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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. $92.02$

edited | 7.7k views
0

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: $B$
by Veteran (57k points)
edited
0
why cant we put 'b' value directly in equation (4)  ??
+1
You can
+1

ok... got it sir...
i didnt used constant 'a' in my equation...

0
What is that general equation about sir @Praveen Saini??
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The number of cycles to failure (y) decrease exponentially with an increase in load (x).
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how did you form this equation??like why are u taking two constants and 'e' into picture??
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it is given as "The number of cycles to failure decrease exponentially with an increase in load."
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sir,why do we need 2 contants??why cant just 'b'??because i am taking only one constant but not getting the answer.
+1
what is the difference between these two equations:

y=ab^x .....(1)

y=ae^bx ....(2)

If they are same then how to solve the question using eq (1)..
+5
And Why $\large y = ka^{-x}$ gives wrong answer? Even this represents exponential decay.
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@PraveenSaini Sir

Is it necessarry to take two constants because we are given 2 equations?
0

@ sir

And Why $y= ka^{-x}$gives wrong answer? Even this represents exponential decay.

It is neccesary to consider constant a ????

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.

by Boss (24.2k points)
edited by
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where havi i written linear decrement ?
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sry i was wrong.

please elaborate how u got 1000 for 60?
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i took mid value of (40 and 80)  and accordingly took mid value between $10000$ and $100$. (it is not exact value ...i have just estimated it using the given part). Just think that if $40$ means $10000$ and $80$ means $100$ so can $46$ mean $1000$ or $60$ means $1000$ keeping in mind that as $40$ increases to $80$ , $10000$ decreases to $100$
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Yes thank u
B) 46.02
by Loyal (9.7k points)