Class 12 Physics Notes Chapter 4 () – Lab Manual (English) Book

Detailed Notes with MCQs of Experiment number 4 from your Physics Lab Manual, which is crucial not just for your practical exams but also often features in competitive government exams. The experiment is "To compare the e.m.f. of two given primary cells using potentiometer."

Here are the detailed notes covering the essential aspects:

Experiment 4: To compare the e.m.f. of two given primary cells using potentiometer.

1. Aim:
To compare the electromotive forces (EMFs) of two given primary cells (like a Leclanche cell and a Daniel cell) using a potentiometer.

2. Apparatus Required:

• Potentiometer (with 4 to 10 wires)
• Two primary cells (E₁, E₂) (e.g., Leclanche cell, Daniel cell)
• A stable power source (driver cell or battery eliminator) E, whose EMF is greater than E₁ and E₂
• Rheostat (Rh)
• Galvanometer (G)
• Jockey (J)
• Two one-way keys (K₁, K₂) or preferably a two-way key
• High Resistance Box (HRB)
• Connecting wires (Copper wires)
• Sandpaper (for cleaning wire ends)

3. Theory and Principle:

• Potentiometer Principle: A potentiometer operates on the principle that the potential drop across any segment of a wire having uniform cross-sectional area and uniform composition is directly proportional to the length of that segment, provided a constant current flows through the wire.
  • Mathematically, V ∝ l, or V = kl
  • Here, 'V' is the potential drop across length 'l' of the wire, and 'k' is the constant of proportionality called the potential gradient (potential drop per unit length).
  • Potential gradient, k = V/L = IR/L, where L is the total length of the potentiometer wire, R is its total resistance, and I is the constant current flowing through it. 'k' is constant if I is constant and the wire is uniform.
• EMF Comparison:
  • When a cell (say E₁) is connected such that its EMF is balanced against the potential drop on the potentiometer wire, no current flows through the galvanometer (null deflection). Let the balancing length be l₁.
  • At balance, the EMF of the cell equals the potential drop across the balancing length:
    E₁ = k * l₁ ---(i)
  • Similarly, if the second cell (E₂) is balanced against a length l₂ (keeping 'k' constant), then:
    E₂ = k * l₂ ---(ii)
  • Dividing equation (i) by (ii):
    E₁ / E₂ = (k * l₁) / (k * l₂)
    E₁ / E₂ = l₁ / l₂
  • By measuring the respective balancing lengths l₁ and l₂, we can determine the ratio of the EMFs of the two cells.
• Essential Conditions:
  • The EMF of the driver cell (E) must be greater than the EMFs of the cells being compared (E > E₁ and E > E₂).
  • All positive terminals (of E, E₁, and E₂) must be connected to the same end (the 'zero' end or high potential end) of the potentiometer wire.
  • The current supplied by the driver cell through the potentiometer wire must remain constant throughout the experiment (especially when measuring l₁ and l₂ for a single set of readings).

4. Circuit Diagram:

(Imagine a standard potentiometer setup)

• The potentiometer wire (AB) is connected in series with the driver cell (E), a key (K₁), and a rheostat (Rh). The positive terminal of E is connected to end A (zero mark).
• End A is also connected to the positive terminals of the two cells E₁ and E₂.
• The negative terminal of E₁ is connected to terminal 1 of a two-way key.
• The negative terminal of E₂ is connected to terminal 3 of the two-way key.
• The common terminal (terminal 2) of the two-way key is connected to one terminal of the galvanometer (G).
• A High Resistance Box (HRB) is connected in series with the galvanometer for safety.
• The other terminal of the galvanometer is connected to the jockey (J).

