GATE CSE 2016 | Set 1 | GA | Question: 10

Question

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 _____________.

• A. $40.00$
• B. $46.02$
• C. $60.01$
• D. $92.02$

Comments

Instead of solving full equations:

1. Take logs: exponential means we interpolate on a log scale.

   • log⁡(100)=2, log⁡(10000)=4.

   • Target: log⁡(5000)≈3.7.

2. This is closer to 10000 than 100, so load must be closer to 40 than 80.

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@Ujjwal_Nikam Best Method

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Can anyone explain in a easier way?

Answer 1

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: $B$

Comments

@PraveenSaini Sir

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

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@Praveen Saini sir

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

It is neccesary to consider constant a ????

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Just to be on the safe side!

Answer 2

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.

Comments

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

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how mid is 1000 ?
did u meant GM?

 

Answer 3

This is an interpolation problem : 

load decreases  --> cycles increase 

two data points (load values) are given and  middle value is asked : (80,40)

by rule of interpolation =>  logbase10(N) = a*L + b

here 'N' will be that value which is increasing exponentially in powers of 10 : 100 --> 1000 --> 10,000

so, logbase10(100) = 80a + b 

2 = 80a + b    --1

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logbase10(10000) = 40a+b

4 = 40a + b    --2

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by 1 & 2 

a = -0.05 ; b = 6

logbase10(5000) = x*a + b

3.69 = -0.05*x + 6 

0.05x = 2.3011

x=46.02

option (b)