Question

A ball bearing operating at a load $F$ has $8000$ hours of life. The life of the bearing, in hours, when the load is doubled to $2F$ is

• A. $8000$
• B. $1000$
• C. $4000$
• D. $6000$

Answer 1

The correct option is B $1000$

Life of bearing
$L={\left(\frac{C}{W}\right)}^K\times {10}^6$
where $K=3$ for ball bearing
$\text{at load}F,L_1={\left(\frac{C}{F}\right)}^K\Rightarrow C=F{L}^{\frac{1}{K}}$
$\text{at load}2F{L}_2={\left(\frac{C}{2F}\right)}^3={\left(\frac{F{L}^{\frac{1}{3}}}{2F}\right)}^3={\left(\frac{L^{\frac{1}{3}}}{2}\right)}^3$
$=1000hours$.