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Tamil Nadu Board of Secondary EducationHSC Commerce Class 12

The following data gives the melting point of a alloy of lead and zinc where ‘t’ is the temperature in degree c and P is the percentage of lead in the alloy. P 40 50 60 70 80 90 T 180 204 22 - Business Mathematics and Statistics

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Question

The following data gives the melting point of a alloy of lead and zinc where ‘t’ is the temperature in degree c and P is the percentage of lead in the alloy.

P 40 50 60 70 80 90
T 180 204 226 250 276 304

Find the melting point of the alloy containing 84 percent lead.

Chart
Sum

Solution

Since the required value is at the end of the table

Apply backward interpolation formula.

To find T at p = 84

`"T"_(("p" = "p"_0 + "nh")) = "T"_"n" + "n"/(1!) ∇"T"_"n" + ("n"("n" + 1))/(2!) ∇^2"T"_0 + ("n"("n" + 1)("n"  2))/(3!) Delta^3"T"_0 + .......`

To find T at P = 84

Pn + nh = 84

90 + n(10) = 84

10n = 84 – 90

10n = – 6

⇒ n = `(-6)/10`

n = – 0.6

P T `Delta"T"` `Delta^2"T"` `Delta^3"T"` `Delta^4"T"` `Delta^5"T"`
40 180          
    24        
50 204   – 2      
    22   4    
60 226   2   – 4  
    24   0   4
70 250   2   0  
    26   0    
80 276   2      
    28        
90 304          

`"T"_((84)) = 304 + ((-0.6))/(1!) (28) + ((-0.6)(-0.6 + 1))/(2!) (2)+ ((-0.6)(-0.6 + 1)(-0.6 + 2))/(3!) (0) + ((-0.6)(-0.6 + 1)(-0.6 + 2)(0.6 + 3))/(4!) (0) + ((-0.6)(-0.6 + 1)(-0.6 +2)(-0.6 + 3)(-0.6 + 4))/(5!) (4) +`

= 304 – 16.8 – 0.24 – 0.091392

= 304 – 17.131392

= 286.86

Hence the melting point of the alloy is 286.86°c.

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Interpolation
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Chapter 5: Numerical Methods - Exercise 5.2 [Page 119]

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Samacheer Kalvi Business Mathematics and Statistics [English] Class 12 TN Board
Chapter 5 Numerical Methods
Exercise 5.2 | Q 7 | Page 119
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