301. In a potentiometer experiment to find internal resistance of a cell, the balance length with the cell open is \(l_1\). When an external resistance \(R\) is connected across the cell, the balance length becomes \(l_2\). Which expression gives the internal resistance \(r\)?
ⓐ. \(r=R\left(\frac{l_1-l_2}{l_2}\right)\)
ⓑ. \(r=R\left(\frac{l_2-l_1}{l_1}\right)\)
ⓒ. \(r=R\left(\frac{l_1}{l_2}\right)\)
ⓓ. \(r=R\left(\frac{l_2}{l_1-l_2}\right)\)
Correct Answer: \(r=R\left(\frac{l_1-l_2}{l_2}\right)\)
Explanation: With the cell open, no current is drawn from it, so the balance length \(l_1\) corresponds to its emf \(\mathcal{E}\). When a resistance \(R\) is connected across the cell, the cell supplies current and the balance length \(l_2\) corresponds to its terminal voltage \(V\). Since the same potentiometer potential gradient is used, \(\frac{\mathcal{E}}{V}=\frac{l_1}{l_2}\). For a cell with external resistance \(R\) and internal resistance \(r\), \(\frac{\mathcal{E}}{V}=\frac{R+r}{R}\). Equating gives \(\frac{l_1}{l_2}=\frac{R+r}{R}\), so \(r=R\left(\frac{l_1-l_2}{l_2}\right)\). A common mistake is to put \(l_1\) in the denominator instead of \(l_2\).
302. A cell gives a balance length of \(100\,\text{cm}\) when open. When a \(4\,\Omega\) resistance is connected across it, the balance length becomes \(80\,\text{cm}\). What is the internal resistance of the cell?
ⓐ. \(0.5\,\Omega\)
ⓑ. \(1.0\,\Omega\)
ⓒ. \(2.0\,\Omega\)
ⓓ. \(5.0\,\Omega\)
Correct Answer: \(1.0\,\Omega\)
Explanation: \( \textbf{Given:} \) \(l_1=100\,\text{cm}\), \(l_2=80\,\text{cm}\), and external resistance \(R=4\,\Omega\).
\( \textbf{Required:} \) Internal resistance \(r\).
\( \textbf{Potentiometer formula:} \) \(r=R\left(\frac{l_1-l_2}{l_2}\right)\).
\( \textbf{Difference of lengths:} \) \(l_1-l_2=100-80=20\,\text{cm}\).
\( \textbf{Substitution:} \) \(r=4\left(\frac{20}{80}\right)\,\Omega\).
\( \textbf{Simplification:} \) \(\frac{20}{80}=\frac{1}{4}\).
\( \textbf{Internal resistance:} \) \(r=4\times\frac{1}{4}=1.0\,\Omega\).
\( \textbf{Final answer:} \) \(r=1.0\,\Omega\); normally \(l_1>l_2\) because terminal voltage under load is less than emf.
303. In the potentiometer method for internal resistance, why is the loaded balance length \(l_2\) usually less than the open-circuit balance length \(l_1\)?
ⓐ. Connecting the external resistance increases the emf of the cell
ⓑ. The potential gradient becomes zero under load
ⓒ. The galvanometer current becomes maximum at balance
ⓓ. The load terminal voltage is less than emf
Correct Answer: The load terminal voltage is less than emf
Explanation: The open-circuit balance length \(l_1\) corresponds to the emf \(\mathcal{E}\) of the cell. When an external resistance is connected, the cell supplies current and some potential difference \(Ir\) is lost inside the cell. The terminal voltage becomes \(V=\mathcal{E}-Ir\), which is less than \(\mathcal{E}\). Since balance length is proportional to the measured potential difference, the loaded balance length \(l_2\) is smaller than \(l_1\). The decrease in balance length is a signature of internal resistance, not a change in emf.
304. In a potentiometer internal-resistance experiment, the open-circuit balance length is \(120\,\text{cm}\) and the loaded balance length is \(90\,\text{cm}\). What is the ratio of terminal voltage under load to emf?
