Potentiometer - comparing emf and internal resistance — NEET & JEE Questions

Question 1 · difficulty L1 · recall

A potentiometer measures emf by a null method, so at balance it draws

• A. maximum current from the test cell
• B. the same current as the primary circuit
• C. no current from the test cell
• D. infinite current

Question 2 · difficulty L1 · understanding

For a uniform potentiometer wire carrying constant current, potential drop along the wire is

• A. proportional to length squared
• B. independent of length
• C. inversely proportional to length
• D. proportional to length

Question 3 · difficulty L2 · understanding

If two cell emfs E1 and E2 balance at lengths l1 and l2 under the same potential gradient,

• A. E1/E2=l1/l2
• B. E1/E2=l2/l1
• C. E1E2=l1/l2
• D. E1-E2=l1+l2

Question 4 · difficulty L2 · application

A cell balances at 80 cm and a reference cell of 1.5 V balances at 60 cm. The unknown emf is

• A. 1.125 V
• B. 2.0 V
• C. 1.5 V
• D. 0.75 V

Question 5 · difficulty L2 · understanding

A potentiometer gives the true emf of a cell whereas a voltmeter gives a slightly smaller value. This is because, at the balance point, the potentiometer

• A. has a very low resistance
• B. uses a longer wire
• C. draws no current from the cell being measured, so there is no drop across the cell's internal resistance
• D. measures the terminal voltage directly

Question 6 · difficulty L3 · application

For determining a cell's internal resistance, one compares balance lengths for

• A. two unrelated batteries only
• B. the wire resistance at two temperatures only
• C. open-circuit emf and terminal voltage with a known external load
• D. current with the key always open

Question 7 · difficulty L3 · application

If emf E, terminal voltage V and external resistance R are known, cell internal resistance is

• A. r=R(V/E-1)
• B. r=EV/R
• C. r=R+E/V
• D. r=R(E/V-1)

Question 8 · difficulty L3 · application

Two cells are compared on a potentiometer. Cell 1 balances at 300 cm and cell 2 at 200 cm of the wire, with the same driver circuit. The ratio of their emfs $E_1$ : $E_2$ is

• A. 2 : 3
• B. 9 : 4
• C. 1 : 1
• D. 3 : 2

Question 9 · difficulty L3 · application

A cell balances at 500 cm on a potentiometer wire when no current is drawn from it. When a resistance of 10 $\Omega$ is connected across the cell, the balance length becomes 400 cm. The internal resistance of the cell is

• A. 2.5 $\Omega$
• B. 12.5 $\Omega$
• C. 8.0 $\Omega$
• D. 1.25 $\Omega$

Question 10 · difficulty L4 · analysis

Why is a potentiometer generally more accurate for emf than a voltmeter?

• A. At null it draws essentially no current from the source being measured
• B. It always has zero wire resistance
• C. It amplifies voltage electronically in the basic experiment
• D. It ignores potential gradients