Correct Answer - Option 2 : 44 kN

__Concept:__

Load-life relationship:

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

\({\rm{L_{10}}} = {\left( {\frac{{\rm{C}}}{{\rm{P}}}} \right)^{\rm{k}}} \)

L10 = Rated bearing life (in million revolutions)

C = Dynamic load capacity.

P = Load acting in bearing.

k = 3 for ball bearing and k = 10/3 for the roller bearing.

The relation between life in million revolutions and life in hours is given by:

\(L_{10}=\frac{L_{10h}\;\times \;N \;\times \;60}{10^6}\)

where, L_{10h }= rated bearing life in hours and N = speed of rotation in rpm.

__Calculation: __

__Given:__

N = 1500 rpm, P = 6 kN, L_{10h} = 8100 hours, roller bearing k = 10/3

We know that;

\({\rm{L_{10}}} = {\left( {\frac{{\rm{C}}}{{\rm{P}}}} \right)^{\rm{k}}} \times {10^6}{\rm\;{revolution}}\) and;

\(L_{10}=\frac{L_{10h}\;\times \;N \;\times \;60}{10^6}\)

Combining both equations:

\(\frac{L_{10h}\;\times \;N \;\times \;60}{10^6}= {\left( {\frac{{\rm{C}}}{{\rm{6}}}} \right)^{{\frac{10}{3}}}} \)

\(\frac{8100\;\times \;1500 \;\times \;60}{10^6}= {\left( {\frac{{\rm{C}}}{{\rm{6}}}} \right)^{{\frac{10}{3}}}} \)

C = 43.34 kN