Potentiometer — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Potentiometer MCQs with step-by-step solutions (20 questions). Part of Current Electricity. Practise online on Prepizo — no login needed.
▶ Practise Potentiometer online (free)
Questions with solutions
Q1 — Potentiometer · easy · theory
A potentiometer works on the principle that, for a uniform wire carrying a steady current:
A. The resistance is independent of length
B. The EMF depends on the cell used to balance it
C. The current is proportional to the length
D. The potential difference across a length is proportional to that length ✓ Correct
Solution: A uniform wire has a constant potential gradient, so $V \propto l$ along it.
Q2 — Potentiometer · easy · theory
The potential gradient along a potentiometer wire is defined as:
A. The potential difference per unit length of the wire ✓ Correct
B. The current per unit length
C. The EMF of the driving cell
D. The resistance per unit length
Solution: $K = \dfrac{V}{L}$, measured in volt per metre.
Q3 — Potentiometer · medium · theory
A potentiometer measures EMF more accurately than a voltmeter because at balance it:
A. Has a very low resistance
B. Requires no driving cell
C. Draws no current from the cell being measured ✓ Correct
D. Uses a more sensitive galvanometer
Solution: With zero current there is no drop across the internal resistance, so the true EMF is measured rather than the terminal voltage.
Q4 — Potentiometer · medium · theory
The sensitivity of a potentiometer is increased by:
A. Increasing the current through the wire
B. Using a longer wire, which lowers the potential gradient ✓ Correct
C. Using a shorter wire
D. Using a driving cell of lower EMF
Solution: A smaller potential gradient spreads a given potential difference over a greater length, so the balance point can be located more precisely.
Q5 — Potentiometer · easy · theory
A potentiometer can be used to:
A. Measure current directly
B. Measure capacitance
C. Compare the EMFs of two cells and find the internal resistance of a cell ✓ Correct
D. Measure magnetic flux
Solution: Both measurements reduce to comparing balancing lengths, which is why the instrument is so versatile.
Q6 — Potentiometer · medium · theory
For a potentiometer to give a balance point, the EMF of the driving cell must be:
A. Greater than the EMF being measured ✓ Correct
B. Zero
C. Smaller than the EMF being measured
D. Exactly equal to the EMF being measured
Solution: Otherwise the potential drop available along the whole wire is less than the unknown EMF and no null point exists.
Q7 — Potentiometer · easy · theory
In a potentiometer experiment, the null point is indicated when the galvanometer shows:
A. A steady small deflection
B. An oscillating deflection
C. No deflection ✓ Correct
D. Maximum deflection
Solution: At balance the potential difference across the tapped length exactly opposes the unknown EMF, so no current flows.
Q8 — Potentiometer · medium · theory
The chief advantage of a potentiometer over a voltmeter is that a potentiometer:
A. Acts as an ideal instrument of infinite resistance at balance ✓ Correct
B. Needs no galvanometer
C. Is cheaper to construct
D. Gives a direct reading without calculation
Solution: Because it draws no current at balance, it behaves exactly like a voltmeter of infinite resistance, so it does not load the circuit.
Q9 — Potentiometer · hard · numerical
A potentiometer wire of length $10\text{ m}$ has resistance $20\,\Omega$. It is connected in series with a $5\text{ V}$ cell of internal resistance $5\,\Omega$ and an external resistance of $15\,\Omega$. The potential gradient is:
A. $0.20\text{ V/m}$
B. $0.25\text{ V/m}$ ✓ Correct
C. $0.50\text{ V/m}$
D. $0.125\text{ V/m}$
Solution: Total resistance $= 20 + 5 + 15 = 40\,\Omega$, so $I = \dfrac{5}{40} = 0.125\text{ A}$. The drop across the wire is $0.125 \times 20 = 2.5\text{ V}$, giving $K = \dfrac{2.5}{10} = 0.25\text{ V/m}$.
Q10 — Potentiometer · hard · numerical
A potentiometer wire of length $4\text{ m}$ has resistance $8\,\Omega$. With a $2\text{ V}$ cell and a series resistance of $32\,\Omega$, the potential gradient is:
A. $0.05\text{ V/m}$
B. $0.2\text{ V/m}$
C. $0.4\text{ V/m}$
D. $0.1\text{ V/m}$ ✓ Correct
Solution: $I = \dfrac{2}{8 + 32} = 0.05\text{ A}$, so the drop across the wire is $0.4\text{ V}$ and $K = \dfrac{0.4}{4} = 0.1\text{ V/m}$.
