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

A vernier caliper has two scales: the main scale and the vernier scale. The main scale divisions (MSD) are usually in millimeters, but sometimes the size of each division can vary depending on the instrument.

Given:

• 50 vernier scale divisions (VSD) = 49 main scale divisions (MSD)
• Zero error (Z.E.) = 0.04 mm (positive or negative depending on direction, here it shifts left, so positive zero error)
• When jaws close, the zero of vernier shifts left, and the 4th vernier division coincides with a main scale division.

Step 1: Find the length of one main scale division (MSD). Since 50 VSD = 49 MSD, then $1\, VSD = \frac{49}{50} \times 1\, MSD$

Step 2: The least count (L.C.) of the vernier caliper is the difference between one main scale division and one vernier scale division: $L.C. = 1\, MSD - 1\, VSD = 1\, MSD - \frac{49}{50} \times 1\, MSD = \frac{1}{50} \times 1\, MSD$

Step 3: Zero error is given as 0.04 mm, which equals the shift of zero on the vernier scale.

Step 4: Since the 4th vernier division coincides with a main scale division, the zero error can also be expressed as: $Z.E. = 4 \times L.C. = 4 \times \frac{1}{50} \times 1\, MSD = \frac{4}{50} \times 1\, MSD = 0.08 \times 1\, MSD$

Step 5: Equate this to the given zero error: $0.08 \times 1\, MSD = 0.04 \, mm$

Step 6: Solve for 1 MSD: $1\, MSD = \frac{0.04}{0.08} = 0.5 \text{ mm}$

Step 7: Now, find how many main scale divisions are there in 1 cm (10 mm): $\text{Number of MSD in 1 cm} = \frac{10 \text{ mm}}{0.5 \text{ mm}} = 20$

So, there are 20 main scale divisions in 1 cm.

Imagine you have a ruler with two scales: a big one (main scale) and a small sliding one (vernier scale). When you close the jaws of the vernier caliper, the zero on the small scale doesn't line up exactly with the zero on the big scale. Instead, it shifts a little to the left. This shift is called zero error. We also know how the small scale divisions relate to the big scale divisions. The question asks: How many big scale divisions make up 1 cm on the main scale? To solve this, we use the information about the zero error and the relationship between the scales to find the size of one main scale division and then find how many such divisions are in 1 cm.

Given:

• 50 vernier divisions = 49 main scale divisions
• Zero error = 0.04 mm
• Zero of vernier shifts left, and 4th vernier division coincides with main scale division

Step 1: Calculate length of one main scale division (MSD): $50 \times VSD = 49 \times MSD$ $1 \times VSD = \frac{49}{50} \times MSD$

Step 2: Calculate least count (L.C.): $L.C. = MSD - VSD = MSD - \frac{49}{50} MSD = \frac{1}{50} MSD$

Step 3: Zero error corresponds to 4 vernier divisions: $Z.E. = 4 \times L.C. = 4 \times \frac{1}{50} MSD = \frac{4}{50} MSD = 0.08 MSD$

Step 4: Given zero error is 0.04 mm, so: $0.08 MSD = 0.04 mm$ $MSD = \frac{0.04}{0.08} = 0.5 mm$

Step 5: Number of main scale divisions in 1 cm (10 mm): $\frac{10 mm}{0.5 mm} = 20$

Final Answer: There are 20 main scale divisions in 1 cm.