ⓐ. \(\frac{1}{4}\)
ⓑ. \(\frac{3}{4}\)
ⓒ. \(\frac{4}{3}\)
ⓓ. \(\frac{1}{3}\)
Correct Answer: \(\frac{3}{4}\)
Explanation: \( \textbf{Known idea:} \) In a potentiometer, measured potential difference is proportional to balance length when the potential gradient is unchanged.
\( \textbf{Open-circuit condition:} \) The balance length \(l_1=120\,\text{cm}\) corresponds to emf \(\mathcal{E}\).
\( \textbf{Loaded condition:} \) The balance length \(l_2=90\,\text{cm}\) corresponds to terminal voltage \(V\).
\( \textbf{Ratio relation:} \) \(\frac{V}{\mathcal{E}}=\frac{l_2}{l_1}\).
\( \textbf{Substitution:} \) \(\frac{V}{\mathcal{E}}=\frac{90}{120}\).
\( \textbf{Simplification:} \) \(\frac{90}{120}=\frac{3}{4}\).
\( \textbf{Final answer:} \) \(\frac{V}{\mathcal{E}}=\frac{3}{4}\); the loaded length belongs in the numerator for terminal-voltage-to-emf ratio.
305. A cell is tested using a potentiometer. With no external load, its balance length is \(150\,\text{cm}\). With a \(10\,\Omega\) external resistor connected across it, the balance length is \(125\,\text{cm}\). Find the internal resistance.
ⓐ. \(1\,\Omega\)
ⓑ. \(4\,\Omega\)
ⓒ. \(2\,\Omega\)
ⓓ. \(5\,\Omega\)
Correct Answer: \(2\,\Omega\)
Explanation: \( \textbf{Given:} \) \(l_1=150\,\text{cm}\), \(l_2=125\,\text{cm}\), and \(R=10\,\Omega\).
\( \textbf{Required:} \) Internal resistance \(r\).
\( \textbf{Formula:} \) \(r=R\left(\frac{l_1-l_2}{l_2}\right)\).
\( \textbf{Length difference:} \) \(l_1-l_2=150-125=25\,\text{cm}\).
\( \textbf{Substitution:} \) \(r=10\left(\frac{25}{125}\right)\,\Omega\).
\( \textbf{Fraction simplification:} \) \(\frac{25}{125}=\frac{1}{5}\).
\( \textbf{Internal resistance:} \) \(r=10\times\frac{1}{5}=2\,\Omega\).
\( \textbf{Final answer:} \) \(r=2\,\Omega\); using \(l_1\) instead of \(l_2\) in the denominator would underestimate the internal resistance.
306. A cell has emf \(2.0\,\text{V}\) and internal resistance \(0.5\,\Omega\). It is connected to an external resistance \(3.5\,\Omega\). In a potentiometer experiment with potential gradient \(0.01\,\text{V cm}^{-1}\), what balance length corresponds to the terminal voltage under load?
ⓐ. \(150\,\text{cm}\)
ⓑ. \(175\,\text{cm}\)
ⓒ. \(200\,\text{cm}\)
ⓓ. \(225\,\text{cm}\)
Correct Answer: \(175\,\text{cm}\)
Explanation: \( \textbf{Given:} \) \(\mathcal{E}=2.0\,\text{V}\), \(r=0.5\,\Omega\), \(R=3.5\,\Omega\), and \(k=0.01\,\text{V cm}^{-1}\).
\( \textbf{Current through load:} \) \(I=\frac{\mathcal{E}}{R+r}\).
\( \textbf{Substitution:} \) \(I=\frac{2.0}{3.5+0.5}=\frac{2.0}{4.0}=0.50\,\text{A}\).
\( \textbf{Terminal voltage:} \) \(V=IR=(0.50)(3.5)=1.75\,\text{V}\).
\( \textbf{Potentiometer balance relation:} \) \(V=kl\).
\( \textbf{Balance length:} \) \(l=\frac{V}{k}=\frac{1.75}{0.01}=175\,\text{cm}\).
