Question

If resistor X were made of manganin and readings for V and I are taken without switching off the circuit, the graph between V and I will be as :

• A. A
• B. B
• C. C
• D. D
• A. due to error in voltmeter reading only.
• B. due to error in ammeter reading only.
• C. equal to the sum of error in voltmeter reading and error in ammeter reading.
• D. equal to error in voltmeter reading divided by the error in ammeter reading.
• A. reading in ammeter decreases and reading in voltmeter increases.
• B. readings in both voltmeter and ammeter increase.
• C. reading in ammeter increases and reading in voltmeter decreases.
• D. readings in both voltmeter and ammeter decrease.
• A. \(v_1 > v_2 > v_3\)
• B. \(v_2 > v_3 > v_1\)
• C. \(v_3 > v_2 > v_1\)
• D. \(v_1 = v_2 = v_3\)
• A. \(E_1 = E_2 = E_3\)
• B. \(E_3 > E_2 > E_1\)
• C. \(E_2 > E_3 > E_1\)
• D. \(E_1 > E_2 > E_3\)

Answer 1

The Correct Option is A

Concept: According to Ohm's Law, the electric current $I$ flowing through a conductor is directly proportional to the potential difference $V$ across its ends, provided the physical conditions such as temperature, tension, and material composition remain constant. Mathematically, this relationship is expressed as: \[ V = I \cdot R \quad \implies \quad I = \left(\frac{1}{R}\right)V \] Here, $R$ represents the electrical resistance of the material, which acts as the constant of proportionality. On an $I$ versus $V$ plot (where $I$ is plotted on the vertical $y$-axis and $V$ is plotted on the horizontal $x$-axis), the slope of the curve is equal to the reciprocal of the resistance ($\text{Slope} = \frac{1}{R}$). When electric current passes through any resistor continuously without switching off the circuit, heat is generated inside the material due to Joule heating, which is given by the formula: \[ H = I^2 \cdot R \cdot t \] This heat energy leads to a rise in the internal temperature of the resistor. For standard conductors, a change in temperature modifies the resistance according to the relationship: \[ R(T) = R_0 [1 + \alpha(T - R_0)] \] where $\alpha$ is the temperature coefficient of resistance.

Step 1: Analyzing the material characteristics of Manganin.
Manganin is a specialized alloy typically composed of approximately 84% copper, 12% manganese, and 4% nickel. It belongs to a unique category of materials known as precision resistance alloys. The defining characteristic of manganin is that it possesses an exceptionally low, nearly negligible temperature coefficient of resistance ($\alpha \approx 0$). This implies that even if the internal temperature of a manganin wire increases significantly due to prolonged current flow or continuous operation without switching off the circuit, its electrical resistance $R$ remains virtually unaltered and stays perfectly stable at its initial value.

Step 2: Determining the behavior of the $I$-$V$ graph based on its material properties.
Since the resistance $R$ of the manganin resistor stays constant throughout the experiment despite the continuous flow of current and subsequent Joule heating: \[ R = \text{constant} \] The relationship between current $I$ and voltage $V$ remains strictly linear at all times: \[ I \propto V \] Because the slope $\frac{1}{R}$ is constant and does not change, the graph between the current $I$ and the potential difference $V$ must be a straight line passing through the origin, as depicted in option (A). There will be no deviation or bending towards either axis because the material does not exhibit significant non-ohmic heating characteristics under normal laboratory conditions.

Answer 2

Concept: The unknown resistance $X$ is determined experimentally by applying Ohm's Law to the recorded values of potential difference ($V$) measured by the voltmeter and electric current ($I$) measured by the ammeter. The governing formula is: \[ X = \frac{V}{I} \] In any physical measurement, the quantities $V$ and $I$ are prone to experimental uncertainties or instrumental errors ($\Delta V$ and $\Delta I$). When calculating a quantity derived via multiplication or division, the absolute errors do not simply add up or divide; instead, the maximum relative (or fractional) error of the result is equal to the sum of the relative errors of the individual measured quantities.

