-
Engineering and Architecture
Exams
Colleges
Predictors
Resources
-
Computer Application and IT
Quick Links
Colleges
-
Pharmacy
Colleges
Resources
-
Hospitality and Tourism
Colleges
Resources
Diploma Colleges
-
Competition
Other Exams
Resources
-
School
Exams
Top Schools
Products & Resources
-
Study Abroad
Top Countries
Student Visas
-
Arts, Commerce & Sciences
Exams
Colleges
Upcoming Events
Resources
-
Management and Business Administration
Exams
Colleges & Courses
Predictors
-
Learn
Online Courses
Engineering Preparation
Medical Preparation
-
Online Courses and Certifications
Top Streams
Specializations
- Digital Marketing Certification Courses
- Cyber Security Certification Courses
- Artificial Intelligence Certification Courses
- Business Analytics Certification Courses
- Data Science Certification Courses
- Cloud Computing Certification Courses
- Machine Learning Certification Courses
- View All Certification Courses
Resources
-
Medicine and Allied Sciences
Colleges
Predictors
Resources
-
Law
Resources
Colleges
-
Animation and Design
Predictors & Articles
Colleges
Resources
-
Media, Mass Communication and Journalism
Exams
Colleges
Resources
-
Finance & Accounts
Top Courses & Careers
Colleges
Get Answers to all your Questions
- Home
- Engineering
- A laboratory experiment involves measuring the thickness of a thin plastic sheet using a screw gauge. The main scale reading is 1.5 mm, and the circular scale is divided into 50 divisions. The pitc
A laboratory experiment involves measuring the thickness of a thin plastic sheet using a screw gauge. The main scale reading is 1.5 mm, and the circular scale is divided into 50 divisions. The pitch of the screw gauge (p) is 0.2 mm. The screw gauge has a negative zero error of −0.01 mm.
Calculate the actual thickness of the plastic sheet.
11.51mm
27.46mm
200 mm
20 mm
Answers (1)
Given: Main scale reading = 1.5 mm
Number of circular scale divisions = 50
Pitch of screw gauge (p) = 0.2 mm
Negative zero error (Z.E.) = −0.01 mm
The actual reading on the circular scale (C.S.R.) is given by:
Plugging in the values:
The corrected reading is the sum of the circular scale reading and the absolute value of the zero error:
Corrected reading = C.S.R. + |Z.E.|
Plugging in the values: Corrected reading = 10 mm + 0.01 mm = 10.01 mm
The total reading is the sum of the main scale reading and the corrected reading:
Total reading = Main scale reading + Corrected reading Plugging in the values:
Total reading = 1.5 mm + 10.01 mm = 11.51 mm
Therefore, the actual thickness of the plastic sheet is 11.51 mm.
Therefore, the correct option is 1.
JEE Main high-scoring chapters and topics
Study 40% syllabus and score up to 100% marks in JEE
Similar Questions
- 5 g of Na2SO4 was dissolved in x g of H2O. The change in freezing point was found to be 3.820C. If Na2SO4 is 81.5% ionised, the value of x (K<img alt="f" src="https://learn.c
- A capacitor is made of two square plates each of side 'a' making a very small angle <img alt="\alpha" src=
- A solution of m-chloroaniline, m-chlorophenol and m-chlorobenzoic acid in ethyl acetate was extracted initially with a saturated solution of NaHCO3 to give fraction A. The leftover organic phase was extracted with d
Trending Articles/News
Latest Question
- A sum of money under compound interest doubles itself in 4 years. In how many years will it become 16 times itself? Option: 1 1
- A certain loan amounts, under compound interest, compounded annually earns an interest of Rs.1980 in the second year and Rs.2178 in the third year. How much interest did it earn in the first year?<
