Q.Why are alloys used for making standard resistance coils?

Showing the 12 most recent of 16 on this concept.

1. COMEDK 2026Set 2026-A1 markMCQ
   Q.The voltage - current graph for a metal wire of uniform area of cross section at two different temp $T$ and $T^{\prime}$ is shown. Then choose the correct statement: (A) Resistivity is independent of temperature (B) Resistance of the conductor at temperature T is greater than resistance of the conductor at temperature $\mathrm{T}^{\prime}$ (C) Temperature $\mathrm{T}^{\prime}$ is greater than temperature T (D) Temperature T is greater than temperature $\mathrm{T}^{\prime}$

   ›Reveal solutionSolution

   The slope of the I–V graph is the conductance (1/R). A steeper slope means lower resistance. Since resistance increases with temperature for a metal, the steeper line (T) corresponds to a lower temperature, so T < T′.

   Concept & Intuition

   For a metallic conductor obeying Ohm’s law, the current $I$ is proportional to the applied voltage $V$:

   $I = \frac{V}{R}$

   Thus the slope of the I–V graph is $1/R$, the conductance. A steeper line means a larger slope → smaller resistance.

   Metals have a positive temperature coefficient: resistance increases as temperature rises. So if one line is steeper (lower resistance), that wire must be at a lower temperature. The problem shows two lines through the origin: the steeper line is labelled $T$, the shallower line is labelled $T'$. At the same voltage, the steeper line gives a larger current, confirming its lower resistance.

   Step-by-step reasoning

   1. Interpret the graph

      Both lines are straight through the origin, so Ohm’s law holds. The slope of each line is $\frac{I}{V} = \frac{1}{R}$. The line labelled $T$ has a larger slope than the line labelled $T'$.

   2. Relate slope to resistance

      Larger slope → smaller resistance. Hence

   $R_T < R_{T'}$

   The conductor at temperature $T$ has lower resistance than at temperature $T'$.

   3. Temperature dependence of resistance in metals For a metal, resistance increases with temperature (positive temperature coefficient). So lower resistance means lower temperature. Therefore $T < T'$ …

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2. COMEDK 2026Set 2026-M1 markMCQ
   Q.A wire of length 1 m has a resistance of $20 \Omega$ at $0^{\circ} \mathrm{C}$. It is uniformly stretched so that its length increases by $21 \%$. Assuming the volume of the wire remains constant, the percentage change in resistance is $n \%$. Alternatively, if the wire is heated [without stretching] through a temperature of $27^{\circ} \mathrm{C}$ and if the temperature coefficient of resistance of the material of wire is $0.004 K^{-1}$, the percentage change in resistance is $m \%$. The values of $m$ and $n$ are: (A) 2.08 and 1.64 (B) 10.8 and 46.4 (C) 20.8 and 16.4 (D) 1.08 and 4.64

   ›Reveal solutionSolution

   The problem asks for the percentage change in resistance due to stretching (volume constant) and due to heating (temperature coefficient given). For stretching, resistance changes as the square of length, giving $n = 46.4\%$. For heating, resistance changes linearly with temperature, giving $m = 10.8\%$. The correct option is (B).

   ---

   Concept & Intuition

   Resistance of a wire depends on its geometry and material resistivity.

   • When you stretch a wire uniformly at constant volume, its length increases and its cross-sectional area decreases. Since $R = \rho \frac{L}{A}$ and volume $V = A L$ is constant, $A \propto 1/L$, so $R \propto L^2$. A small percentage change in length leads to a much larger percentage change in resistance.
   • When you heat a wire without stretching, only the resistivity changes (geometry changes negligibly). The resistance change is directly proportional to the temperature change via the temperature coefficient of resistance.

   We compute each case separately.

   ---

   Step-by-step solution

   1. Stretching case (constant volume) Initial length $L_0 = 1 \, \text{m}$, initial resistance $R_0 = 20 \, \Omega$. Length increases by $21\%$, so new length $L = L_0 \times (1 + 0.21) = 1.21 L_0$. Volume constant: $A_0 L_0 = A L$ ⇒ $A = A_0 \frac{L_0}{L} = \frac{A_0}{1.21}$. New resistance:

   $R = \rho \frac{L}{A} = \rho \frac{1.21 L_0}{A_0 / 1.21} = \rho \frac{L_0}{A_0} \times (1.21)^2 = R_0 \times (1.21)^2.$

   So $R = 20 \times 1.4641 = 29.282 \, \Omega$.