Step 1: Applying the theory of error propagation to the quotient formula.
To mathematically determine how errors combine, we take the natural logarithm ($\ln$) on both sides of the equation $X = \frac{V}{I}$: \[ \ln(X) = \ln\left(\frac{V}{I}\right) = \ln(V) - \ln(I) \] Differentiating both sides to find the relationship between small fractional increments yields: \[ \frac{dX}{X} = \frac{dV}{V} - \frac{dI}{I} \] For estimating the maximum possible fractional error (worst-case uncertainty), the negative signs are replaced with positive signs because errors can accumulate destructively in the same direction: \[ \frac{\Delta X}{X} = \frac{\Delta V}{V} + \frac{\Delta I}{I} \]

Step 2: Evaluating the total error composition.
The expression clearly shows that the total relative error in finding $X$ is bounded by the combined contributions of the fractional error from the voltmeter measurement ($\frac{\Delta V}{V}$) and the fractional error from the ammeter measurement ($\frac{\Delta I}{I}$). Thus, any variation or error observed in $X$ from different sets of readings is due to both devices, and the mathematical framework for maximum limits dictates that it is equal to the combined sum of these fractional errors. Hence, option (C) is the most accurate choice among the given qualitative options.

Answer 3

Concept: A rheostat acts as a variable resistor in a circuit. The resistance of a uniform conductor is directly proportional to its effective length connected in the path of the current: \[ R_{\text{rheo}} = \rho \frac{l}{A} \] The total equivalent resistance of a series circuit consisting of a fixed resistance $X$ and a variable rheostat resistance $R_{\text{rheo}}$ connected to a source of constant electromotive force ($V_{\text{total}}$) is given by: \[ R_{\text{total}} = X + R_{\text{rheo}} \] According to Ohm's Law for the entire circuit, the total circuit current ($I$), which is measured directly by the ammeter, is: \[ I = \frac{V_{\text{total}}}{R_{\text{total}}} = \frac{V_{\text{total}}}{X + R_{\text{rheo}}} \]

Step 1: Analyzing the variation in rheostat resistance when moving towards terminal P.
When the sliding contact (movable end) of the rheostat is shifted towards the terminal labeled P, the length ($l$) of the resistance wire actively included within the closed circuit is reduced. Since resistance is linearly proportional to the active length of the wire: \[ l \downarrow \quad \implies \quad R_{\text{rheo}} \downarrow \] Therefore, moving the slider towards P decreases the electrical resistance contribution of the rheostat.

Step 2: Determining the effect on ammeter and voltmeter readings.
As the rheostat resistance $R_{\text{rheo}}$ drops, the overall net resistance $R_{\text{total}}$ of the series network diminishes: \[ R_{\text{total}} = X + R_{\text{rheo}} \downarrow \] Since the supply voltage $V_{\text{total}}$ remains fixed, a lower overall resistance causes the total current $I$ flowing through the circuit to rise: \[ I = \frac{V_{\text{total}}}{R_{\text{total}}} \uparrow \] Thus, the ammeter reading increases. The voltmeter is connected in parallel across the unknown fixed resistance $X$ and measures the localized potential drop $V_X$ across it. By Ohm's Law, this localized potential drop is: \[ V_X = I \cdot X \] Since the resistance value $X$ is constant and the passing current $I$ has increased, the product $I \cdot X$ must increase: \[ V_X \uparrow \] Consequently, the voltmeter reading also increases. Therefore, both instrument readings increase simultaneously, corresponding to option (B).

Answer 4

Concept: When multiple segments of a conductor are joined end-to-end (in a series arrangement), the principle of conservation of charge requires that the total electrical current ($I$) passing through each cross-section per unit time remains strictly identical. Therefore, the same current $I$ flows through segment AC, segment CD, and segment DB: \[ I_{\text{AC}} = I_{\text{CD}} = I_{\text{DB}} = I \] The relationship between macroscopic electric current $I$ and microscopic electron drift velocity $v_d$ inside a conductor is given by the formula: \[ I = n \cdot e \cdot A \cdot v_d \] where:
• $n$ is the number density of free conduction electrons (dependent only on the type of metal).
• $e$ is the elementary charge of a single electron.
• $A$ is the cross-sectional area of the specific segment ($A = \pi \cdot r_{\text{wire}}^2$).
• $v_d$ is the average drift velocity of electrons.