Q11 — Potentiometer · medium · numerical
In a potentiometer experiment a standard cell of $1.2\text{ V}$ balances at $60\text{ cm}$. Another cell balances at $90\text{ cm}$. Its EMF is:
A. $1.8\text{ V}$ ✓ Correct
B. $0.8\text{ V}$
C. $1.5\text{ V}$
D. $2.0\text{ V}$
Solution: $\dfrac{E_2}{E_1} = \dfrac{l_2}{l_1} \Rightarrow E_2 = 1.2 \times \dfrac{90}{60} = 1.8\text{ V}$.
Q12 — Potentiometer · easy · numerical
The potential gradient along a potentiometer wire is $0.2\text{ V/m}$. A cell balances at $2.5\text{ m}$. Its EMF is:
A. $12.5\text{ V}$
B. $0.08\text{ V}$
C. $0.5\text{ V}$ ✓ Correct
D. $5\text{ V}$
Solution: $E = Kl = 0.2 \times 2.5 = 0.5\text{ V}$.
Q13 — Potentiometer · easy · numerical
Two cells balance at $40\text{ cm}$ and $60\text{ cm}$ on a potentiometer. The ratio of their EMFs is:
A. $2 : 3$ ✓ Correct
B. $1 : 1$
C. $4 : 9$
D. $3 : 2$
Solution: EMF is proportional to balancing length, so $\dfrac{E_1}{E_2} = \dfrac{40}{60} = \dfrac{2}{3}$.
Q14 — Potentiometer · hard · numerical
A cell balances at $60\text{ cm}$ on open circuit and at $50\text{ cm}$ when shunted by $5\,\Omega$. Its internal resistance is:
A. $2\,\Omega$
B. $0.5\,\Omega$
C. $1\,\Omega$ ✓ Correct
D. $5\,\Omega$
Solution: $r = R\left(\dfrac{l_1 - l_2}{l_2}\right) = 5 \times \dfrac{60 - 50}{50} = 1\,\Omega$.
Q15 — Potentiometer · easy · numerical
A potential difference of $2\text{ V}$ is maintained across a $5\text{ m}$ potentiometer wire. The potential gradient is:
A. $2.5\text{ V/m}$
B. $0.4\text{ V/m}$ ✓ Correct
C. $0.2\text{ V/m}$
D. $10\text{ V/m}$
Solution: $K = \dfrac{V}{L} = \dfrac{2}{5} = 0.4\text{ V/m}$.
Q16 — Potentiometer · medium · numerical
The potential gradient on a potentiometer is $0.3\text{ V/m}$. The balancing length for a cell of EMF $1.5\text{ V}$ is:
A. $2\text{ m}$
B. $5\text{ m}$ ✓ Correct
C. $0.2\text{ m}$
D. $0.45\text{ m}$
Solution: $l = \dfrac{E}{K} = \dfrac{1.5}{0.3} = 5\text{ m}$.
Q17 — Potentiometer · easy · numerical
A potentiometer has a potential gradient of $0.2\text{ V/m}$. A cell of EMF $1\text{ V}$ balances at a length of:
A. $5\text{ m}$ ✓ Correct
B. $10\text{ m}$
C. $0.2\text{ m}$
D. $2\text{ m}$
Solution: $l = \dfrac{1}{0.2} = 5\text{ m}$.
Q18 — Potentiometer · hard · numerical
If the length of a potentiometer wire is doubled while the current through it is unchanged, the potential gradient:
A. Remains unchanged ✓ Correct
B. Becomes one-fourth
C. Halves
D. Doubles
Solution: Doubling the length also doubles the wire resistance and hence the total drop, so the drop per unit length is unaltered.
Q19 — Potentiometer · medium · numerical
A potentiometer wire of length $6\text{ m}$ has a potential gradient of $0.5\text{ V/m}$. The largest EMF it can measure is:
A. $0.5\text{ V}$
B. $3\text{ V}$ ✓ Correct
C. $12\text{ V}$
D. $6\text{ V}$
Solution: The maximum measurable EMF corresponds to the full length: $E = 0.5 \times 6 = 3\text{ V}$.
Q20 — Potentiometer · medium · numerical
A cell of EMF $1\text{ V}$ balances at $50\text{ cm}$ on a potentiometer. The potential gradient is:
A. $1\text{ V/m}$
B. $2\text{ V/m}$ ✓ Correct
C. $0.02\text{ V/m}$
D. $0.5\text{ V/m}$
Solution: $K = \dfrac{E}{l} = \dfrac{1}{0.5} = 2\text{ V/m}$.