5. Procedure:

1. Clean the ends of connecting wires with sandpaper and set up the circuit as described above. Ensure tight connections.
2. Close the main key K₁. Check that the EMF of the driver cell E is greater than E₁ and E₂.
3. Test Connections: Include a high resistance (e.g., 2000 Ω) from the HRB. Touch the jockey at end A (0 cm); note the galvanometer deflection direction. Touch the jockey at end B (last wire end); the deflection should be in the opposite direction. If not, check connections or driver cell voltage. Adjust the rheostat if the deflection goes out of scale or if opposite deflections are not obtained.
4. Measure l₁: Connect cell E₁ into the circuit using the two-way key (plug between terminals 1 & 2). Find the approximate null point on the wire by sliding the jockey. Now, make the HRB resistance zero and find the exact point J₁ where the galvanometer shows zero deflection. Measure the balancing length AJ₁ = l₁.
5. Measure l₂: Without changing the rheostat setting, disconnect E₁ and connect cell E₂ (plug between terminals 3 & 2). Find the new null point J₂. Measure the balancing length AJ₂ = l₂.
6. Repeat: Take 2-3 more sets of readings for l₁ and l₂ by slightly changing the rheostat setting (this changes 'k' but verifies the ratio l₁/l₂).
7. Record all observations systematically.

6. Observations:

• EMF of driver cell E ≈ ______ V

• Approximate EMF of cell E₁ ≈ ______ V

• Approximate EMF of cell E₂ ≈ ______ V

• Length of one wire of potentiometer = ______ cm

• Total length of potentiometer wire L = ______ cm

• Least count of the scale = ______ cm

• Observation Table:

  S. No.	Balancing length for E₁ (l₁ cm)	Balancing length for E₂ (l₂ cm)	Ratio E₁/E₂ = l₁/l₂
  1
  2
  3
  Mean	Mean l₁ =	Mean l₂ =	Mean Ratio =

7. Calculations:

• Calculate the ratio l₁/l₂ for each set.
• Find the mean value of the ratio E₁/E₂.
  Mean E₁/E₂ = (Mean l₁) / (Mean l₂) OR Mean (l₁/l₂)

8. Result:
The ratio of the EMFs of the two given primary cells is E₁ : E₂ = ______ : 1 (or E₁/E₂ = ______).

9. Precautions:

1. Connections must be clean and tight.
2. Connect all positive terminals to the zero end (A) of the potentiometer.
3. Driver cell EMF (E) must exceed cell EMFs (E₁, E₂).
4. Maintain constant current in the potentiometer wire for each set of (l₁, l₂) measurements.
5. Do not slide the jockey; lift and touch gently to avoid scraping the wire and ensure point contact.
6. Use the High Resistance Box initially to protect the galvanometer. Reduce resistance to zero for final null point determination.
7. Ensure the null point is obtained on the potentiometer wire itself (not outside the ends). Aim for balance points on the latter half of the wire length for better precision if possible.
8. Avoid passing current through the primary cells for extended periods.

10. Sources of Error:

1. Non-uniformity in the potentiometer wire (area or composition).
2. Variation in the driver cell EMF or current during the experiment.
3. End resistances (contact resistances where the wire connects to terminals).
4. Inaccurate measurement of balancing lengths (parallax error, thick jockey point).
5. Galvanometer sensitivity might be too low.
6. EMF of primary cells might change slightly if they are not fresh or if used for too long.

11. Key Concepts for Exams (Viva Voce points):

• Potentiometer Principle: V ∝ l for constant current and uniform wire.
• Why Potentiometer is Superior to Voltmeter for EMF: It measures EMF accurately because it draws no current from the cell at the null point (measures open circuit potential difference). A voltmeter always draws some current, measuring terminal voltage (V = E - Ir), which is less than EMF.
• Potential Gradient (k): Potential drop per unit length (V/m or V/cm). Lower 'k' means higher sensitivity.
• Sensitivity: Ability to detect small changes in potential difference. Increased by increasing wire length (L) or decreasing current (I) (making 'k' smaller).
• Conditions for Balance: E_driver > E_cell; Correct polarity connections.
• Role of Rheostat: To adjust current 'I' and thus potential gradient 'k'.
• Primary vs. Secondary Cells: Primary (non-rechargeable, e.g., Leclanche, Daniel), Secondary (rechargeable, e.g., Lead-Acid Accumulator).

