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NEET
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Physics
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Experimental Physics

NEET · Physics · Experimental Physics

Data recorded during an experiment to measure the diameter of a...

Data recorded during an experiment to measure the diameter of a sphere using a Vernier calliper (11 main scale division = 11 mm, 1010 Vernier scale divisions = 99 main scale divisions) is given below:

Condition 1: Jaws fully closed
Main Scale Reading: 00
Vernier Scale Alignment: Vernier zero is to the right of main scale zero; 6th6^{th} division coincides.

Condition 2: Measuring sphere diameter
Main Scale Reading: 2.42.4 cm
Vernier Scale Alignment: 8th8^{th} division coincides.

The actual diameter of the sphere is:

Options

  1. A2.542.54 cm
  2. B1.881.88 cm
  3. C2.482.48 cm
  4. D2.422.42 cm

Correct answer

D. 2.422.42 cm

Step-by-step solution

The least count (LC) of the Vernier calliper is calculated as:

LC=1 MSD−1 VSD=1 mm−0.9 mm=0.1 mm=0.01 cm\text{LC} = 1 \text{ MSD} - 1 \text{ VSD} = 1 \text{ mm} - 0.9 \text{ mm} = 0.1 \text{ mm} = 0.01 \text{ cm}

Since the Vernier zero is to the right of the main scale zero, the instrument has a positive zero error.

Zero error=+n×LC=+6×0.01 cm=+0.06 cm\text{Zero error} = +n \times \text{LC} = +6 \times 0.01 \text{ cm} = +0.06 \text{ cm}

The observed reading for the sphere's diameter is:

Observed reading=MSR+(VSR×LC)=2.4 cm+(8×0.01 cm)=2.48 cm\text{Observed reading} = \text{MSR} + (\text{VSR} \times \text{LC}) = 2.4 \text{ cm} + (8 \times 0.01 \text{ cm}) = 2.48 \text{ cm}

The actual diameter is obtained by subtracting the zero error from the observed reading:

Actual diameter=Observed reading−Zero error\text{Actual diameter} = \text{Observed reading} - \text{Zero error}

Actual diameter=2.48 cm−(+0.06 cm)=2.42 cm\text{Actual diameter} = 2.48 \text{ cm} - (+0.06 \text{ cm}) = 2.42 \text{ cm}

Answer: 2.422.42 cm
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