Discussion
In a process, the number of cycles to failure decreases exponentially with an increase in load. At a load of 80 units, it takes 100 cycles for failure. When the load is halved, it takes 10000 cycles for failure. The load for which the failure will happen in 5000 cycles is ________.
Answer: B
The load is 46.02.
Write the cycles to failure as \(N = Ae^{-kL}\). From the two data points:
\[ \begin{aligned} \frac{N(40)}{N(80)} &= e^{40k} = \frac{10000}{100} = 100 \\ k &= \frac{\ln 100}{40} \end{aligned} \]
For 5000 cycles, half of the 10000 at \(L = 40\):
\[ \begin{aligned} e^{-k(L - 40)} &= \tfrac{1}{2} \\ L &= 40 + \frac{40\ln 2}{\ln 100} = 40 + 6.02 = 46.02 \end{aligned} \]
5000 cycles lies between 10000 (at 40 units) and 100 (at 80 units), so the load must lie between 40 and 80: 40.00 itself gives 10000 cycles and 92.02, above 80 units, would give fewer than 100. 60.01 is far too high because the exponential falls fastest at first.
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