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
65 views
in General by (111k points)
closed by
Equilibrium cant for B.G. (with usual notations) is:-
1. 1.315 ν2/R
2. 0.80 ν2​/R
3. 0.60 ν2​/R
4. 0.70 ν2​/R

1 Answer

0 votes
by (108k points)
selected by
 
Best answer
Correct Answer - Option 1 : 1.315 ν2/R

Concept:

Equilibrium cant \(= \frac{{{\rm{G}}{{\rm{V}}^2}}}{{127{\rm{R}}}}\)

V - Design speed (kmph)

R - Radius of curvature (m)

For B.G track G = 1.676

\(\begin{array}{l} {\rm{Cant}} = \frac{{1.676{\rm{\;}}{{\rm{V}}^2}}}{{127{\rm{\;R}}}}{\rm{m}}\\ {\rm{e}} = \frac{{1.315{\rm{\;}}{{\rm{V}}^2}}}{{\rm{R}}}{\rm{cm}} \end{array}\)

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

...