Author: DiplomaMitra

 Practical No. 13: Determination of Coefficient of thermal conductivity.

• The value of coefficient of thermal conductivity (K) for the material of the rod indicates the ability of the material to conduct heat. Comparing it to standard values can help identify the material (e.g., copper, aluminum).

• It is concluded that Searle’s apparatus provides an effective method for determining the thermal conductivity of good conductors.
• It is recommended to ensure a steady state is achieved (constant θ₁, θ₂, θ₃, θ₄) before collecting water to get accurate results.

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• Answer: The copper rod is covered with insulating material to prevent heat loss to the surroundings through radiation and convection. This ensures that all the heat passing through section AB is carried away by the water flowing in the spiral tube, making the calculations accurate.

• Answer: (c) Does not change with time and it remain same at all points of the body.
  • Clarification: In a steady state, the temperature at each specific point does not change with time. However, there is a temperature gradient across the body (temperature is different at different points, like θ₁ and θ₂).

• Answer: This indicates that a steady state has been achieved. The rate at which heat is entering any section of the rod is equal to the rate at which it is leaving that section. Therefore, the temperature at any fixed point does not change with time.

• Answer: The rate of heat conduction (Q/t) is directly proportional to the area of cross-section (A). Therefore, if the area is doubled, the amount of heat conducted per second will also double.

• Answer: No, this method is not suitable for bad conductors.
  • Reason: Bad conductors have very low thermal conductivity. It would be very difficult to achieve a steady state and a measurable temperature gradient across a practical length of a bad conductor using this setup. Methods like Lee’s Disc are used for bad conductors.

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 Practical No. 12: Determination of the rate of heat loss due to convection by Newton’s law of cooling.

• It is concluded that the rate of cooling of a body is directly proportional to the temperature difference between the body and its surroundings.
• It is recommended to use a double-walled enclosure to minimize heat loss due to radiation and drafts, ensuring the cooling is primarily due to convection.

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• Answer: Newton’s Law of Cooling states that the rate of loss of heat of a body is directly proportional to the difference in temperature between the body and its surroundings, provided the temperature difference is small.

• Answer: The three modes of heat transfer are:
  1. Conduction
  2. Convection
  3. Radiation

• Answer: Convection is the mode of heat transfer in fluids (liquids and gases) by the actual movement of the heated matter from one place to another.

• Answer:
  • Example 1: Heat from the sun reaching the earth.
  • Example 2: Heat felt from a campfire.

• Answer: A thermos flask is designed to minimize heat transfer by all three modes.
  • It has a vacuum to prevent conduction and convection.
  • Its walls are silvered to minimize heat loss by radiation.

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 Practical No. 11: Determination of pressure-volume relation using Boyle’s law.

• The product of pressure and volume (P × V) was found to be approximately constant for each observation. The graph of P vs. 1/V was a straight line passing through the origin.

• It is concluded that Boyle’s law is verified for the enclosed air at constant temperature.
• It is recommended to ensure that the temperature remains constant throughout the experiment by not handling the tube excessively and taking readings promptly.

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• Answer:
  • Formula: According to Boyle’s Law, P1 × V1 = P2 × V2
  • Given Data: P1 = 720 mm Hg, V1 = 350 cm³, P2 = 600 mm Hg
  • Calculation:
    • V2 = (P1 × V1) / P2
    • V2 = (720 × 350) / 600
    • V2 = 252000 / 600
    • V2 = 420 cm³

• Answer: Yes, indirectly. When the levels of mercury in both limbs are equal, the pressure on the trapped air is equal to the atmospheric pressure (H). This value of H can be read from the barometer provided with the apparatus.

• Answer: The gas enclosed in the closed tube is air.

• Answer: Boyle’s law states that for a fixed mass of an ideal gas at constant temperature, the pressure of the gas is inversely proportional to its volume.
  • Mathematical expression: P ∝ 1/V or PV = constant.

• Answer:
  • Example 1: Inhaling and exhaling (lungs expand, pressure decreases allowing air to flow in; lungs contract, pressure increases forcing air out).
  • Example 2: A syringe (pulling the plunger increases volume, decreasing pressure, drawing fluid in).

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 Practical No. 10: Determination of Joule’s mechanical equivalent of heat by Joule’s law.

• The experiment successfully demonstrates the conversion of electrical energy into heat energy and verifies Joule’s law.
• It is recommended to stir the water continuously for uniform temperature distribution.

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• Answer: No, there will be no change in the amount of heat produced.
  • Reason: The heating effect of current (H = I²Rt) depends on the square of the current. Since the square of both positive and negative values is always positive, the heat produced is independent of the direction of current flow.

• Answer:
  • Formula: Heat (H) = Work done (W) / J
  • Given Data: W = 5000 J, J = 4186 J/kcal
  • Calculation:
    • H = 5000 / 4186
    • H ≈ 1.194 kcal

• Answer:
  • Explanation: From Joule’s law, H = I²Rt. If the current (I) is doubled (becomes 2I), the new heat produced will be (2I)² = 4I². Therefore, the heat produced will increase by four times.

