $I_g = I \cdot \frac{R_s}{R_g + R_s}, \qquad I_s = I \cdot \frac{R_g}{R_g + R_s}$

   This works because the voltage across both is the same, so $I_g R_g = I_s R_s$.

   3. Plug in the numbers

   $I_g = I \cdot \frac{10}{90 + 10} = I \cdot \frac{10}{100} = \frac{I}{10}$

   $$$$

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7. KCET 2023Set A-31 markMCQ
   Q.A proton and an alpha-particle moving with the same velocity enter a uniform magnetic field with their velocities perpendicular to the magnetic field. The ratio of radii of their circular paths is (A) $1:4$ (B) $4:1$ (C) $1:2$ (D) $2:1$

   ›Reveal solutionSolution

   When a charged particle moves perpendicular to a uniform magnetic field, the radius of its circular path is $r = \frac{mv}{qB}$. For the same velocity and field, the ratio of radii equals the ratio of $m/q$. For a proton and an alpha-particle, this ratio is $1:2$.

   The key idea here is that the magnetic force provides the centripetal force for circular motion. When a charged particle enters a magnetic field perpendicularly, it experiences a force that is always perpendicular to its velocity, causing it to move in a circle. The radius of that circle depends on the particle's mass, charge, speed, and the field strength.

   For any charged particle moving perpendicular to a uniform magnetic field $B$ with velocity $v$, the magnetic force is $F = qvB$. This force acts as the centripetal force needed for circular motion: $F_c = \frac{mv^2}{r}$. Setting them equal gives $qvB = \frac{mv^2}{r}$, which simplifies to $r = \frac{mv}{qB}$.

   Since both particles have the same velocity $v$ and are in the same magnetic field $B$, the radius is directly proportional to $m/q$. So the ratio of their radii is simply the ratio of their mass-to-charge ratios.

   1. Identify the particles and their properties. A proton has mass $m_p$ and charge $+e$. An alpha-particle is a helium nucleus with mass $4m_p$ and charge $+2e$. …

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8. KCET 2023Set A-31 markMCQ
   Q.The current in a coil changes from 2A to 5A in 0.3s. The magnitude of emf induced in the coil is 1.0 V. The value of self-inductance of the coil is (A) 100 mH (B) 0.1 mH (C) 10 mH (D) 1.0 mH

   ›Reveal solutionSolution

   Apply $|\varepsilon| = L\,|dI/dt|$ and solve for $L$.

   1. The concept — self-induction.

   When the current through a coil changes, its own magnetic flux changes, and by Faraday's law an emf is induced in the coil itself that opposes the change (Lenz's law). The flux linkage is proportional to the current, $N\Phi = LI$, where the constant $L$ is the self-inductance. Differentiating and applying Faraday's law:

   $\varepsilon = -\frac{d(N\Phi)}{dt} = -L\frac{dI}{dt} \quad\Longrightarrow\quad |\varepsilon| = L\left|\frac{dI}{dt}\right|$

   This is exactly the relation we need: it ties together the three quantities in the question ($\varepsilon$, the rate of current change, and $L$).

   2. Rate of change of current.

   $\frac{dI}{dt} = \frac{\Delta I}{\Delta t} = \frac{5\,\text{A} - 2\,\text{A}}{0.3\,\text{s}} = \frac{3}{0.3} = 10\ \text{A s}^{-1}$ …

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9. COMEDK 2023Set 2023-M1 markMCQ
   Q.A galvanometer having a resistance of $8 \Omega$ is shunted by a wire of resistance $2 \Omega$. If the total current is $1 \mathrm{~A}$, the part of it passing through the shunt will be (A) 0.25 A (B) 0.8 A (C) 0.2 A (D) 0.5 A

   ›Reveal solutionSolution

   By the current-divider rule the shunt (2 Ω) carries $0.8$ A of the $1$ A total.

   Galvanometer $G=8\,\Omega$ and shunt $S=2\,\Omega$ are in parallel, so they share the same voltage: $I_g G=I_s S$. With $I_g+I_s=I=1$ A, …

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10. KCET 2022Set B-31 markMCQ
    Q.A circular coil of wire of radius ‘r’ has ‘n’ turns and carries a current ‘I’. The magnetic induction ‘B’ at a point on the axis of the coil at a distance $\sqrt{3} r$ from its centre is (A) $\dfrac{\mu_0 n I}{16 r}$ (B) $\dfrac{\mu_0 n I}{4 r}$ (C) $\dfrac{\mu_0 n I}{32 r}$ (D) $\dfrac{\mu_0 n I}{8 r}$

    ›Reveal solutionSolution

    The magnetic field on the axis of a circular coil depends on the distance from the centre. Substituting $x = \sqrt{3}r$ into the standard formula gives $B = \dfrac{\mu_0 n I}{16 r}$, so option (A) is correct.

