A student sets up the circuit as shown in the figure to find the value of unknown resistance X and records a set of readings of the voltmeter and the ammeter by using the rheostat.
Question: 1
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 :
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Alloys like Manganin and Constantan are heavily used in standard resistance coils specifically because their resistance does not drift with Joule heating.
This guarantees strict adherence to Ohm's Law over a wide range of operational currents.
Concept: • Manganin is a specialized alloy consisting primarily of copper, manganese, and nickel.
• A defining characteristic of manganin is its incredibly low temperature coefficient of resistivity ($\alpha \approx 0$).
• This means its electrical resistance remains practically constant even when subjected to significant temperature changes.
• Ohm's law states that $V \propto I$, provided the physical conditions (most notably temperature and resistance) remain constant.
Step 1: Analyze the physical scenario
When taking readings without ever switching off the circuit, a continuous steady current heavily flows through the resistor X.
According to Joule's law of heating ($H = I^2 R t$), this continuous current inevitably causes the resistor wire to heat up and increase in temperature.
For standard metals like copper or iron, this rise in temperature would drastically increase their electrical resistance. Step 2: Apply the properties of Manganin
However, the resistor X is explicitly specified to be made of manganin.
Because manganin has a temperature coefficient of resistivity that is nearly negligible, the generated Joule heating does not affect its resistance value.
The resistance $R$ of the manganin wire will stay absolutely constant regardless of how long the current is kept flowing. Step 3: Determine the correct V-I graph
Since the resistance $R$ is strictly constant, the relationship between the potential difference $V$ and the current $I$ follows a perfect linear proportionality ($V = IR$).
In graphical terms, a direct proportionality is always represented by a perfectly straight line passing through the origin.
Graphs that curve upwards or downwards represent non-ohmic components (like filaments or semiconductors) where resistance changes with temperature. Step 4: Conclusion
The V-I characteristic will be a perfectly straight, linear line originating from zero, which is depicted in option (A).
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Question: 2
Error in the value of X obtained from different sets of voltmeter and ammeter readings, is :
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Always remember that whether quantities are multiplied ($Z = AB$) or divided ($Z = A/B$), their fractional errors always add up to give the maximum possible error in the result.
Never subtract or divide errors.
equal to the sum of error in voltmeter reading and error in ammeter reading.
equal to error in voltmeter reading divided by the error in ammeter reading.
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The Correct Option isC
Solution and Explanation
Concept: • In experimental physics, the calculation of an unknown resistance relies on Ohm's Law: $X = \frac{V}{I}$.
• Whenever a quantity is derived from the division or multiplication of two measured variables, the maximum fractional (or relative) error in the derived quantity is the sum of the fractional errors of the individual variables.
• Mathematically, for $Z = \frac{A}{B}$, the relative error is given by $\frac{\Delta Z}{Z} = \frac{\Delta A}{A} + \frac{\Delta B}{B}$.
Step 1: Identify the measurement formula
The student determines the unknown resistance X by measuring the voltage drop $V$ across it and the current $I$ passing through the main circuit.
The fundamental equation used for the calculation is $X = \frac{V}{I}$.
Both $V$ and $I$ are independently measured quantities subject to their own instrumental and observational errors, denoted as $\Delta V$ and $\Delta I$. Step 2: Apply error propagation rules
According to the principles of error propagation for quotients, the fractional errors simply add up.
The maximum possible fractional error in the calculated value of resistance X is formulated as:
\[ \frac{\Delta X}{X} = \frac{\Delta V}{V} + \frac{\Delta I}{I} \]
This equation clearly demonstrates that the total error in X is contributed to by both the voltmeter's error and the ammeter's error. Step 3: Interpret the given options
Option (A) and (B) wrongly isolate the error to only one instrument, ignoring the dual dependency of the calculation.
Option (D) incorrectly suggests dividing the errors, which violates mathematical error combination laws entirely.
Option (C) states the error is equal to the sum of the error in voltmeter reading and error in ammeter reading. In standard experimental contexts, this phrasing specifically implies the summation of their respective relative/fractional errors as shown in the derived formula. Step 4: Conclusion
The cumulative error in resistance $X$ is inherently the combined sum of the individual measurement errors from both meters.
