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
133 views
in General by (114k points)
closed by
The speed of a journal in a journal bearing is 1500 r.p.m. The absolute viscosity of the lubricant is 1.2 kg-m/s and the bearing pressure on the projected bearing area is 1N/m2. Determine the bearing characteristics number
1. 30
2. 25
3. 15
4. 20

1 Answer

0 votes
by (115k points)
selected by
 
Best answer
Correct Answer - Option 1 : 30

Concept:

The bearing characteristic number is calculated by,

\(\frac{{μ N_s}}{P}=\rm Bearing~characteristic~number\)

Where, μ = absolute viscosity of lubricant (kg/m.s), N = speed of the journal (rps), P = bearing pressure (N/mm2)

Calculation:

Given:

μ = 1.2 kg-m/s, N = 1500 r.p.m, P = 1 N/m2

\(\rm Bearing~characteristic~number=\frac{{μ N_s}}{P}\;\)

\(= \frac{1.2\times1500}{60\times1} = 30\;\)

The bearing characteristics number is 30.

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

...