Question

the length of an elastic string is x metre when the tension is 8 N and y metre when the tension is 10 N.The length in metre when the tension is 18N is?

Answer 1

we have to apply hooke's law here...

F = Kx
where F is the tension and x is the increased in length..k is a constant

lets say when tension is 8N..the increased in length is x1..then

x = l + x1
l is the length of the string when there is no tention

in the same way when tension is 10N ..lets take x2..then

y = l + x2

now for the above instance we can also write

8 =kx1
10 = kx2
( hooke's law)

and for the final instance..when tension is 18N

18 = kx3

now we have five equations and we should find the length of the rope(Z) , when tension is 18 N

Z = l + x3

simplifications seems a littlle but complicated..but i strongly recomend you to try to find Z on your own before reading that part :)

8= kx1 ---(1)
10 = kx2 ---(2)

(1) / (2)

8/10 = x1/x2

x2= x1(10/8) ---(3)

afterwards

x = l + x1 ---(4)

y = l + x2
substituting (3)

y = l + x1(10/8) ---- (5)

now

(5) - (4)

y - x = x1(10/8) - x1
= x1(10/8 -1)
=x1(2/8)
x1 = 4( y - x)

now considering

18 = kx3 ---(6)

(6)/(1)

18/8 = x3/x1

x3 = x1(18/8)
= 4(y - x)(18/8) ---(7)

now to find "l"

from (4)
x1 = x - l ----(8)

from (5)
(10/8)x1 = y -l ---(9)

(9)/(8)

10/8 =(y -l) / (x-l)
10x - 10l = 8y -8l
2l = 10x - 8y
l = 5x -4y ---(10)

from (10) & (7)

Z = 5x - 4y +9(y-x)
= 5y - 4x

hope this will help.you :)

Answer 2

please refer to the attachment file given.

hope this answer will be helpful for you mate.

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please mark me as a brainliest......xD