Correct Answer - Option 3 : 27 times
Concept:
The approximate rating of the service life of a ball or roller bearing is based on the given fundamental equation.
\({\rm{L}} = {\left( {\frac{{\rm{C}}}{{\rm{P}}}} \right)^{\rm{k}}} \times {10^6}~{\rm{revolution}}\)
where
L is rating life, C is a basic dynamic load, P is the equivalent dynamic load
k = 3 for ball bearing
k = 10/3 for roller bearing
Calculation:
Given:
\(L = {\left( {\frac{C}{P}} \right)^3}⇒ L \propto { {\frac{1}{P^3}} }\\ \therefore \;\frac{{{L_1}}}{{{L_2}}} ={\left( {\frac{P_2}{P_1}} \right)^3}= {{\frac {1}{3}}^3}\\ \therefore\; {L_1} = \frac{{{L_2}}}{{27}}\)
⇒ L2 = 27 times L1
