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  • Practical No. 03: Determination of resistance by Ohm’s law.

    Practical No. 03: Determination of resistance by Ohm’s law.
    XIII. Interpretation of Results
    • The straight-line graph of Voltage (V) vs. Current (I) passing through the origin confirms that the wire obeys Ohm’s law.
    • The resistance value calculated from the slope of the graph matches the average resistance calculated from the observations, validating the experiment.

    XIV. Conclusions and Recommendations
    • Ohm’s law (V = IR) is successfully verified for the given wire.
    • It is highly recommended to switch on the circuit only while taking readings to prevent heating of the wire, which can change its resistance.

    XV. Practical Related Questions & Answers
    Q1. State the function of rheostat in this lab experiment.
    • Answer: The rheostat acts as a variable resistor. Its main function is to control and vary the amount of electric current flowing through the circuit, allowing us to take multiple pairs of voltage (V) and current (I) readings.

    Q2. A potential difference of 15V appears across the ends of a resistor when 5A of current flows through it. Find resistance of the resistor?
    • Answer:
      • Formula: Resistance (R) = Voltage (V) / Current (I)
      • Given Data:
        • Voltage (V) = 15 V
        • Current (I) = 5 A

      • Calculation:
        • R = 15 V / 5 A
        • R = 3 Ohms (Ω)

    Q3. If the voltage across a fixed value of resistance is increased five times, what will be the variation in current? Does the resistance depend on the temperature?
    • Answer:
      • Variation in Current: From Ohm’s law, Current (I) = Voltage (V) / Resistance (R). If resistance (R) is constant and voltage is increased five times, the current will also increase five times.
      • Dependence on Temperature: Yes, the resistance of a conductor depends on temperature. For most metals, resistance increases with an increase in temperature.

    Q4. Calculate the voltage if a resistance of 25 Ω produces a current of 250 amperes.
    • Answer:
      • Formula: Voltage (V) = Current (I) × Resistance (R)
      • Given Data:
        • Resistance (R) = 25 Ω
        • Current (I) = 250 A

      • Calculation:
        • V = 250 A × 25 Ω
        • V = 6250 V

    Q5. State Ohm’s Law.
    • Answer: Ohm’s Law states that the electric current (I) flowing through a conductor is directly proportional to the potential difference (V) applied across its ends, provided the physical conditions (like temperature) remain constant.
    • Mathematical expression: V = IR (where R is the resistance).

  • Practical No. 02: Measurements of dimensions of given objects by micrometer screw gauge.

    Practical No. 02: Measurements of dimensions of given objects by micrometer screw gauge.
    XIV. Conclusions and Recommendations
    • It is concluded that consistent and accurate readings can be achieved by using the ratchet to avoid over-tightening.
    • It is highly recommended to always determine and apply the zero error correction for every measurement session.

    XV. Practical Related Questions & Answers
    Q1. Are micrometers used in telescopes or microscopes?
    • Answer: The measuring tool called a micrometer is not typically a direct part of a telescope or microscope. However, the precision screw mechanism, which is the core principle behind the micrometer, is widely used in these instruments for precise focusing.

    Q2. State the main types of screw gauges.
    • Answer: The main types are:
      • Outside Micrometer: For measuring external dimensions.
      • Inside Micrometer: For measuring internal dimensions.
      • Depth Micrometer: For measuring depths of holes and slots.

    Q3. State the range of given micrometer screw gauge used in experiment.
    • Answer: The range of the given micrometer screw gauge is 0 to 2.5 cm.

    Q4 & 5. The circular scale of a screw gauge contains 100 divisions and its pitch is 1 mm. Give the least count of the screw gauge.
    • Answer:
      • Formula:
        Least Count (L.C.) = Pitch / Number of circular divisions
      • Given Data:
        • Pitch = 1 mm
        • Number of circular divisions (n) = 100

      • Calculation:
        • L.C. = 1 mm / 100
        • L.C. = 0.01 mm

    Q6. In ten rotations of the screw, distance travelled on main scale is 20 mm. If number of divisions on circular scale are 50. Find L.C. of micrometer screw gauge.
    • Answer:
      • Step 1: Find the Pitch
        • Pitch = Distance moved / Number of rotations
        • Pitch = 20 mm / 10 = 2 mm

      • Step 2: Find the Least Count (L.C.)
        • L.C. = Pitch / Number of circular divisions (n)
        • L.C. = 2 mm / 50
        • L.C. = 0.04 mm

  • Practical No. 01: Measurements of dimensions of given object by Vernier caliper.

     Practical No. 01: Measurements of dimensions of given object by Vernier caliper.

    XIV. Conclusions and Recommendations
    • The Vernier caliper is a precise instrument for measuring linear dimensions up to 0.002 cm.
    • It is highly recommended to always note the zero error and apply its proper correction for every single reading.

    XV. Practical Related Questions & Answers
    Q1. Give the value of smallest division on main scale and the total no. of divisions on Vernier scale of the instrument used in experiment.
    • Answer:
      • The smallest division on the main scale (m) = 0.1 cm
      • The total number of divisions on the Vernier scale (n) = 50

    Q2. Write the range of the Vernier caliper used in the experiment. Write two applications of a vernier caliper.
    • Answer:
      • Range: The range of the Vernier caliper is 0 to 15 cm.
      • Application 1: Measuring the internal diameter of a pipe, tube, or beaker.
      • Application 2: Measuring the exact depth of a small hole, slot, or recess.

    Q3. Determine the least count (L.C.) of a Vernier caliper if the smallest division on the main scale is 1 mm and the number of divisions on the Vernier scale is 20.
    • Answer:
      • Formula:
        Least Count (L.C.) = (Smallest division on main scale) / (Total divisions on Vernier scale)
      • Given Data:
        • Smallest main scale division (m) = 1 mm = 0.1 cm
        • Total Vernier divisions (n) = 20

      • Calculation:
        • L.C. = 0.1 cm / 20
        • L.C. = 0.005 cm (or 0.05 mm)

    Q4 & 5. A student has used a Vernier caliper with L.C. 0.01 cm, to measure the length of his pencil. He found that the main scale showed a reading of 7.3 cm and the Vernier scale coincides with the 5th position to the main scale. Give the correct measurement of the length of the pencil neglecting zero error.
    • Answer:
      • Given Data:
        • Main Scale Reading (MSR) = 7.3 cm
        • Vernier Scale Division (VSD) = 5
        • Least Count (L.C.) = 0.01 cm

      • Step-by-Step Calculation:
        1. Vernier Scale Reading (VSR):
          • VSR = VSD × L.C.
          • VSR = 5 × 0.01 cm = 0.05 cm

        2. Total Reading (Correct Measurement):
          • Total Reading = MSR + VSR
          • Total Reading = 7.3 cm + 0.05 cm = 7.35 cm

      • Conclusion: The correct length of the pencil is 7.35 cm.