JEE Challenger
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PhyMedAdv 2021 • Paper 1

Measurement of Sphere Diameter Using Vernier Calipers with Zero Error

The smallest main scale division of a Vernier caliper measures 0.1 cm0.1\text{ cm}. Ten divisions on the Vernier scale coincide with nine divisions of the main scale. The diagram on the left displays the reading when the jaws are closed without any gap. The diagram on the right displays the reading when a solid sphere is held between the jaws. Determine the correct diameter of the sphere.

Question Diagram 1

Options

A

3.07 cm3.07\text{ cm}

B

3.11 cm3.11\text{ cm}

C

3.15 cm3.15\text{ cm}

Correct
D

3.17 cm3.17\text{ cm}

Topics & Concepts

Units and MeasurementsMeasuring Instruments

Step-by-Step Solution

To find the correct diameter of the solid sphere, we calculate the least count, the zero error of the Vernier calipers, and the observed reading step-by-step.

Step 1: Calculate the Least Count (LC)

Given:

  • 1 Main Scale Division (MSD)=0.1 cm1\text{ Main Scale Division (MSD)} = 0.1\text{ cm}
  • 10 Vernier Scale Divisions (VSD)=9 MSD10\text{ Vernier Scale Divisions (VSD)} = 9\text{ MSD}

Thus, the value of 1 VSD1\text{ VSD} is: 1 VSD=910×0.1 cm=0.09 cm1\text{ VSD} = \frac{9}{10} \times 0.1\text{ cm} = 0.09\text{ cm}

The Least Count (LC) of the instrument is: LC=1 MSD−1 VSD=0.1 cm−0.09 cm=0.01 cm\text{LC} = 1\text{ MSD} - 1\text{ VSD} = 0.1\text{ cm} - 0.09\text{ cm} = 0.01\text{ cm}

Step 2: Determine the Zero Error

From the diagram on the left (when jaws are closed without any object):

  • The 00 mark of the Vernier scale lies to the left of the 00 mark of the main scale, which indicates a negative zero error.
  • The 6th6\text{th} division of the Vernier scale coincides with a division on the main scale (the 0.5 cm0.5\text{ cm} mark).

Using the formula for negative zero error: Zero Error=−(10−6)×LC=−4×0.01 cm=−0.04 cm\text{Zero Error} = -(10 - 6) \times \text{LC} = -4 \times 0.01\text{ cm} = -0.04\text{ cm}

Step 3: Determine the Observed Reading

From the diagram on the right (when the solid sphere is held between the jaws):

  • The zero mark of the Vernier scale lies just beyond 3.1 cm3.1\text{ cm} on the main scale. Main Scale Reading (MSR)=3.1 cm\text{Main Scale Reading (MSR)} = 3.1\text{ cm}
  • The 1st1\text{st} division of the Vernier scale coincides with a line on the main scale. Vernier Scale Reading (VSR)=1\text{Vernier Scale Reading (VSR)} = 1

The observed reading is: Observed Reading=MSR+(VSR×LC)\text{Observed Reading} = \text{MSR} + (\text{VSR} \times \text{LC}) Observed Reading=3.1 cm+(1×0.01 cm)=3.11 cm\text{Observed Reading} = 3.1\text{ cm} + (1 \times 0.01\text{ cm}) = 3.11\text{ cm}

Step 4: Calculate the Correct Diameter

The correct diameter of the sphere is obtained by subtracting the zero error from the observed reading: Correct Diameter=Observed Reading−Zero Error\text{Correct Diameter} = \text{Observed Reading} - \text{Zero Error} Correct Diameter=3.11 cm−(−0.04 cm)=3.11 cm+0.04 cm=3.15 cm\text{Correct Diameter} = 3.11\text{ cm} - (-0.04\text{ cm}) = 3.11\text{ cm} + 0.04\text{ cm} = 3.15\text{ cm}

Conclusion

strategies lead to the correct option (C).

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