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

Step 1: Find the least count.
One main scale division is $1$ mm and there are $10$ vernier divisions, so the least count is $\dfrac{1\text{ mm}}{10} = 0.1$ mm $= 0.01$ cm.
Step 2: Spot the zero error.
With the jaws closed, the vernier scale has shifted to the left and the 4th vernier line matches a main-scale line, so the zero of the vernier sits to the left of the main-scale zero.
Step 3: Find the size of the zero error.
The zero error is $-4 \times 0.01 = -0.04$ cm (negative because the shift is to the left).
Step 4: Note the measured reading.
The reading taken is $1.00$ cm.
Step 5: Apply the correction.
Actual length = measured length + zero error = $1.00 + (-0.04) = 0.96$ cm.
Step 6: Choose the option.
The corrected length is $0.96$ cm, which is option C.
\[ \boxed{ 0.96 \text{ cm} } \]