Step 1: Expressing drift velocity in terms of the variable radius.
Since all three segments (AC, CD, and DB) are made from the exact same metal, the electron concentration $n$ is identical across all parts. Isolating the drift velocity $v_d$ from our fundamental equation gives: \[ v_d = \frac{I}{n \cdot e \cdot A} = \frac{I}{n \cdot e \cdot (\pi \cdot r_{\text{wire}}^2)} \] Given that $I$, $n$, $e$, and $\pi$ are completely constant throughout this specific setting, we can see that the drift velocity is inversely proportional to the square of the segment's radius: \[ v_d \propto \frac{1}{r_{\text{wire}}^2} \]

Step 2: Comparing the drift velocities for the three sections.
Let us write down the explicit radii given for each of the three sequential parts:
• For part AC: \( r_1 = r \)
• For part CD: \( r_2 = \frac{r}{3} \)
• For part DB: \( r_3 = \frac{r}{2} \) Comparing the dimensions, we observe the following strict order for the radii: \[ r > \frac{r}{2} > \frac{r}{3} \quad \implies \quad r_1 > r_3 > r_2 \] Squaring these terms maintains the same order inequality: \[ r_1^2 > r_3^2 > r_2^2 \] Since the drift velocity is inversely proportional to the square of the radius ($v_d \propto \frac{1}{r_{\text{wire}}^2}$), reversing the inequality gives: \[ \frac{1}{r_2^2} > \frac{1}{r_3^2} > \frac{1}{r_1^2} \quad \implies \quad v_2 > v_3 > v_1 \] Hence, the drift velocity is highest where the wire is narrowest (part CD) and lowest where the wire is widest (part AC). This matches option (B).

Answer 5

Concept: The localized electric field $E$ inside a current-carrying cylindrical conductor of uniform cross-section and length $L$ is related to the potential difference $V$ across its boundaries by the electrostatic relationship: \[ E = \frac{V}{L} \] According to Ohm's law, the potential difference across a specific segment is determined by the constant series current $I$ and the segment's individual resistance $R$: \[ V = I \cdot R \] The resistance of a uniform conductor is defined by its dimensions and material properties as: \[ R = \rho \frac{L}{A} \] where $\rho$ is the electrical resistivity of the metal and $A$ is the cross-sectional area ($A = \pi \cdot r_{\text{wire}}^2$).

Step 1: Deriving the electric field equation as a function of radius.
Let us substitute the expression for resistance $R$ directly into Ohm's Law to find the voltage drop: \[ V = I \cdot \left(\rho \frac{L}{A}\right) \] Now, substituting this expression for $V$ back into our electric field relationship yields: \[ E = \frac{V}{L} = \frac{I \cdot \rho \cdot L}{A \cdot L} = \frac{I \cdot \rho}{A} \] Replacing the cross-sectional area $A$ with $\pi \cdot r_{\text{wire}}^2$, we obtain: \[ E = \frac{I \cdot \rho}{\pi \cdot r_{\text{wire}}^2} \] Because the wire segments are coupled in a series configuration, the current $I$ is identical in all parts. Furthermore, since they are composed of the exact same metal, the resistivity $\rho$ is also identical. This shows that the internal electric field is inversely proportional to the square of the radius: \[ E \propto \frac{1}{r_{\text{wire}}^2} \]

Step 2: Comparing the electric fields for the three regions.
The given dimensions for the radii of sections AC, CD, and DB are: \[ r_1 = r, \quad r_2 = \frac{r}{3}, \quad r_3 = \frac{r}{2} \] Arranging these values in descending order gives: \[ r_1 > r_3 > r_2 \] Squaring the radii yields: \[ r^2 > \frac{r^2}{4} > \frac{r^2}{9} \quad \implies \quad r_1^2 > r_3^2 > r_2^2 \] Since the electric field magnitude shares an inverse square relationship with the radius ($E \propto \frac{1}{r_{\text{wire}}^2}$), taking the reciprocals reverses the inequalities: \[ \frac{1}{r_2^2} > \frac{1}{r_3^2} > \frac{1}{r_1^2} \quad \implies \quad E_2 > E_3 > E_1 \] Thus, the electric field is strongest in the thinnest region (CD) and weakest in the thickest region (AC). This corresponds to option (C).