\( \textbf{Final answer:} \) The loaded terminal voltage balances at \(175\,\text{cm}\); using emf instead would give the open-circuit balance length.
307. Use the graph description below. The vertical axis represents balance length \(l\), and the horizontal axis represents emf \(\mathcal{E}\) for cells measured on the same potentiometer wire. The graph is a straight line through the origin. What is the slope of this graph?
ⓐ. \(k\)
ⓑ. \(\frac{1}{k}\)
ⓒ. \(k^2\)
ⓓ. \(\frac{\mathcal{E}}{r}\)
Correct Answer: \(\frac{1}{k}\)
Explanation: \( \textbf{Potentiometer relation:} \) At balance, \(\mathcal{E}=kl\).
\( \textbf{Rearrange for plotted variables:} \) \(l=\frac{\mathcal{E}}{k}\).
\( \textbf{Graph axes:} \) The vertical axis is \(l\), and the horizontal axis is \(\mathcal{E}\).
\( \textbf{Straight-line form:} \) Comparing \(l=\left(\frac{1}{k}\right)\mathcal{E}\) with \(y=mx\), the slope is \(\frac{1}{k}\).
\( \textbf{Unit check:} \) If \(k\) is in \(\text{V cm}^{-1}\), then \(\frac{1}{k}\) is in \(\text{cm V}^{-1}\), matching \(\frac{l}{\mathcal{E}}\).
\( \textbf{Final answer:} \) The slope is \(\frac{1}{k}\); if the axes were reversed, the slope would be \(k\).
308. Which graph description corresponds to an ohmic conductor if the vertical axis is \(I\) and the horizontal axis is \(V\)?
ⓐ. Straight line through the origin with slope equal to conductance
ⓑ. Straight line through the origin with slope equal to resistance
ⓒ. Curve bending away from the origin with constant resistance
ⓓ. Horizontal line showing current independent of voltage
Correct Answer: Straight line through the origin with slope equal to conductance
Explanation: For an ohmic conductor, \(V=IR\), so \(I=\frac{V}{R}\). If \(I\) is plotted on the vertical axis and \(V\) on the horizontal axis, the graph is a straight line through the origin. Its slope is \(\frac{\Delta I}{\Delta V}=\frac{1}{R}\), which is conductance. If \(V\) were plotted against \(I\), the slope would be resistance. The graph-axis order is the main source of wrong answers in \(V\)-\(I\) and \(I\)-\(V\) questions.
309. A cell has terminal voltage \(V\) plotted on the vertical axis and current \(I\) on the horizontal axis. The graph is a straight line cutting the \(V\)-axis at \(3.0\,\text{V}\) and the \(I\)-axis at \(6.0\,\text{A}\). What are the emf and internal resistance?
ⓐ. \(\mathcal{E}=6.0\,\text{V}\), \(r=0.50\,\Omega\)
ⓑ. \(\mathcal{E}=3.0\,\text{V}\), \(r=2.0\,\Omega\)
ⓒ. \(\mathcal{E}=6.0\,\text{V}\), \(r=2.0\,\Omega\)
ⓓ. \(\mathcal{E}=3.0\,\text{V}\), \(r=0.50\,\Omega\)
Correct Answer: \(\mathcal{E}=3.0\,\text{V}\), \(r=0.50\,\Omega\)
Explanation: \( \textbf{Cell graph relation:} \) For a discharging cell, \(V=\mathcal{E}-Ir\).
\( \textbf{Vertical intercept:} \) At \(I=0\), \(V=\mathcal{E}\), so the \(V\)-axis intercept gives emf.
\( \textbf{Emf:} \) The intercept is \(3.0\,\text{V}\), so \(\mathcal{E}=3.0\,\text{V}\).
\( \textbf{Current-axis intercept:} \) At \(V=0\), \(0=\mathcal{E}-Ir\), so \(I=\frac{\mathcal{E}}{r}\).
\( \textbf{Given intercept:} \) The \(I\)-axis intercept is \(6.0\,\text{A}\).