    The magnetic field on the axis of a current-carrying circular coil is a classic application of the Biot–Savart law. The key idea is that each small current element contributes a field that, due to symmetry, has only an axial component at points on the axis. The formula that results is clean and directly usable.

    The standard result for the magnetic induction at a point on the axis of a single-turn coil of radius $r$, carrying current $I$, at a distance $x$ from the centre is:

    $B_{\text{single}} = \frac{\mu_0 I r^2}{2 (r^2 + x^2)^{3/2}}$

    For a coil with $n$ turns, the fields simply add (since the turns are closely wound and carry the same current), so:

    $B = n \cdot \frac{\mu_0 I r^2}{2 (r^2 + x^2)^{3/2}}$

    Now we substitute the given distance $x = \sqrt{3}\, r$.

    1. Substitute $x = \sqrt{3}r$ into the denominator.

       Compute $r^2 + x^2 = r^2 + (\sqrt{3}r)^2 = r^2 + 3r^2 = 4r^2$.

    2. Raise to the power $3/2$.

       $(r^2 + x^2)^{3/2} = (4r^2)^{3/2}$.

       $(4r^2)^{3/2} = 4^{3/2} \cdot (r^2)^{3/2} = (\sqrt{4})^3 \cdot r^3 = 2^3 \cdot r^3 = 8r^3$.

    3. Plug into the formula. …

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11. COMEDK 2022Set 20221 markMCQ
    Q.A galvanometer having a resistance of 4 $\Omega$ is shunted by a wire of resistance 2 $\Omega$. If the total current is 1.5 A, the current passing through shunt is (A) 1.25 A (B) 1 A (C) 0.75 A (D) 0.5 A

    ›Reveal solutionSolution

    (Then I_g = 0.5 A; check: 1 A × 2 Ω = 2 V = 0.5 A × 4 Ω ✓.)

    Concept: Current divider between the galvanometer (R_g) and its shunt (R_s) in parallel — the current splits inversely as the resistances.

    I_s / I_g = R_g / R_s, and I_s + I_g = I. …

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12. KCET 2021Set B-21 markMCQ
    Q.In an oscillating LC circuit, $L = 3.00~\text{mH}$ and $C = 2.70~\mu\text{F}$. At $t = 0$ the charge on the capacitor is zero and the current is $2.00~\text{A}$. The maximum charge that will appear on the capacitor will be (A) $1.8 \times 10^{-5}~\text{C}$ (B) $18 \times 10^{-5}~\text{C}$ (C) $9 \times 10^{-5}~\text{C}$ (D) $90 \times 10^{-5}~\text{C}$

    ›Reveal solutionSolution

    Conserve the LC energy: at $t=0$ all of it is magnetic ($\tfrac12 Li^2$), and at maximum charge all of it is electric ($q_{\max}^2/2C$), so $q_{\max}=i\sqrt{LC}$.

    1. The physics — energy conservation in an LC oscillator

    An ideal LC circuit has no resistance, so no energy is dissipated. The total energy

    $U = \underbrace{\frac{q^2}{2C}}_{\text{electric (capacitor)}} + \underbrace{\frac{1}{2}Li^2}_{\text{magnetic (inductor)}} = \text{constant}$

    oscillates back and forth between the two forms at the angular frequency $\omega = 1/\sqrt{LC}$.

    2. Evaluate the constant at $t = 0$

    Given: at $t=0$, $q = 0$ and $i = 2.00\ \mathrm{A}$. Since $q=0$, the capacitor stores nothing, so the current at that instant is the maximum current $i_{\max}=2.00\ \mathrm{A}$, and

    $U = 0 + \frac{1}{2}L\,i_{\max}^{2}$

    3. Evaluate the same constant at maximum charge

    At the instant the charge peaks, $\dfrac{dq}{dt}=i=0$, so the whole energy is back in the capacitor:

    $U = \frac{q_{\max}^{2}}{2C} + 0$

    4. Equate and solve for $q_{\max}$

    $\frac{q_{\max}^{2}}{2C} = \frac{1}{2}L\,i_{\max}^{2}$

    $q_{\max}^{2} = L\,C\,i_{\max}^{2}$

    $\boxed{q_{\max} = i_{\max}\sqrt{LC}}$

    5. Put in the numbers (SI units throughout)

    $L = 3.00\ \mathrm{mH} = 3.00\times10^{-3}\ \mathrm{H}, \qquad C = 2.70\ \mu\mathrm{F} = 2.70\times10^{-6}\ \mathrm{F}$ …

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