Correct Answer - Option 1 : 65 kN

__Concept:__

**Equivalent bearing load (W)** is given by:

**W = XVFr + YFa**

where X = radial factor, V = service factor, F_{r} = radial load, Y = thrust factor, F_{a} =thrust load

**Life of a bearing is given by:**

\({L_{10}} = {\left( {\frac{C}{W}} \right)^n}\)

where

L_{10} = basic life rating in million revolutions

C = dynamic load-carrying capacity

W = equivalent dynamic load

n = constant = 3 for ball bearing and \(\frac{{10}}{3}\) for roller bearing

__Calculation:__

__Given:__

F_{r} = 7000 N, F_{a} = 2100 N, X = 0.65, V = 1, Y = 3.5, L_{10} = 160 million revolutions

Equivalent bearing load (W) is given by:

**W = XVFr + YFa**

W = (0.65 × 1 × 7000) + (3.5 × 2100) = 11900 N

Now,

Life of a bearing is:

\({L_{10}} = {\left( {\frac{C}{W}} \right)^n}\)

\(C = W \times {\left( {{L_{10}}} \right)^{\frac{1}{n}}}\)

C = 11900 × (160)^{1/3} = 64603.14 N = **64.6 kN ≈ 65 kN**