\( \textbf{Internal resistance:} \) \(r=\frac{\mathcal{E}}{I}=\frac{3.0}{6.0}=0.50\,\Omega\).
\( \textbf{Final answer:} \) \(\mathcal{E}=3.0\,\text{V}\) and \(r=0.50\,\Omega\); the slope is negative, but internal resistance is positive.
310. The power dissipated in a resistor is plotted against current. If the resistor has constant resistance \(R\), what is the shape of the \(P\)-versus-\(I\) graph?
ⓐ. Straight line through the origin
ⓑ. Horizontal straight line above the current axis
ⓒ. Rectangular hyperbola in the first quadrant
ⓓ. An upward-opening parabola through the origin
Correct Answer: An upward-opening parabola through the origin
Explanation: For a resistor of constant resistance, power is \(P=I^2R\). Since \(R\) is constant, \(P\propto I^2\). A graph of \(P\) against \(I\) is therefore a parabola passing through the origin. Doubling the current makes the power four times, not two times. A straight-line graph would correspond to direct proportionality, which is not the case here. A common mistake is to remember \(P=VI\) and assume power is always directly proportional to current, without checking whether \(V\) changes with \(I\).
311. Use the graph description below. The vertical axis represents resistance \(R\), and the horizontal axis represents temperature \(T\). For a metallic wire, the graph is approximately a straight line. If the slope is larger for material \(P\) than for material \(Q\), assuming equal initial resistance at \(T_0\), what can be inferred?
ⓐ. Material \(P\) has a smaller temperature coefficient of resistance
ⓑ. Material \(P\) has a larger temperature coefficient of resistance
ⓒ. Material \(P\) must be a semiconductor
ⓓ. Material \(Q\) has zero resistance at all temperatures
Correct Answer: Material \(P\) has a larger temperature coefficient of resistance
Explanation: For a metal over a moderate range, resistance varies approximately as \(R_T=R_0[1+\alpha(T-T_0)]\). If \(R_0\) is the same for two materials, the slope of the \(R\)-versus-\(T\) graph is proportional to \(R_0\alpha\). A larger slope therefore means a larger value of \(\alpha\). This indicates that the resistance of material \(P\) changes more rapidly with temperature than that of material \(Q\). A semiconductor would usually show a negative slope over ordinary conditions, not a larger positive metallic slope. The graph slope must be interpreted together with the material behaviour and the initial resistance condition.
312. A resistor of constant resistance \(R\) is connected across different potential differences. Which graph description is correct for power \(P\) plotted against voltage \(V\)?
ⓐ. A straight line through the origin because \(P\propto V\)
ⓑ. A horizontal line because \(P\) is independent of \(V\)
ⓒ. A rectangular hyperbola because \(P\propto \frac{1}{V}\)
ⓓ. A parabola through the origin because \(P\propto V^2\)
Correct Answer: A parabola through the origin because \(P\propto V^2\)
Explanation: For a resistor of constant resistance, power can be written as \(P=\frac{V^2}{R}\). Since \(R\) is constant, \(P\) is proportional to \(V^2\). Therefore the graph of \(P\) against \(V\) is a parabola passing through the origin. Doubling the voltage makes the power four times, not two times. A straight-line graph would be expected only for direct proportionality. The formula must be chosen according to the quantity varied and the quantity kept constant.
313. The terminal voltage \(V\) of a cell is plotted against current \(I\). The straight line has equation \(V=2.0-0.25I\), where \(V\) is in \(\text{V}\) and \(I\) is in \(\text{A}\). What are the emf and internal resistance?
ⓐ. \(\mathcal{E}=2.0\,\text{V}\), \(r=0.25\,\Omega\)
ⓑ. \(\mathcal{E}=0.25\,\text{V}\), \(r=2.0\,\Omega\)
ⓒ. \(\mathcal{E}=2.25\,\text{V}\), \(r=0.25\,\Omega\)
ⓓ. \(\mathcal{E}=2.0\,\text{V}\), \(r=4.0\,\Omega\)
Correct Answer: \(\mathcal{E}=2.0\,\text{V}\), \(r=0.25\,\Omega\)
Explanation: \( \textbf{Cell graph relation:} \) For a discharging cell, \(V=\mathcal{E}-Ir\).
