Question

The life of a ball bearing at a load of 10 kN is 8000 hour. Its life in hours, If the load is increased to 20kN, keeping all other conditions same, is

• A. 500
• B. 1000
• C. 2000
• D. 4000

Answer 1

The correct option is B 1000

$L=\left(\frac{C}{P}\right)[k=3forballbearingandk=\frac{10}{3}forrollerbearing]$

$\frac{L_2}{L_1}=(\frac{P_1}{P_2}{)}^3=(\frac{10}{20}{)}^3=\frac{1}{8}$

$L_2=\frac{8000}{8}=1000hours$