Question

One cm on the main scale of the vernier caliper is divided into 20 equal parts and the number of vernier scale divisions is 10. The Z.E of the instrument is –0.45 mm. If an object is placed between the jaws of vernier calipers, the M.S.R and V.C.D are 12 M.S.D and 9, respectively. What is the length of the object?

• A. 69 mm
• B. 6.90 mm
• C. 6.90 cm
• D. 0.690 mm

Answer 1

The correct option is B

6.90 mm

Step 1, Given data

Number of divisions (N) = 10

1 M.S.D = $\frac{1cm}{20}=0.05cm=0.5mm$

Zero error (Z.E) = -0.45 mm

So, correction = +0.45 mm

M.S.R = 12 M.S.D

V.C.D = 9

Step 2, Finding the length

We know,

$L.C=\frac{1M.S.D}{N}=\frac{0.5}{10}=0.05mm=0.005cm$

it is given that

$M.S.R=12M.S.D=12\times \frac{1}{2}mm=6mm$

V.C.D = 9

Now we can write

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

Putting all the values

$Length=6+9\times 0.05+\left(+0.45\right)$

After solving

$Length=6.90mm$

Therefore the length of the object is 6.90 mm

Hence the correct option is (B)