Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 21: Problem 2762

What is the least count of commonly available vernier? (A) \(0.01 \mathrm{~cm}\) (B) \(0.001 \mathrm{~cm}\) (C) \(0.0001 \mathrm{~cm}\) (D) \(0.1 \mathrm{~cm}\)

Short Answer

Expert verified
The least count of a commonly available vernier is (A) \(0.01 \,cm\).

Step by step solution

01

Recognize the information given

In this problem, we are given four choices for the least count of a commonly available vernier. We need to find the correct one by knowing the actual least count of the vernier scale.
02

Calculate the least count of the vernier scale

To find the least count, we need to know the difference between the main scale and the vernier scale. The least count of the vernier scale can be calculated using the formula: \(Least \, Count = \frac{1 \, MSD}{10}\) where MSD is the least count of the main scale, which is usually 1 mm or 0.1 cm.
03

Compare the calculated least count with the given choices

Using the formula from Step 2, we can calculate the least count: \(Least \, Count = \frac{1 \, MSD}{10} = \frac{0.1 \, cm}{10} = 0.01 \, cm\)
04

Choose the correct answer

By comparing the calculated least count with the given choices, we can see that the correct answer is: (A) \(0.01 \,cm\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

If observed reading is OR, corrected reading is CR, zero error in \(\mathrm{ZE}\) and zero correction in \(\mathrm{ZC}\), then what will be the possibility? (A) \(\mathrm{CR}=\mathrm{OR}+\mathrm{ZC}\) and \(\mathrm{ZE}=\mathrm{CR}-\mathrm{OR}\) (B) \(\mathrm{CR}=\mathrm{OR}+\mathrm{ZE}\) and \(\mathrm{ZC}=\mathrm{CR}-\mathrm{OR}\) (C) \(\mathrm{CR}=\mathrm{OR}-\mathrm{ZC}\) and \(\mathrm{ZE}=\mathrm{OR}-\mathrm{CR}\) (D) \(\mathrm{CR}=\mathrm{OR}-\mathrm{ZE}\) and \(\mathrm{ZC}=\mathrm{CR}-\mathrm{OR}\)

What is the least count of the vernier calipers? (A) Smallest division on the vernier scale. (B) difference of the smallest division on the main scale and the smallest division on the vernier scale. (C) sum of the smallest division on the main scale and the smallest division on the vernier scale. (D) smallest division on the main scale.

When the zero mark on the vernier scale lies towards the left side of the zero mark of the main scale, when the jaws are connect, then what will be the zero error? (A) zero error is positive (B) zero error is negative (C) zero correction is positive (D) zero error does not exist

In an usual vernier, 10 vernier scale divisions, coin side with 8 main scale divisions, then what is the least count of the vernier? (A) \(0.1 \mathrm{~mm}\) (B) \(0.2 \mathrm{~mm}\) (C) \(0.8 \mathrm{~mm}\) (D) \((1 / 8) \mathrm{mm}\)

In an unusual vernier, 9 vernier scale divisions coincide with 8 main scale division, then what is the least count of the vernier? (A) \((8 / 9) \mathrm{mm}\) (B) \((1 / 9) \mathrm{mm}\) (C) \((1 / 17) \mathrm{mm}\) (D) \((1 / 8) \mathrm{mm}\)
See all solutions

Recommended explanations on English Textbooks

Essay Writing Skills

Read Explanation

Pragmatics

Read Explanation

Essay Prompts

Read Explanation

Language and Social Groups

Read Explanation

Research and Composition

Read Explanation

Rhetoric

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free