Correct Answer - Option 2 : 103.4 kN

__Concept:__

**Load life equation:**

**F**^{a}**L = Constant**

where a = 3 For ball bearing and a = 10/3 For roller bearing.

For 90% reliability, L = 10^{6} cycles.

**Load life in cycles** is represented as

**L = 60 × L**_{H}** × N**, where L = Life in cycles, L_{H} = Life in hour, N = RPM.

__Calculation:__

__Given:__

Application factor (k) = 1.7, N = 1600 rpm, L_{H} = 18000 hours, F_{r} = 6.5 kN

Now, we know that

**F**^{a}L = Constant

i.e. \({\left( {\frac{F}{k}} \right)^{\frac{{10}}{3}}} \times {L_{90}} = {\left( {{F_r}} \right)^{\frac{{10}}{3}}} \times Lr\;\)

\({\left( {\frac{{\bf{F}}}{{1.7}}} \right)^{\frac{{10}}{3}}} \times {10^6} = {\left( {6.5} \right)^{\frac{{10}}{3}}} \times 60 \times 18000 \times 1600\)

**∴ F = 103.4 kN**