Question

One main scale division (MSD) of a Vernier calliper is \(1\) mm and the Vernier scale has \(10\) divisions. When the jaws touch, the Vernier scale shifts to the left and the \(4^{th}\) Vernier division coincides with a main scale division. If the measured length is \(1\) cm, the actual length is:

• A. 1.04 cm
• B. 0.60 cm
• C. 0.96 cm
• D. 1.00 cm

Hint:

If Vernier zero lies to the left of main scale zero, the instrument has negative zero error. Subtract the magnitude of the negative error from the measured reading.

Answer 1

The Correct Option is C

Concept:

• Least Count of Vernier Calliper: \[ LC=1\text{ MSD}-1\text{ VSD} \]

• For a 10-division Vernier, \[ LC=0.1\text{ mm}=0.01\text{ cm} \]

Step 1: Determine the zero error.
The Vernier zero lies to the left of the main scale zero. Hence the instrument has negative zero error. \[ \text{Zero Error} = -4\times0.01 \] \[ = -0.04\text{ cm} \]

Step 2: Apply zero correction.
Measured length \[ =1.00\text{ cm} \] Actual length \[ = \text{Measured Length} + \text{Zero Error} \] \[ = 1.00-0.04 \] \[ = 0.96\text{ cm} \]

Step 3: Write the final answer.
\[ \boxed{0.96\text{ cm}} \] Hence, \[ \boxed{\text{Option (C)}} \]