   Percentage change:

   $n = \frac{R - R_0}{R_0} \times 100 = (1.21^2 - 1) \times 100 = (1.4641 - 1) \times 100 = 46.41\%.$

   Thus $n \approx 46.4\%$.

   2. Heating case (no stretching) …

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3. KCET 2026Set C21 markMCQ
   Q.Given below are two statements: Statement I: The resistivity of a conductor is independent of its temperature Statement II: The resistivity of a semiconductor decreases with increase in temperature Select the correct option. (A) Both Statement I and Statement II are false (B) Both Statement I and Statement II are true (C) Statement I is true but Statement II is false (D) Statement I is false but Statement II is true

   ›Reveal solutionSolution

   Resistivity of a conductor rises with temperature (more lattice collisions reduce the relaxation time $\tau$), while resistivity of a semiconductor falls with temperature (more thermally-generated charge carriers dominate over the reduced relaxation time).

   Step 1 — Examine Statement I (conductor)

   For a conductor, resistivity is given by $\rho = \dfrac{m}{ne^2\tau}$, where $n$ (free-electron density) stays essentially constant but the relaxation time $\tau$ decreases as temperature rises (more frequent collisions with vibrating lattice ions). Hence $\rho$ increases with temperature — it is NOT independent of temperature. Statement I is false.

   Step 2 — Examine Statement II (semiconductor) …

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4. COMEDK 2025Set 2025-E1 markMCQ
   Q.The resistance of a heating element is found to be $120 \Omega$ at room temperature which is $20^{\circ} \mathrm{C}$. If the temperature coefficient of the material of the resistor is $1.6 \times 10^{-4} C^{-1}$ and the resistance is found to be $160 \Omega$, the temperature of the element is: (A) $1203^{\circ} \mathrm{C}$ (B) $2083^{\circ} \mathrm{C}$ (C) $2310^{\circ} \mathrm{C}$ (D) $2013^{\circ} \mathrm{C}$

   ›Reveal solutionSolution

   Using the linear resistance–temperature relation $R_T = R_0(1 + \alpha \Delta T)$, we solve for the temperature change and add it to the reference temperature. The final temperature is $2083^\circ \mathrm{C}$, so the correct option is (B).

   The key idea is that for most metals, resistance increases approximately linearly with temperature over a wide range. The formula $R_T = R_0(1 + \alpha \Delta T)$ directly connects the change in resistance to the change in temperature, where $\alpha$ is the temperature coefficient of resistance. Here we know $R_0$, $R_T$, and $\alpha$, so we can solve for $\Delta T$ and then find the actual temperature.

   1. Identify the known quantities

      • Reference temperature: $T_0 = 20^\circ \mathrm{C}$
      • Resistance at $T_0$: $R_0 = 120 \, \Omega$
      • Resistance at unknown temperature $T$: $R_T = 160 \, \Omega$
      • Temperature coefficient: $\alpha = 1.6 \times 10^{-4} \, \mathrm{C}^{-1}$

   2. Write the resistance–temperature relation

      The standard formula is:

   $R_T = R_0 \left[ 1 + \alpha (T - T_0) \right]$

   This assumes $\alpha$ is constant over the temperature range, which is a good approximation here.

   3. Solve for the temperature change $\Delta T = T - T_0$ Rearranging:

   $\frac{R_T}{R_0} = 1 + \alpha \Delta T$

   $\alpha \Delta T = \frac{R_T}{R_0} - 1$

   $\Delta T = \frac{1}{\alpha} \left( \frac{R_T}{R_0} - 1 \right)$

   4. Plug in the numbers

   $\frac{R_T}{R_0} = \frac{160}{120} = \frac{4}{3} \approx 1.3333$

   $\frac{R_T}{R_0} - 1 = \frac{4}{3} - 1 = \frac{1}{3}$

   $\Delta T = \frac{1}{1.6 \times 10^{-4}} \times \frac{1}{3}$

   $$$$

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5. COMEDK 2025Set 2025-M1 markMCQ
   Q.The resistance of a wire is 5 ohm at $25^{\circ} \mathrm{C}$ and 7 ohm at $100^{\circ} \mathrm{C}$. The resistance of the wire at $0^{\circ} \mathrm{C}$ is (A) $\frac{13}{3} \mathrm{ohm}$ (B) $\frac{5}{3} \mathrm{ohm}$ (C) $\frac{2}{3}$ ohm (D) 0.1 ohm

