Correct Answer - Option 1 : below 0.5

__Concept:__

The approximate rating of service life of a ball or roller bearing is based on the fundamental equation.

\({{L}} = {\left( {\frac{{{C}}}{{{W}}}} \right)^{{k}}} \times {10^6}\ {{revolution}}\)

where L is rating life, C is basic dynamic load, W is equivalent dynamic load

k = 3 for ball bearing

k = 10/3 for roller bearing

The relationship between the life in revolutions (L) and the life in working hours (LH) is given by:

L = 60 N.L_{H } revolutions

where N is the speed in rpm

__Calculation:__

Given: C = 35 kN, W = 45 kN

\({\rm{L}} = {\left( {\frac{{\rm{C}}}{{\rm{W}}}} \right)^{\rm{k}}} \times {10^6}{\rm{\;rev}} = {\left( {\frac{{35}}{{45}}} \right)^3} \times {10^6} = 0.47 \times {10^6}\;rev\)