This set of Class 12 Physics Chapter 3 Multiple Choice Questions & Answers (MCQs) focuses on “Current Electricity – Potentiometer”.

1. A potentiometer wire of length 100 cm has a resistance of 30 ohms. It is connected in series with a resistance of 20 ohms and accumulator of emf 8V having negligible internal resistance. A source of 1.2V is balanced against a length L of the potentiometer wire. What is the value of L?

a) 20

b) 25

c) 30

d) 35

View Answer

Explanation: The current passing through the potentiometer wire:

I = \(\frac {8}{(30 + 20)} = \frac {8}{50}\) = 0.16A

The potential difference across the potentiometer wire:

V = current × resistance = 0.16 × 30 = 4.8V

Length of the wire = 100 cm

k = \(\frac {V}{l} = \frac {4.8}{100}\) = 0.048

The emf 1.2V is balanced against the length L of the wire, i.e. 1.2 = kL

Length = \(\frac {1.2}{k} = \frac {1.2}{0.048}\) = 25 cm

Therefore, the length L is 25 cm.

2. A potentiometer using cell C of emf 5V and internal resistance 0.2 ohms is connected to a wire AB in the figure below. A standard cell C_{0} of a constant emf of 1.10 V gives a balance point at 55 cm of the wire. When C_{0} is replaced by a cell of emf E, the balance point is obtained at 85 cm. What is the value of E?

a) 1.4 V

b) 1.5 V

c) 1.7 V

d) 1.9 V

View Answer

Explanation: l

_{1}= 55 cm; I

_{2}= 85

\(\frac {E}{C_0} = \frac {l_2}{l_1}\) → E = C

_{0}× \(\frac {l_2}{l_1}\)

E = 1.10 × \(\frac {85}{55}\) = 1.7 V

Therefore, the value of E is 1.7 V.

3. A potentiometer wire of length 20 m has a resistance of 50 ohms. It is connected in series with a resistance box and a 5 V storage cell. If the potential gradient along the wire is 0.5 mV/cm, what is the resistance unplugged in the box?

a) 450 ohms

b) 400 ohms

c) 405 ohms

d) 500 ohms

View Answer

Explanation: Potential gradient along the potentiometer wire = \(\frac {potential \, difference \, along \, wire}{length \, of \, wire}\).

0.5 × 10

^{-3}= I × \(\frac {50}{1000}\)

I = 0.5 × 10

^{-3}× \(\frac {1000}{50}\)

I = \(\frac {1}{100}\)

So, \(\frac {5}{(50 + R)} = \frac {1}{100}\)

R + 50 = 500

R = 450 ohms

Therefore, the resistance unplugged in the box is 450 ohms.

4. In a potentiometer of 5 wires, the balance point is obtained on the 2^{nd} wire. To shift the balance point to the 4^{th}, we should decrease the current of the main circuit.

a) True

b) False

View Answer

Explanation: If we have to shift the balance point of a potentiometer to a higher length, the potential gradient of the wire is to be decreased. This can also be achieved by increasing the resistance in series with the potentiometer wire, and this is possible by decreasing the current of the main circuit. So, this is a true statement.

5. The balancing length of a potentiometer is at 120 cm. On shunting the cell with a resistance of 4 ohms, the balancing point shifts to a length of 60 cm. Then, find the internal resistance of the cell.

a) 2 ohms

b) 5 ohms

c) 3 ohms

d) 4 ohms

View Answer

Explanation: l

_{1}= 120 cm; l

_{2}= 60 cm; R = 4 ohms

Internal resistance (r) = \([ \frac {(l2 – l1)}{I_2} ]\) × R

= \([\frac {(120 – 60)}{60} ]\) × 4

= \(\frac {60}{60}\) × 4

= 4 ohms

Therefore, the internal resistance is 4 ohms.

6. A potentiometer has uniform potential gradient. The specific resistance of the material of the potentiometer wire is 10^{-9}Ωm and the current passing through it is 0.5A and the cross sectional area of the wire is 10^{-8}m^{2}. Calculate the potential gradient along the potentiometer wire.

a) 0.5 × 10^{-9} V/m

b) 0.5 × 10^{-8} V/m

c) 0.5 × 10^{-1} V/m

d) 0.5 × 10^{-15} V/m

View Answer

Explanation: Resistance of a wire = \(\frac {Ꝭl}{A}\), where Ꝭ is the specific resistance of the material of the wire.

Potential gradient = \(\frac {V}{l} \rightarrow \frac {IR}{l}\) → \(\frac {I(\frac {Ꝭl}{A})}{l}\)

= \(\frac {Ꝭl}{A}\)

= 0.5 × \(\frac {10^{-9}}{10^{-8}}\)

= 0.5 × 10

^{-1}V/m

Therefore, the potential gradient is equal to 0.5 × 10

^{-1}V/m.

7. Pick out the application of potentiometer from the following.

a) It measures current

b) It measures internal resistance

c) It measures external resistance

d) It is used to compare two currents

View Answer

Explanation: Potentiometer is an electric instrument used to measure internal resistance, electromotive force (emf), and it is also used for comparing the emfs of different cells. Potentiometer can be used as a variable resistor as well.

**Sanfoundry Global Education & Learning Series – Physics – Class 12**.

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