   ›Reveal solutionSolution

   The resistance of a metal wire changes linearly with temperature (over moderate ranges). Using the two given data points to find the temperature coefficient and then extrapolating back to 0°C gives a resistance of $\frac{13}{3}$ ohm, which corresponds to option (A).

   Concept & Intuition

   For most metals, resistance increases approximately linearly with temperature over a limited range. This is described by the formula

   $R_T = R_0 (1 + \alpha T)$

   where $R_T$ is resistance at temperature $T$ (in °C), $R_0$ is resistance at 0°C, and $\alpha$ is the temperature coefficient of resistance.

   We are given two points: $(25^\circ\text{C}, 5\,\Omega)$ and $(100^\circ\text{C}, 7\,\Omega)$. We can use these to solve for $R_0$ and $\alpha$, then read off the answer directly.

   Step-by-step solution

   1. Set up the linear relation Let $R_0$ be the resistance at $0^\circ\text{C}$. Then at any temperature $T$ (in °C):

   $R_T = R_0 (1 + \alpha T)$

   This is a straight line with intercept $R_0$ and slope $R_0 \alpha$.

   2. Write equations for the two given points At $T = 25^\circ\text{C}$:

   $5 = R_0 (1 + 25\alpha) \quad \text{(1)}$

   At $T = 100^\circ\text{C}$:

   $7 = R_0 (1 + 100\alpha) \quad \text{(2)}$

   3. Eliminate $R_0$ by dividing the equations Divide (2) by (1):

   $\frac{7}{5} = \frac{1 + 100\alpha}{1 + 25\alpha}$

   Cross-multiply:

   $7(1 + 25\alpha) = 5(1 + 100\alpha)$

   $7 + 175\alpha = 5 + 500\alpha$

   $2 = 325\alpha$

   $\alpha = \frac{2}{325} \quad (\text{per }^\circ\text{C})$

   4. Find $R_0$ using equation (1) Substitute $\alpha$ back: $5 = R_0 \left(1 + 25 \cdot \frac{2}{325}\right) = R_0 \left(1 + \frac{50}{325}\right) = R_0 \left(1 + \frac{2}{13}\right)$ …

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6. KCET 2024Set D-21 markMCQ
   Q.The I–V graph for a conductor at two different temperatures $100^\circ \mathrm{C}$ and $400^\circ \mathrm{C}$ is as shown in the figure. The temperature coefficient of resistance of the conductor is about (in per degree Celsius) (A) $3 \times 10^{-3}$ (B) $6 \times 10^{-3}$ (C) $9 \times 10^{-3}$ (D) $12 \times 10^{-3}$

   ›Reveal solutionSolution

   Read each resistance off the graph as the reciprocal of the line's slope, then put the two $(R,T)$ pairs into the temperature-coefficient formula.

   Step 1 — What the graph's slope means

   For an ohmic conductor $V = IR$, so plotting $I$ (vertical) against $V$ (horizontal) gives a straight line through the origin with

   $\text{slope} = \frac{I}{V} = \frac{1}{R} \qquad\Longrightarrow\qquad R = \frac{1}{\text{slope}}$

   Since the angles are measured from the V-axis (the horizontal axis), slope $=\tan\theta$ and therefore

   $R \;\propto\; \cot\theta$

   A steeper line (bigger $\theta$) means a smaller resistance.