• Answer: Yes, the amount of heat produced will change.
  • Reason: The heat produced depends directly on resistance (H = I²Rt). If the coil is replaced by a copper wire of the same dimensions, its resistance will be much lower because copper has very low resistivity. Therefore, the heat produced will be significantly less.

• Answer: Joule’s law states that the heat (H) produced in a resistor is directly proportional to the square of the current (I) flowing through it, the resistance (R) of the resistor, and the time (t) for which the current flows.
  • Mathematical expression: H = I²Rt
  • Explanation: Electrical work done (W = VIt) is converted into heat. Using Ohm’s law (V = IR), this work is expressed as W = I²Rt, which appears entirely as heat energy.

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 Practical No. 09: Determination of forbidden energy band gap in semiconductors.

• It is concluded that the reverse saturation current in a semiconductor diode increases exponentially with temperature.
• It is recommended to ensure the diode is properly immersed in the oil bath for uniform temperature distribution.

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• Answer: Germanium will have more conductivity at room temperature.
  • Reason: A smaller energy band gap means electrons in the valence band need less energy to jump to the conduction band. Therefore, at a given temperature, germanium will have more electron-hole pairs available for conduction than silicon.

• Answer: The oil bath ensures uniform and gradual heating of the diode. It also prevents direct exposure to flame, which could cause sudden, non-uniform temperature changes or damage the diode.

• Answer: Yes, it does.
  • Reason: A larger forbidden energy gap means fewer charge carriers (electrons and holes) are available for conduction at a given temperature. Fewer charge carriers directly lead to higher resistivity.

• Answer: Yes, it is highly dependent on temperature.
  • Explanation: Reverse saturation current (I_s) is due to minority carriers. As temperature increases, more electron-hole pairs are generated thermally, significantly increasing the number of minority carriers. Hence, I_s increases exponentially with temperature.

• Answer: The forbidden energy gap (Eg) is the minimum energy required to excite an electron from the valence band to the conduction band. It is the energy difference between the top of the valence band and the bottom of the conduction band.

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 Practical No. 08: Determination of static and dynamic resistance of given P N junction diode.

• The forward resistance (both static and dynamic) was low, while the reverse resistance was very high. The knee voltage was identified, and the values of static and dynamic resistance were calculated from the graph.

• It is concluded that a PN junction diode conducts easily in forward bias and offers very high resistance in reverse bias.
• It is recommended to carefully limit the current in forward bias and the voltage in reverse bias to avoid damaging the diode.

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• Answer: For a Silicon diode (e.g., IN4007), the knee voltage is approximately 0.7 V. For a Germanium diode, it is approximately 0.3 V.

• Answer:
  1. Set the multimeter to diode test mode.
  2. Connect the red probe to one terminal and the black probe to the other.
  3. If the meter shows a reading (between 0.5V to 0.8V for Si), the terminal connected to the red probe is the P-side (Anode) and the black is the N-side (Cathode).
  4. If the meter shows “OL” or “1”, the connections are reversed.

• Answer: The current in reverse bias is very small (in microamperes or even less). Using an ammeter with a higher range (e.g., milliammeter) would not give a measurable or accurate reading for such a small current.

• Answer: The diode used was IN4007. Its specifications are:
  • Maximum Average Forward Current: 1 A
  • Peak Repetitive Reverse Voltage: 1000 V
  • Forward Voltage Drop: ~1.1 V at 1 A

• Answer:
  • If forward biased: The diode is ON. Assuming a knee voltage (V_k = 0.7V for Si) drops across it, the current is:
    • Formula: I = (V_supply – V_k) / R
    • Given circuit simple calculation (as per manual text short note): I = V / R = 5 / 1000 = 0.005 A

  • If reverse biased: The diode is OFF. The current is approximately zero (only a very small leakage current flows).

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 Practical No. 07: Determination of neutral points by magnetic compass.

• The neutral points were successfully located on the equatorial line when the North pole of the magnet faced geographic North, and on the axial line when the South pole faced geographic North.

• It is concluded that neutral points exist where the magnetic field due to the magnet exactly cancels the Earth’s horizontal magnetic field.
• It is recommended to use a large sheet of paper and mark points carefully for accurate plotting of field lines.

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• Answer: The compass needle deflects because it experiences a torque due to the magnetic field of the bar magnet, which is much stronger than the Earth’s magnetic field at that specific point.

• Answer: Two neutral points are obtained for any orientation of the bar magnet.

• Answer: The bar magnet is made of steel and its length is 5 cm (Note: You can write the exact length of the magnet used in your college lab, e.g., 5 cm or 7 cm).

• Answer: When the South Pole points to geographic north, the neutral points lie on the axial line of the bar magnet.

• Answer: At a neutral point, the magnetic field due to the magnet (B) is equal to the Earth’s horizontal magnetic field (B_H).
  • For a point on the axial line: B = (μ₀ / 4π) × (2M / d³) = B_H
  • For a point on the equatorial line: B = (μ₀ / 4π) × (M / d³) = B_H
    (where M is the magnetic moment and d is the distance from the center).

