Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
91 views
in Physics by (55.3k points)

A given screw gauge has a least count of 0.002 cm and there are 50 divisons on its circular scale. When the jaws, of this screw guage, are just in contact with each other, the 7th division of the circular scale lies along its ‘reference line’.

When this scrw gauge, is used to measure the thickness of a thin mica sheet, the main scale reads 0.7 cm and the 21st division, of its circular scale, coincides with the ‘reference line’.

The ‘pitch’ of this screw gauge, and the thickness of the mica sheet, are respectively, equal to 

(1) 0.2 cm and 0.717 cm 

(2) 0.1 cm and 0.756 cm 

(3) 0.2 cm and 0.728 cm 

(4) 0.2 cm and 0.756 cm

Please log in or register to answer this question.

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...