   Step 2 — Assign each line to its temperature

   • $\theta = 45^\circ$: $R \propto \cot 45^\circ = 1$
   • $\theta = 30^\circ$: $R \propto \cot 30^\circ = \sqrt{3} \approx 1.732$

   For a conductor, resistance increases with temperature (more lattice vibrations → more scattering of electrons → shorter relaxation time). So the larger resistance must belong to the higher temperature:

   $R_1 = 1 \ \text{(at } T_1 = 100^\circ\text{C)}, \qquad R_2 = \sqrt3 \ \text{(at } T_2 = 400^\circ\text{C)}$

   Step 3 — Apply the temperature coefficient of resistance

   Taking $R_1$ as the reference resistance over the interval $\Delta T = T_2 - T_1 = 300^\circ\text{C}$:

   $R_2 = R_1\left[1 + \alpha\,\Delta T\right] \quad\Longrightarrow\quad \alpha = \frac{R_2 - R_1}{R_1\,\Delta T}$ …

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7. COMEDK 2024Set 2024-A1 markMCQ
   Q.The resistance of a wire at room temperature $20^{\circ} \mathrm{C}$ is found to be $10 \Omega$. If resistance of the wire increases by $10 \%$, then the temperature of the wire will be (The temperature coefficient of the material of the wire is $0.002 /{ }^{\circ} \mathrm{C}$) (A) $52^0 \mathrm{C}$ (B) $22^0 \mathrm{C}$ (C) $62^0 \mathrm{C}$ (D) $72^0 \mathrm{C}$

   ›Reveal solutionSolution

   The problem uses the linear temperature dependence of resistance: $R = R_0(1 + \alpha \Delta T)$. Given a 10% increase in resistance, we solve for the new temperature. The final temperature is $70^\circ\mathrm{C}$ above room temperature, so the answer is $72^\circ\mathrm{C}$.

   Concept & Intuition

   For most metals, resistance increases nearly linearly with temperature over modest ranges. The formula $R = R_0(1 + \alpha \Delta T)$ captures this: $R_0$ is the resistance at a reference temperature, $\alpha$ is the temperature coefficient (fractional change per degree), and $\Delta T$ is the temperature change. Here, a 10% increase means $R = 1.1 R_0$. We simply plug in and solve for the new temperature.

   Step-by-step solution

   1. Identify given values

      • Initial resistance at $20^\circ\mathrm{C}$: $R_0 = 10\,\Omega$
      • Resistance increases by 10%, so final resistance: $R = 10\,\Omega \times 1.10 = 11\,\Omega$
      • Temperature coefficient: $\alpha = 0.002\,/^\circ\mathrm{C}$
      • Initial temperature: $T_0 = 20^\circ\mathrm{C}$

   2. Write the resistance-temperature relation

   $R = R_0 \left[1 + \alpha (T - T_0)\right]$

   Here $T$ is the final temperature we need.

   3. Substitute known values

   $11 = 10 \left[1 + 0.002 (T - 20)\right]$

   4. Solve for $T - 20$ Divide both sides by 10:

   $1.1 = 1 + 0.002 (T - 20)$

   Subtract 1:

   $0.1 = 0.002 (T - 20)$ …

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8. COMEDK 2024Set 2024-M1 markMCQ
   Q.The current through a conductor is '$\mathrm{a}$' when the temperature is $0^{\circ} \mathrm{C}$. It is '$\mathrm{b}$' when the temperature is $100^{\circ} \mathrm{C}$. The current through the conductor at $220^{\circ} \mathrm{C}$ is (A) $\frac{5 a b}{11 b-6 a}$ (B) $\frac{5 a b}{6 a-11 b}$ (C) $\frac{5 a b}{11 a-6 b}$ (D) $\frac{11 a b}{5 a-6 b}$

   ›Reveal solutionSolution

   The current changes with temperature due to a linear change in resistance; using the temperature coefficient of resistance and the given currents at 0°C and 100°C, the current at 220°C is found to be $\frac{5ab}{6a - 11b}$, which corresponds to option (B).

   The key idea is that the conductor’s resistance changes linearly with temperature (over a moderate range). Since the voltage across the conductor is constant (implied by the problem’s setup), the current is inversely proportional to resistance. We can use the two given current values to find the temperature coefficient of resistance, then compute the current at 220°C.

   1. Relate current and resistance

      If the voltage $V$ is constant, then $I = V/R$. At 0°C, current $I_0 = a$ and resistance $R_0 = V/a$. At 100°C, current $I_{100} = b$ and resistance $R_{100} = V/b$.

   2. Use the linear resistance-temperature relation

      For most conductors, $R_t = R_0 (1 + \alpha t)$, where $\alpha$ is the temperature coefficient of resistance.

