Question

When an object is placed between the jaws of a vernier calliper, the M.S.R. is 3 mm and V.C.D. is 3 divisions. The maximum number of divisions on the vernier scale is 10 and the zero error of the instrument is +0.12 cm. What is the length of the object (in cm)?

• A. 21
• B. 0.021
• C. 2.1
• D. 0.21

Answer 1

The correct option is D

0.21

Step 1. Given data:

M.S.R = 3 mm

V.C.D = 3

Number of divisions (N) = 10

Zero error (Z.E) = +0.12 cm = +1.2 mm

Correction measure = -1.2 mm

Step 2. Finding the length of the object:

We know,

1 M.S.D = 1 mm

Now,

$L.C=\frac{1M.S.D}{N}=\frac{1}{10}=0.1mm$

Now we can write

$Length=M.S.R+V.C\times L.C+Corr$

Putting all the values,

$Length=3mm+3\times 0.1+\left(-1.2mm\right)$

After solving

$Length=2.1mm=0.21cm$

Therefore the length of the object is 0.21 cm

Hence, the correct option is (D).