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In a vernier calliper, when both jaws touch each other,... - Solution & Explanation | iExplain

Detailed Explanation

Key ideas you must know\n\n1. Relation between scales\n Given: $50\,\text{VSD} = 49\,\text{MSD}$ , where VSD = vernier-scale division and MSD = main-scale division.\n\n2. Least Count (LC)\n The smallest length you can read is\n $LC = 1\,\text{MSD} - 1\,\text{VSD}$ \n Substituting $1\,\text{VSD} = \frac{49}{50}\,\text{MSD}$ ,\n $LC = 1\,\text{MSD} - \frac{49}{50}\,\text{MSD} = \frac{1}{50}\,\text{MSD}$ \n\n3. Zero error from coincidence\n The vernier zero is left of the main zero, so reading is negative.\n If the 4th vernier division coincides, magnitude of zero error is\n $ZE = 4 \times LC$ \n Given: actual zero error magnitude = $0.04\,\text{mm}$ .\n\n4. Find one MSD\n $4 \times \frac{1}{50}\,\text{MSD} = 0.04\,\text{mm}$ \n Solving gives $1\,\text{MSD} = 0.5\,\text{mm}$ .\n\n5. Convert to divisions per cm\n 1 cm = 10 mm.\n $\text{Number of MSD in 1 cm} = \frac{10\,\text{mm}}{0.5\,\text{mm}} = 20$ \n\nThat’s why the main scale must have 20 divisions in every centimetre.

Simple Explanation (ELI5)

🧒 Imagine a ruler-with-a-magnifying-glass 📏🔍\n\n1. A vernier calliper is just a fancy ruler. The big scale is the main scale and the tiny sliding scale is the vernier.\n2. When you shut its jaws, the two zeros should kiss each other. In our tool, the vernier zero sits a bit to the left, so there is a zero error.\n3. We notice that the 4th tiny mark (4th vernier division) lines up perfectly with a mark on the main scale.\n4. We are also told "50 of those tiny marks are as long as 49 big marks".\n5. Using these two clues, we first find exactly how big one big mark is.\n6. Once we know that size, we simply ask, "How many of those big marks can fit inside 1 cm (which is 10 mm)?"\n7. The maths says the answer is 20 main-scale divisions in 1 cm.

Step-by-Step Solution

Step-by-step calculation\n\n1. Given relation\n\n $50\,\text{VSD} = 49\,\text{MSD}$ \n\nSo\n\n $1\,\text{VSD} = \frac{49}{50}\,\text{MSD}$ \n\n2. Least Count (LC)\n\n $LC = 1\,\text{MSD} - 1\,\text{VSD} = 1\,\text{MSD} - \frac{49}{50}\,\text{MSD} = \frac{1}{50}\,\text{MSD}$ \n\n3. Use zero error information\n\nThe 4th vernier mark lines up, hence\n\n $ZE = 4 \times LC = 0.04\,\text{mm}$ \n\nSubstitute $LC$ :\n\n $4 \times \frac{1}{50}\,\text{MSD} = 0.04\,\text{mm}$ \n\n4. Find one main-scale division (MSD)\n\n $\frac{4}{50}\,\text{MSD} = 0.04\,\text{mm}\ \Rightarrow\ \text{MSD} = 0.5\,\text{mm}$ \n\n5. Number of MSD in 1 cm\n\n $\text{Count} = \frac{10\,\text{mm}}{0.5\,\text{mm}} = 20$ \n\n\n\n\nFinal Answer: (\boxed{20}) main-scale divisions per centimetre.

Examples

Example 1

Measuring the diameter of a thin copper wire using a vernier calliper

Example 2

Checking the thickness of a credit card for ATM machine specifications

Example 3

Finding the internal diameter of a small test tube in a chemistry lab

Example 4

Calibrating the gap of a spark plug in automotive service

Visual Representation

References

[1]NCERT Physics Class XI – Chapter 2: Units and Measurements
[2]Concepts of Physics Vol 1 by H. C. Verma – ‘Measurement’ section
[3]Practical Physics by S. L. Arora – Vernier Calliper experiment
[4]IIT JEE Previous Year Papers – Experimental Physics questions