A Vernier caliper with a least count of 0.01\text{ cm} is used to determine the outer diam… — JAMB UTME

A Vernier caliper with a least count of $0.01\text{ cm}$ is used to determine the outer diameter of a cylindrical tube. When the measuring jaws are brought together without any object between them, the zero mark of the vernier scale lies to the right of the main scale zero mark, with the $3\text{rd}$ vernier division coinciding with a main scale line. When measuring the tube, the main scale reads $4.20\text{ cm}$ and the $6\text{th}$ vernier division coincides with a main scale line. What is the actual outer diameter of the tube?

$4.23\text{ cm}$ Answer
B
$4.29\text{ cm}$
C
$4.26\text{ cm}$
D
$4.17\text{ cm}$

Answer

The actual outer diameter of the tube is

4.23\text{ cm}

The observed reading of the Vernier caliper is calculated as Main Scale Reading + (Coinciding Division × Least Count) =

4.20\text{ cm} + (6 \times 0.01\text{ cm}) = 4.26\text{ cm}

. Because the vernier zero lies to the right when closed, it has a positive zero error of

+0.03\text{ cm}

. Subtracting this zero error from the observed reading gives the true measurement of

4.23\text{ cm}

Step-by-Step Solution

Determine the zero error of the Vernier caliper

Zero Error

= + (3 \times 0.01\text{ cm}) = +0.03\text{ cm}

Since the zero of the vernier scale lies to the right of the main scale zero, the instrument has a positive zero error.

Calculate the observed reading

Observed Reading

= 4.20\text{ cm} + (6 \times 0.01\text{ cm}) = 4.26\text{ cm}

The observed measurement is the main scale reading plus the product of the coinciding vernier division and the least count.

Calculate the actual (corrected) reading

Actual Reading

= \text{Observed Reading} - \text{Zero Error} = 4.26\text{ cm} - (+0.03\text{ cm}) = 4.23\text{ cm}

Zero error must always be subtracted algebraically from the observed reading to obtain the accurate measurement.

Key Concept

Zero Error Correction in Vernier Calipers

Estimated Time:1m 30s

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