      At $t = 100^\circ\text{C}$:

   $R_{100} = R_0 (1 + 100\alpha)$

   Substitute $R_{100} = V/b$ and $R_0 = V/a$:

   $\frac{V}{b} = \frac{V}{a} (1 + 100\alpha)$

   Cancel $V$ (nonzero):

   $\frac{1}{b} = \frac{1}{a} (1 + 100\alpha)$

   Solve for $\alpha$:

   $1 + 100\alpha = \frac{a}{b} \quad\Rightarrow\quad 100\alpha = \frac{a}{b} - 1 = \frac{a - b}{b}$

   $\alpha = \frac{a - b}{100b}$

   3. Find resistance at 220°C

   $R_{220} = R_0 (1 + 220\alpha) = \frac{V}{a} \left(1 + 220 \cdot \frac{a - b}{100b}\right)$

   Simplify the bracket:

   $1 + \frac{220(a - b)}{100b} = 1 + \frac{11(a - b)}{5b} = \frac{5b + 11a - 11b}{5b} = \frac{11a - 6b}{5b}$

   So:

   $R_{220} = \frac{V}{a} \cdot \frac{11a - 6b}{5b}$

   4. Compute current at 220°C

   $I_{220} = \frac{V}{R_{220}} = V \div \left( \frac{V}{a} \cdot \frac{11a - 6b}{5b} \right) = \frac{a \cdot 5b}{11a - 6b}$

   That is:

   $I_{220} = \frac{5ab}{11a - 6b}$

   But note: The denominator is $11a - 6b$. However, looking at the options, (B) is $\frac{5ab}{6a - 11b}$. These are not the same unless we check sign. Since current decreases as temperature rises (for a metal), we expect $b < a$, so $11a - 6b > 0$ and $6a - 11b > 0$ as well? Let’s check: If $a > b$, then $6a - 11b$ could be positive or negative depending on values. But the expression we derived is correct algebraically. Let’s re-check the algebra carefully.

   Actually, from step 2:

   $\frac{1}{b} = \frac{1}{a}(1 + 100\alpha) \Rightarrow 1 + 100\alpha = \frac{a}{b} \Rightarrow \alpha = \frac{a/b - 1}{100} = \frac{a - b}{100b}$

   That’s fine.

   Then at 220°C:

   $$$$

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9. COMEDK 2024Set 2024-M1 markMCQ
   Q.On increasing the temperature of a conductor, its resistance increases because (A) Electron density decreases (B) Relaxation time increases (C) Number of collisions between electrons decreases (D) Relaxation time decreases

   ›Reveal solutionSolution

   The resistance of a conductor increases with temperature because the relaxation time (average time between electron collisions) decreases, making it harder for electrons to drift. The correct option is (D).

   The key concept here is relaxation time ($\tau$), which is the average time an electron travels freely between collisions with lattice ions. Resistance $R$ is inversely proportional to $\tau$:

   $R \propto \frac{1}{\tau}$

   When temperature rises, lattice ions vibrate more vigorously, increasing the chance of electron-ion collisions. This reduces the relaxation time, thereby increasing resistance. Let’s examine each option.

   1. Option (A): Electron density decreases

      Electron density (number of free electrons per unit volume) in a conductor is essentially constant with temperature — it depends on the material’s atomic structure, not on thermal agitation. So this is false.

   2. Option (B): Relaxation time increases

      If relaxation time increased, electrons would travel longer between collisions, making conduction easier and resistance lower. But we know resistance rises with temperature, so this is the opposite of what happens.

   3. Option (C): Number of collisions between electrons decreases

      Fewer collisions would mean less obstruction to electron flow, again lowering resistance. In reality, more collisions occur at higher temperatures, so this is incorrect.

   4. Option (D): Relaxation time decreases …

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10. KCET 2023Set A-31 markMCQ
    Q.A radioactive sample has half-life of $3$ years. The time required for the activity of the sample to reduce to $\dfrac{1}{5}$th of its initial value is about (A) $7$ years (B) $15$ years (C) $5$ years (D) $10$ years

    ›Reveal solutionSolution

    Solve $\left(\tfrac12\right)^{t/T}=\tfrac15$ for $t$, i.e. $t=T\,\dfrac{\ln 5}{\ln 2}$.

    Step 1 — The law.