\( \textbf{Given equation:} \) \(V=2.0-0.25I\).
\( \textbf{Compare constants:} \) The constant term \(2.0\) corresponds to the emf \(\mathcal{E}\).
\( \textbf{Emf:} \) \(\mathcal{E}=2.0\,\text{V}\).
\( \textbf{Compare coefficient of } I\textbf{:} \) The coefficient of \(I\) is \(-r\).
\( \textbf{Internal resistance:} \) Since \(-r=-0.25\), \(r=0.25\,\Omega\).
\( \textbf{Unit check:} \) The slope of a \(V\)-versus-\(I\) graph has unit \(\text{V A}^{-1}=\Omega\).
\( \textbf{Final answer:} \) \(\mathcal{E}=2.0\,\text{V}\) and \(r=0.25\,\Omega\); the negative sign indicates falling terminal voltage, not negative internal resistance.
314. A meter bridge experiment is repeated with different values of \(\frac{R}{S}\). Which graph description is expected if the vertical axis represents \(\frac{l}{100-l}\) and the horizontal axis represents \(\frac{R}{S}\)?
ⓐ. A straight line through the origin with slope \(1\)
ⓑ. A straight line through the origin with slope \(100\)
ⓒ. A parabola through the origin
ⓓ. A horizontal line
Correct Answer: A straight line through the origin with slope \(1\)
Explanation: At balance in a meter bridge, \(\frac{R}{S}=\frac{l}{100-l}\). If \(\frac{l}{100-l}\) is plotted on the vertical axis and \(\frac{R}{S}\) on the horizontal axis, the two plotted quantities are equal. This gives a straight line through the origin with slope \(1\). The factor \(100\) is already included in the expression \(100-l\); it does not become the graph slope. The graph checks the proportionality between resistance ratio and length ratio. The relation is valid only if the bridge wire is uniform.
315. In an \(I\)-\(V\) graph of an ohmic conductor, the slope is \(0.50\,\text{S}\). If another ohmic conductor has twice the resistance, what is the slope of its \(I\)-\(V\) graph?
ⓐ. \(0.50\,\text{S}\)
ⓑ. \(1.00\,\text{S}\)
ⓒ. \(2.00\,\text{S}\)
ⓓ. \(0.25\,\text{S}\)
Correct Answer: \(0.25\,\text{S}\)
Explanation: \( \textbf{Graph meaning:} \) In an \(I\)-versus-\(V\) graph, the slope is conductance \(G=\frac{1}{R}\).
\( \textbf{Initial slope:} \) The first conductor has \(G_1=0.50\,\text{S}\).
\( \textbf{Resistance relation:} \) \(G=\frac{1}{R}\), so conductance is inversely proportional to resistance.
\( \textbf{New resistance:} \) The second conductor has \(R_2=2R_1\).
\( \textbf{New conductance:} \) \(G_2=\frac{1}{2R_1}=\frac{G_1}{2}\).
\( \textbf{Calculation:} \) \(G_2=\frac{0.50}{2}=0.25\,\text{S}\).
\( \textbf{Final answer:} \) The new slope is \(0.25\,\text{S}\); doubling resistance halves the slope of an \(I\)-\(V\) graph.
316. In a potentiometer, the balance length \(l\) is plotted against terminal voltage \(V\) for several cells using the same potential gradient. If the slope of the \(l\)-versus-\(V\) graph is \(80\,\text{cm V}^{-1}\), what is the potential gradient?
ⓐ. \(\frac{1}{80}\,\text{V cm}^{-1}\)
ⓑ. \(80\,\text{V cm}^{-1}\)
ⓒ. \(\frac{1}{80}\,\text{cm V}^{-1}\)
ⓓ. \(80\,\text{cm V}^{-1}\)
Correct Answer: \(\frac{1}{80}\,\text{V cm}^{-1}\)
Explanation: \( \textbf{Potentiometer relation:} \) At balance, \(V=kl\).
