Question

In a vernier calliper, when both jaws touch each other, zero of the vernier scale shifts towards left and its $4^\text{th}$ division coincides exactly with a certain division on the main scale. If 50 vernier scale divisions equal to 49 main scale divisions and zero error in the instrument is 0.04 mm, then how many main scale divisions are there in 1 cm?

• A. 40
• B. 20
• C. 5
• D. 10

Answer 1

Answer: (C)

5

Answer 2

Since the 4th division coincides with the 3rd division on the main scale, we have:
$0.004 \, \text{cm} = 4 \, \text{VSD} - 3 \, \text{MSD}$
Given that $49 \, \text{MSD} = 50 \, \text{VSD}$.
Now, calculate the length of 1 MSD:
$1 \, \text{MSD} = \frac{1}{N} \, \text{cm}$
Using the zero error formula:
$0.004 = 4 \left( \frac{49}{50} \, \text{MSD} \right) - 3 \, \text{MSD}$
$0.004 = \frac{196}{50} \, \text{MSD} - 3 \, \text{MSD}$
Simplifying further:
$0.004 = \left( \frac{196 - 150}{50} \right) \, \text{MSD}$
$0.004 = \frac{46}{50} \times \frac{1}{N}$
Solve for $N$:
$N = \frac{46 \times 1000}{4 \times 50} = 230$
Therefore, there are 20 main scale divisions in 1 cm.