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 _____________. A. 81.00 B. 46.02 C. 60.01 D. 92.02

GATE 2016 · General Aptitude · Ratio Proportion · medium

Answer: The load for failure in 5000 cycles is L = 46.02 units (Answer B).

1. Find decay constant k by dividing the two equations: 100/10000 = exp(-k*(80-40)) => 0.01 = exp(-40k) => ln(0.01) = -40k => k = -ln(0.01)/40 = ln(100)/40 = 4.6052/40 = 0.11513
2. Find constant A using one data point: A = 10000 * exp(0.11513 * 40) = 10000 * exp(4.6052) = 10000 * 100 = 1,000,000
3. Solve for L when N = 5000: L = ln(1000000/5000) / 0.11513 = ln(200) / 0.11513 = 5.2983 / 0.11513 = 46.02