\( \textbf{Rearrange for graph:} \) \(l=\frac{V}{k}\).
\( \textbf{Slope of } l\textbf{-versus-}V\textbf{:} \) The slope is \(\frac{l}{V}=\frac{1}{k}\).
\( \textbf{Given slope:} \) \(\frac{1}{k}=80\,\text{cm V}^{-1}\).
\( \textbf{Find } k\textbf{:} \) \(k=\frac{1}{80}\,\text{V cm}^{-1}\).
\( \textbf{Calculation:} \) \(k=0.0125\,\text{V cm}^{-1}\).
\( \textbf{Final answer:} \) The potential gradient is \(0.0125\,\text{V cm}^{-1}\); the slope is reciprocal of \(k\) because length is on the vertical axis.
317. A resistor of \(10\,\Omega\) is connected across a variable voltage source. The voltage is increased from \(4\,\text{V}\) to \(8\,\text{V}\). By what factor does the power change?
ⓐ. \(2\)
ⓑ. \(4\)
ⓒ. \(\frac{1}{2}\)
ⓓ. \(\frac{1}{4}\)
Correct Answer: \(4\)
Explanation: \( \textbf{Condition:} \) The resistance remains constant at \(10\,\Omega\).
\( \textbf{Power formula for fixed resistance:} \) \(P=\frac{V^2}{R}\).
\( \textbf{Voltage change:} \) Voltage changes from \(4\,\text{V}\) to \(8\,\text{V}\), so it becomes twice.
\( \textbf{Power proportionality:} \) Since \(P\propto V^2\), power changes by the square of the voltage factor.
\( \textbf{Factor:} \) \(2^2=4\).
\( \textbf{Final answer:} \) The power becomes \(4\) times; power is not directly proportional to \(V\) for a fixed resistor.
318. An ammeter should be connected in ________ with the circuit element whose current is to be measured, and an ideal ammeter has ________ resistance.
ⓐ. series, zero
ⓑ. series, infinite
ⓒ. parallel, zero
ⓓ. parallel, infinite
Correct Answer: series, zero
Explanation: An ammeter measures the current through a circuit element. To measure that same current, it must be placed in series with the element. An ideal ammeter has zero resistance so that it does not change the current it is meant to measure. If an ammeter had large resistance, it would reduce the circuit current and disturb the measurement. It should not be connected in parallel across a source or component because its low resistance could create a large current path. The connection rule comes from the fact that series elements carry the same current.
319. A voltmeter should be connected in ________ across the circuit element whose potential difference is to be measured, and an ideal voltmeter has ________ resistance.
ⓐ. series, negligible resistance
ⓑ. series, infinite resistance
ⓒ. parallel, negligible resistance
ⓓ. parallel, infinite resistance
Correct Answer: parallel, infinite resistance
Explanation: A voltmeter measures potential difference between two points. Therefore it must be connected in parallel across the circuit element. An ideal voltmeter has infinite resistance so that it draws no current from the circuit. If a voltmeter drew significant current, it would change the potential difference it is trying to measure. This unwanted disturbance is called loading effect. The connection rule comes from the fact that parallel elements share the same potential difference.
320. A student connects an ideal ammeter directly across an ideal battery. What is the main danger in this connection?
ⓐ. No current flows because the ammeter has infinite resistance
ⓑ. It acts like a short circuit, so a very large current may flow
ⓒ. The battery emf becomes permanently zero during measurement
ⓓ. The ammeter measures potential difference accurately in parallel
Correct Answer: It acts like a short circuit, so a very large current may flow
Explanation: An ideal ammeter has zero resistance. If it is connected directly across a battery, it provides a nearly zero-resistance path between the battery terminals. This is a short circuit. A very large current may flow, limited only by internal resistance and practical circuit resistance. Such a connection can damage the ammeter, battery, or wires. The ammeter must be placed in series with a load, not directly in parallel across a source.
