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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 $10000 \ \text{cycles}$ for failure.The load for which the failure will happen in $5000 \ \text{cycles}$ is _____________.

  1. $40.00$
  2. $46.02$
  3. $60.01$
  4. $92.02$
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Best answer
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The number of cycles to failure decrease exponentially with an increase in load.

So, we have general equation

$y=ae^{-bx}$
where $y$ is number of cycles to failure, and $x$ is load.

At load of 80 units , it takes 100 cycles for failure.

$100=ae^{-80b} \qquad \to(1)$

when load is halved it takes $10000$ cycles for failure.

$10000=ae^{-40b} \qquad \to(2)$

Divide $(2)$ by $(1)$

$\implies e^{40b}=100$

$\implies b=\dfrac{\log_{e}100}{40} \qquad \to (3)$

At $5000$ cycles to failure

$5000=\large ae^{-xb} \qquad \to(4)$

divide $(2)$ by $(4)$

$\implies e^{b(x-40)}=2$

$\implies b(x-40)=\log_{e}2$

$\implies \dfrac{\log_{e}100}{40}\times(x-40)=\log_{e}2$  $($using $(3))$

$\implies x= 40 \times\dfrac{(\log_{e}2 +\log_{e}100 )}{\log_{e}100}$

$\qquad = 40 \times\dfrac{\log_{e}200}{\log_{e}100}$

$\qquad= 46.02$

Correct Answer: $B$
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Elimination method

 

Given :-

$\rightarrow$ for $80$ units it takes $100$ cycles

$\rightarrow$ for $40$ units(load is halved) it takes $10000$ cycles.

$\rightarrow$ for $5000$ cycles$\rightarrow$ __ units ?

 

My Approach :-

$\rightarrow$ Since we have to calculate for $5000$ cycles the answer would lie between $40$ and $80$.

$\rightarrow$ so, option a and d are eliminated.

$\rightarrow$ now if we see the mid values i.e.

$\rightarrow$ for $40$ and $80$ it would be $60$ and for $100$ and $10000$ it would be $1000$.

$\rightarrow$ so we can eliminate option c.

$\rightarrow$ the only option left is b which is the required answer.

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