Therefore, option (C) is the most scientifically accurate statement among the choices.
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Question: 3
If the movable end of rheostat is moved towards P, then :
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Always trace the path of the current from the positive terminal to the negative terminal to clearly identify which section of a rheostat is "active".
Reducing the active length of a rheostat always boosts the overall circuit current.
reading in ammeter decreases and reading in voltmeter increases.
readings in both voltmeter and ammeter increase.
reading in ammeter increases and reading in voltmeter decreases.
readings in both voltmeter and ammeter decrease.
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The Correct Option isB
Solution and Explanation
Concept: • A rheostat is a variable resistor used to actively control the flow of electric current by manually changing the resistance in the circuit.
• The total resistance of a simple series circuit dictates the total current supplied by the voltage source ($I = \frac{V_{total}}{R_{eq}}$).
• The voltage across a fixed component in a series circuit is directly proportional to the current flowing through it ($V = IR$).
Step 1: Analyze the circuit diagram
Carefully observing the provided circuit diagram reveals the connections: The positive terminal of the battery connects to the ammeter, which then connects to the unknown fixed resistance X.
The other end of resistance X connects to the movable slider (wiper) of the rheostat.
The fixed upper end of the rheostat, labeled P, is directly connected to the negative terminal of the battery.
This configuration means that the active resistance of the rheostat currently included in the circuit is simply the portion of the wire spanning from the fixed point P to the slider's current position. Step 2: Determine the effect of moving the slider
The question states that the movable end (slider) is physically pushed towards point P.
As the slider gets closer to P, the physical length of the rheostat wire included in the current path dramatically shortens.
Since resistance is directly proportional to length ($R = \rho \frac{l}{A}$), a shorter length means the rheostat introduces significantly less resistance into the circuit.
Consequently, the total equivalent resistance of the entire series circuit ($R_{eq} = X + R_{rheo}$) effectively decreases. Step 3: Determine the impact on meter readings
According to Ohm's law applied to the entire circuit, total current is $I = \frac{V_{battery}}{R_{eq}}$.
Because the total equivalent resistance $R_{eq}$ has decreased while the battery voltage remains constant, the overall current $I$ must increase.
Thus, the reading on the ammeter will visibly increase.
Next, consider the voltmeter, which is connected directly in parallel across the fixed resistor X.
The voltage drop across X is determined by $V_X = I \cdot X$.
Since the resistance $X$ is perfectly constant and the current $I$ has just increased, the voltage drop $V_X$ must identically increase.
Thus, the reading on the voltmeter also increases. Step 4: Conclusion
Both the ammeter and the voltmeter will show increased readings when the slider is moved towards P.
This precisely matches the statement in option (B).
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Question: 4
Suppose the unknown resistance X is replaced by a wire made of the same metal. This wire consists of three parts, of the same length L but has radii r, r/3 and r/2 as shown in the figure.
For a particular setting of the rheostat, let MATH_dea7480ed0df482d9f5cc3c8d0160263, MATH_21e9b49b4cb34232b03ff9434387e0c5 and MATH_281d234b316a44d7828748b2b7f13fed be the value of drift velocities in parts AC, CD and DB. Then :
Show Hint
Think of drift velocity exactly like the speed of water flowing through a pipe.
Where the pipe (wire) narrows, the water (electrons) must speed up dramatically to maintain the same continuous flow rate (current).
Concept: • The relationship between macroscopic electric current $I$ and microscopic drift velocity $v_d$ is governed by the relation $I = n e A v_d$.
• In a strict series combination, the exact same electric current $I$ flows continuously through every segment, regardless of changes in cross-sectional area.
• Therefore, drift velocity is universally inversely proportional to the cross-sectional area for a constant current ($v_d \propto \frac{1}{A}$).
Step 1: Establish the mathematical relationship
Since the three varied segments AC, CD, and DB are connected sequentially in series, the steady state current $I$ is identical in all three parts.
