If number of divisions on circular scale is 50 in a screw gauge and pitch of screw is 1 mm then find least count of screw gauge.
If number of divisions on circular scale is 50 in a screw gauge and pitch of screw is 1 mm then find least count of screw gauge.
Option 1 - <p>0.01 mm</p>
Option 2 - <p>0.002 mm</p>
Option 3 - <p>0.2 mm</p>
Option 4 - <p>0.02 mm</p>
1 Views|Posted 3 months ago
Asked by Shiksha User
1 Answer
A
Answered by
3 months ago
Correct Option - 4
Detailed Solution:
Kindly go through the solution
Similar Questions for you
From A to B the process is isobaric

= W = 2 × R (600 - 350)
= 500 R
Taking an Exam? Selecting a College?
Get authentic answers from experts, students and alumni that you won't find anywhere else.
On Shiksha, get access to
66K
Colleges
|
1.2K
Exams
|
6.9L
Reviews
|
1.8M
Answers
Learn more about...

Physics Motion in Straight Line 2025
View Exam DetailsMost viewed information
SummaryDidn't find the answer you were looking for?
Search from Shiksha's 1 lakh+ Topics
or
Ask Current Students, Alumni & our Experts
Have a question related to your career & education?
or
See what others like you are asking & answering