Multiple Choice Questions (MCQs):

1. A potentiometer measures the potential difference more accurately than a voltmeter because:
   a) It has a wire of high resistance.
   b) It has a wire of low resistance.
   c) It does not draw any current from the source of unknown EMF.
   d) It draws a large current from the source of unknown EMF.

2. In comparing the EMFs of two cells E₁ and E₂ using a potentiometer, the respective balancing lengths were found to be 250 cm and 400 cm. The ratio E₁/E₂ is:
   a) 8/5
   b) 5/8
   c) 25/16
   d) 16/25

3. To increase the sensitivity of a potentiometer, one should:
   a) Increase the resistance in the rheostat in the main circuit.
   b) Decrease the resistance in the rheostat in the main circuit.
   c) Decrease the length of the potentiometer wire.
   d) Replace the driver cell with one of lower EMF.

4. Which condition is essential for obtaining a balance point on the potentiometer wire?
   a) The EMF of the driver cell must be less than the EMF of the cell being measured.
   b) The positive terminals of all cells must be connected to the zero end.
   c) The jockey must be slid continuously.
   d) A high resistance must always be kept in series with the galvanometer.

5. The potential gradient of a potentiometer wire depends on:
   a) Only the current flowing through the wire.
   b) Only the resistance per unit length of the wire.
   c) Both the current and the resistance per unit length.
   d) The length of the wire only.

6. If the driver cell in a potentiometer experiment is replaced by another cell of lower EMF, the balancing length will:
   a) Increase.
   b) Decrease.
   c) Remain unchanged.
   d) Become infinite (no balance point).

7. In a potentiometer experiment, if the connections of the cell E₁ (whose EMF is to be measured) are reversed, the balance point will:
   a) Shift towards end A (0 cm).
   b) Shift towards end B.
   c) Remain unchanged.
   d) Not be obtained on the wire.

8. What is the purpose of the rheostat in the potentiometer's primary circuit?
   a) To measure the current accurately.
   b) To protect the galvanometer.
   c) To vary the potential gradient across the wire.
   d) To reverse the direction of the current.

9. A potentiometer wire is 10 m long and has a resistance of 20 Ω. It is connected in series with a battery of EMF 3 V and negligible internal resistance, and a resistance box with 10 Ω resistance. The potential gradient along the wire is:
   a) 0.01 V/m
   b) 0.02 V/m
   c) 0.1 V/m
   d) 0.2 V/m

10. While comparing EMFs E₁ and E₂ (E₁ > E₂), the balancing length l₁ is obtained. If E₂ is now connected, the balancing length l₂ will be:
    a) Greater than l₁
    b) Less than l₁
    c) Equal to l₁
    d) Dependent on the rheostat setting only.

Answer Key for MCQs:

1. c
2. b (E₁/E₂ = l₁/l₂ = 250/400 = 25/40 = 5/8)
3. a (Increasing resistance decreases current I, thus decreasing k = IR/L, increasing sensitivity)
4. b
5. c (k = I * (R/L))
6. d (If driver EMF E becomes less than the cell EMF E₁, potential drop across the entire wire EL = kL might be less than E₁, so no balance point exists where k*l = E₁)
7. d (The potential drop along the wire and the cell EMF will add up instead of opposing, no balance possible)
8. c
9. d (Total R = 20Ω + 10Ω = 30Ω. Current I = V/R_total = 3V / 30Ω = 0.1 A. Potential drop across wire V_wire = I * R_wire = 0.1 A * 20 Ω = 2 V. Potential gradient k = V_wire / Length = 2 V / 10 m = 0.2 V/m)
10. b (Since E₁ > E₂, and E = kl, for the same k, l₁ must be greater than l₂)

Study these notes thoroughly. Understanding the principle and precautions is key. Let me know if any part needs further clarification.