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Units and Measurements question

2013 · Shift 1 · Q52
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Units and Measurements question

2013 · Shift 1 · Q52

JEE AdvancedPhysicsUnits and MeasurementsMCQ+4 / −1
The diameter of a cylinder is measured using a Vernier callipers with no zero error. It is found that the zero of the Vernier scale lies between 5.10 cm and 5.15 cm of the main scale. The Vernier scale has 50 divisions equivalent to 2.45 cm. The 24th division of the Vernier scale exactly coincides with one of the main scale divisions. The diameter of the cylinder is
  1. A
    5.112 cm
  2. B
    5.124 cm
  3. C
    5.136 cm
  4. D
    5.148 cm
View written solutionFree

Correct answer: B

Step-by-step derivation:

  1. Determine the value of one Main Scale Division (MSD) and one Vernier Scale Division (VSD).

    • The problem states that the Vernier scale has 50 divisions (VSD) which are equivalent to 2.45 cm.
    • So, 50 VSD = 2.45 cm.
    • This allows us to calculate the length of one Vernier scale division: 1 VSD=2.45 cm50=0.049 cm1 \text{ VSD} = \frac{2.45 \text{ cm}}{50} = 0.049 \text{ cm}1 VSD=502.45 cm​=0.049 cm
    • For a standard Vernier callipers, N divisions on the Vernier scale coincide with (N-1) divisions on the main scale. Let's check if this is the case here with N=50.
    • If 50 VSD = 49 MSD, then 49 MSD = 2.45 cm.
    • Let's calculate the value of one main scale division from this relationship: 1 MSD=2.45 cm49=0.05 cm1 \text{ MSD} = \frac{2.45 \text{ cm}}{49} = 0.05 \text{ cm}1 MSD=492.45 cm​=0.05 cm
    • This is consistent with the information that the zero of the Vernier scale lies between 5.10 cm and 5.15 cm, which implies the main scale is marked in intervals of 0.05 cm.

  2. Calculate the Least Count (LC) of the Vernier callipers.

    • The least count is the smallest measurement that can be accurately made with the instrument. It is defined as the difference between one main scale division and one Vernier scale division.
    • LC = 1 MSD - 1 VSD
    • Substituting the values we found: LC=0.05 cm−0.049 cm=0.001 cmLC = 0.05 \text{ cm} - 0.049 \text{ cm} = 0.001 \text{ cm}LC=0.05 cm−0.049 cm=0.001 cm
    • Alternatively, using the formula LC = (value of 1 MSD) / (Number of divisions on Vernier scale): LC=0.05 cm50=0.001 cmLC = \frac{0.05 \text{ cm}}{50} = 0.001 \text{ cm}LC=500.05 cm​=0.001 cm

  3. Determine the Main Scale Reading (MSR).

    • The question states that the zero of the Vernier scale lies between 5.10 cm and 5.15 cm on the main scale.
    • The Main Scale Reading (MSR) is the reading on the main scale immediately to the left of the zero of the Vernier scale.
    • Therefore, MSR = 5.10 cm.

  4. Identify the Vernier Scale Coincidence (VSC).

    • The problem states that the 24th division of the Vernier scale exactly coincides with one of the main scale divisions.
    • So, the Vernier Scale Coincidence, VSC = 24.

  5. Calculate the diameter of the cylinder.

    • The formula for the measurement, with no zero error, is: Total Reading = MSR + (VSC × LC)
    • Substituting the values: Diameter=5.10 cm+(24×0.001 cm)\text{Diameter} = 5.10 \text{ cm} + (24 \times 0.001 \text{ cm})Diameter=5.10 cm+(24×0.001 cm) Diameter=5.10 cm+0.024 cm\text{Diameter} = 5.10 \text{ cm} + 0.024 \text{ cm}Diameter=5.10 cm+0.024 cm Diameter=5.124 cm\text{Diameter} = 5.124 \text{ cm}Diameter=5.124 cm

  6. Conclusion.

    • The calculated diameter is 5.124 cm. This matches option B.
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