From the established formula $I = n e A v_d$, we can rearrange it to isolate drift velocity:
\[ v_d = \frac{I}{n e A} \]
Assuming the wire is completely uniform in material, the free electron density $n$ and elemental charge $e$ are strictly constant.
Since the cross-section is circular, area $A = \pi \cdot r_{segment}^2$.
This yields the powerful proportionality: $v_d \propto \frac{1}{r_{segment}^2}$. Step 2: Calculate relative drift velocities for each segment
Let's analyze segment AC:
Radius $r_{AC} = r$.
Drift velocity $v_1 \propto \frac{1}{r^2}$.
Let's analyze segment CD:
Radius $r_{CD} = \frac{r}{3}$.
Drift velocity $v_2 \propto \frac{1}{(r/3)^2} = \frac{9}{r^2}$.
This clearly indicates $v_2 = 9 \cdot v_1$.
Let's analyze segment DB:
Radius $r_{DB} = \frac{r}{2}$.
Drift velocity $v_3 \propto \frac{1}{(r/2)^2} = \frac{4}{r^2}$.
This clearly indicates $v_3 = 4 \cdot v_1$. Step 3: Compare and arrange the magnitudes
By comparing the calculated proportionality coefficients:
$v_1$ has a factor of 1.
$v_2$ has an enormous factor of 9.
$v_3$ has a factor of 4.
Arranging them purely in strictly descending order of magnitude gives:
$v_2 > v_3 > v_1$. Step 4: Conclusion
The highest drift velocity occurs in the thinnest section, and the lowest in the thickest section.
This logical ordering matches option (B).
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Question: 5
Consider the same wire, as shown in figure in question (iv) (a) connected in place of X. For a particular setting of rheostat, let $E_1$, $E_2$ and $E_3$ be the value of electric fields in part AC, CD and DB. Then :
Show Hint
Current density $J$, drift velocity $v_d$, and internal electric field $E$ all inherently share the exact same inverse proportionality to the cross-sectional area ($1/A$) in series configurations.
If you know the sequence for one, you automatically know the sequence for all three.
Concept: • According to the microscopic form of Ohm's Law, the electric field $E$ inside a conductor is proportional to the current density $J$.
• The explicit mathematical relation is $E = \rho J$, where $\rho$ is the electrical resistivity of the material.
• Current density $J$ is strictly defined as the total current $I$ divided by the cross-sectional area $A$ ($J = \frac{I}{A}$).
Step 1: Derive the electric field proportionality
Because the varied segments are arranged in series, the total electric current $I$ is universally constant across the entire wire structure.
Substitute the definition of current density into the microscopic Ohm's law:
\[ E = \rho \cdot \left(\frac{I}{A}\right) \]
Since the wire is made of one uniform metal, the resistivity $\rho$ is constant everywhere.
Because $I$ is also constant, the internal electric field strictly depends on the area:
\[ E \propto \frac{1}{A} \]
Substituting the formula for circular area $A = \pi \cdot r_{segment}^2$:
\[ E \propto \frac{1}{r_{segment}^2} \] Step 2: Evaluate the field in each distinct segment
For the first segment AC:
Radius is $r$.
So, $E_1 \propto \frac{1}{r^2}$.
For the middle segment CD:
Radius is extremely narrow at $\frac{r}{3}$.
So, $E_2 \propto \frac{1}{(r/3)^2} = \frac{9}{r^2}$.
For the final segment DB:
Radius is $\frac{r}{2}$.
So, $E_3 \propto \frac{1}{(r/2)^2} = \frac{4}{r^2}$. Step 3: Rank the electric fields
By comparing the numerical scaling factors computed above:
The field $E_1$ acts as the baseline (factor of 1).
The field $E_2$ is 9 times stronger than $E_1$.
The field $E_3$ is exactly 4 times stronger than $E_1$.
Placing these systematically in descending order gives:
$E_2 > E_3 > E_1$. Step 4: Conclusion
The electric field is most intense in the narrowest part of the wire to drive the same amount of current through the restricted bottleneck.
This fully aligns with option (C).