    Activity falls with the same exponential law as the number of nuclei (since $A=\lambda N$):

    $A(t) = A_0 e^{-\lambda t} = A_0\left(\frac{1}{2}\right)^{t/T_{1/2}},$

    where $T_{1/2}=3$ years. Using the half-life form avoids ever having to compute $\lambda$.

    Step 2 — Impose the required drop.

    We want $\dfrac{A}{A_0}=\dfrac{1}{5}$:

    $\left(\frac{1}{2}\right)^{t/3} = \frac{1}{5}.$

    Step 3 — Take logarithms.

    $\frac{t}{3}\ln\frac{1}{2} = \ln\frac{1}{5}\ \Longrightarrow\ \frac{t}{3} = \frac{\ln 5}{\ln 2} = \frac{1.609}{0.693} = 2.322.$ …

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11. KCET 2023Set A-31 markMCQ
    Q.For a given electric current the drift velocity of conduction electrons in a copper wire is $v_d$ and their mobility is $\mu$. When the current is increased at constant temperature (A) $v_d$ remains the same, $\mu$ increases (B) $v_d$ decreases, $\mu$ remains the same (C) $v_d$ remains the same, $\mu$ decreases (D) $v_d$ increases, $\mu$ remains the same

    ›Reveal solutionSolution

    $I=neAv_d$ makes $v_d\propto I$; but $\mu=e\tau/m$ is a material property fixed by temperature, so it does not change.

    Step 1 — Drift velocity.

    The current in a metal is carried by electrons drifting with speed $v_d$:

    $I = n e A v_d \;\Longrightarrow\; v_d = \frac{I}{neA}.$

    For a given copper wire, the free-electron density $n$, the electronic charge $e$ and the cross-section $A$ are all fixed. Hence

    $v_d \propto I.$

    Increase the current and the drift velocity increases proportionally. (This must be so — the drift of charge is the current.)

    Step 2 — Mobility.

    Mobility is defined as drift velocity per unit applied field:

    $\mu = \frac{v_d}{E}.$

    From the Drude picture, $v_d = \dfrac{eE\tau}{m}$, so

    $\mu = \frac{e\tau}{m}.$ …

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12. KCET 2022Set B-31 markMCQ
    Q.A galvanometer of resistance $50 \Omega$ is connected to a battery of 3 V along with a resistance $2950 \Omega$ in series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 divisions, the resistance in series should be (A) $5050 \Omega$ (B) $4450 \Omega$ (C) $6050 \Omega$ (D) $5550 \Omega$

    ›Reveal solutionSolution

    Deflection $\propto$ current, so scale the current down in the ratio 20/30, find the new total circuit resistance from Ohm's law, and subtract the galvanometer's own $50\ \Omega$.

    Step 1 — The key physical fact.

    In a moving-coil galvanometer the deflection is directly proportional to the current through the coil:

    $\theta = \left(\frac{NAB}{k}\right) I \qquad\Longrightarrow\qquad \theta \propto I .$

    So fewer divisions simply means proportionally less current.

    Step 2 — Current for full-scale (30 divisions).

    The galvanometer ($G=50\ \Omega$) and the series resistance ($R_1=2950\ \Omega$) are in series across the 3 V battery:

    $I_1=\frac{\mathcal{E}}{G+R_1}=\frac{3}{50+2950}=\frac{3}{3000}=1\times10^{-3}\ \mathrm{A}=1\ \mathrm{mA}.$

    So the galvanometer's full-scale (30 div) current is 1 mA, i.e. its figure of merit is

    $\frac{1\ \mathrm{mA}}{30\ \text{div}}=\frac{1}{30}\ \mathrm{mA/div}.$

    Step 3 — Current needed for 20 divisions.

    $I_2 = I_1\times\frac{20}{30}=1\ \mathrm{mA}\times\frac{2}{3}=\frac{2}{3}\ \mathrm{mA}=6.67\times10^{-4}\ \mathrm{A}.$

    Step 4 — New total resistance.

    The emf is unchanged at 3 V, so by Ohm's law the whole loop must now present

    $R_{\text{total}}=\frac{\mathcal{E}}{I_2}=\frac{3}{\tfrac{2}{3}\times10^{-3}}=\frac{3\times 3}{2}\times10^{3}=4500\ \Omega .$ …

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