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
+1 vote
121 views
in Physics by (88.6k points)
closed by
A carbon resistor of `(47 +- 4.7) k Omega` is to be marked with rings of different colours for its identification. The colour code sequence will be
A. Yellow-Green-Violet-Gold
B. Green-Orange-Violet-Gold
C. Yellow-Violet-Orange-Silver
D. Violet-Yellow-Orange-Silver

1 Answer

0 votes
by (90.6k points)
selected by
 
Best answer
Correct Answer - C
`(47+-4.7) k Omega=47xx10^(3) +-10%`
Colour code sequence: Yellow-Violet-Orange-Silver

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

...