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• The experimentally found equivalent resistance for the parallel combination (Rp) was less than the smallest individual resistance.

• It is concluded that for a parallel combination, the voltage remains the same across all resistors, but the current is divided.
• It is recommended to ensure all connections are tight to prevent any unintended resistance in the circuit.

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• Answer: The range of the ammeter used was 0 to 5 Amperes.

• Answer:
  • Step 1: For the parallel part (3 Ohms, 3 Ohms, 3 Ohms):
    • 1/Rp = 1/3 + 1/3 + 1/3 = 3/3
    • Therefore, Rp = 1 Ohm (Ω)

  • Step 2: For the entire circuit in series with the 4 Ohms resistor:
    • Rs = R1 + Rp = 4 + 1
    • Rs = 5 Ohms (Ω)

• Answer:
  • Formula: 1/Rp = 1/R1 + 1/R2
  • Given Data: R1 = 3 Ω, R2 = 4 Ω
  • Calculation:
    • 1/Rp = 1/3 + 1/4 = (4 + 3) / 12 = 7/12
    • Rp = 12/7 Ω
    • Rp ≈ 1.71 Ohms (Ω)

• Answer: The effective (equivalent) resistance decreases.

• Answer: The law states that the reciprocal of the equivalent resistance of a parallel combination is equal to the sum of the reciprocals of the individual resistances.
• Formula: 1/Rp = 1/R1 + 1/R2 + 1/R3 + …

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 Practical No. 05: Determination of equivalent resistance in series connection of resistors.

• The experimental value of the equivalent resistance in series (Rs) was found to be approximately equal to the sum of the individual resistances (R1 + R2).

• It is concluded that for a series combination, the current remains the same through all resistors, but the voltage is shared.
• It is recommended to use resistors whose values are sufficiently different to clearly observe the voltage division.

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• Answer: The rheostat is used to vary the total current in the circuit. This allows us to obtain multiple sets of voltage and current readings for each resistor configuration, which improves the accuracy of the calculated resistance values.

• Answer:
  • Formula: Voltage (V) = Current (I) × Resistance (R)
  • Given Data:
    • Current (I) = 5 A
    • Resistance (R) = 10 Ω

  • Calculation:
    • V = 5 A × 10 Ω
    • V = 50 V

• Answer:
  • Formula for series: R_effective = R1 + R2 + R3
  • Given Data: R1 = 25 Ω, R2 = 50 Ω, R3 = 75 Ω
  • Calculation:
    • R_effective = 25 + 50 + 75
    • R_effective = 150 Ω

• Answer:
  • Explanation: From Ohm’s law, Current (I) = Voltage (V) / Resistance (R). If resistance (R) is constant and voltage is increased to 5V, the new current will also increase five times.

• Answer:
  • Formula: R_equivalent = R1 + R2 + R3
  • Given Data: R1 = 15 Ω, R2 = 10 Ω, R3 = 20 Ω
  • Calculation:
    • R_equivalent = 15 + 10 + 20
    • R_equivalent = 45 Ω

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 Practical No. 04: Determination of specific resistance of given wire.

• The specific resistance calculated from the formula and from the graph are consistent.
• This value is a constant for the material of the wire, as it remained unchanged for different lengths of the wire, confirming the theory.

• It is concluded that the specific resistance is an intrinsic property of the material.
• It is highly recommended to measure the diameter of the wire very precisely using a micrometer, as the area (A) is very sensitive to errors in diameter measurement.

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• Answer: The specific resistance depends only on the material of the wire and its temperature. It is completely independent of the length and cross-sectional area of the wire.

• Answer: Yes, the resistance changes.
  • Reason: Resistance is inversely proportional to the cross-sectional area (A = 3.142 × r²). Therefore, the wire with the larger radius (larger area) will have a lower resistance.

• Answer:
  1. Ammeter-Voltmeter method (using Ohm’s law).
  2. Metre Bridge method (Wheatstone bridge principle).
  3. Using a digital multimeter in resistance mode.
  4. Carey Foster’s bridge method.

• Answer:
  • Given Data:
    • Specific resistance (rho) = 1.678 × 10⁻⁶ Ω-cm
    • Length (L) = 20 m = 2000 cm
    • Diameter (d) = 0.05 cm -> Radius (r) = 0.025 cm

  • Step 1: Calculate Cross-Sectional Area (A)
    • Area (A) = 3.142 × r²
    • Area (A) = 3.142 × (0.025)² = 0.00196375 cm²

  • Step 2: Calculate Resistance (R)
    • Formula: R = (rho × L) / Area
    • R = (1.678 × 10⁻⁶ × 2000) / 0.00196375
    • R = 0.003356 / 0.00196375
    • R = 1.71 Ohms (Ω)

• Answer: No, the specific resistance will not change.
  • Explanation: Specific resistance is an intrinsic property of the material itself, not its shape or size. Doubling the radius changes the wire’s dimensions (increases Area), which changes its total resistance (R), but the material’s specific resistance remains constant.

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