Length of an elastic string is x metre when longitudinal tension is 3Nand y metre when it is 5 N. If length of string is found to be 2. By - 1.5 x, calculate tension in the string in N.

Answer

8

Explanation

Let L be original length of string and k be force constant of string. New final length = initial length + elongation
`L. = L + F/K`
When tension of 3N is applied,
`x = L + 3/k`....... (i)
When tension of 5N is applied,
`x = L + 3/k`....(ii)
Subtracting equation (ii) from (i),
`x -y = 3/k - 5/k`
`therefore k = (-2)/(x-y) = 2/(y-x)`
Substituting value of k in equation (i),
`x = L + 3/(2//y-x) = L + (3(y-z))/2`
`therefore L = x -(3(y-x))/2 = (2x-3y + 3z)/2 = (5x-3y)/2`
Let T be tension when length is `2.5 y- 1.5 x`
`therefore 2.5y - 1.5 x = L + T/k`
`2.5 y - 1.5 x - (5x - 3y)/2 + ((y-x)T)/2`
`therefore 5y - 3x = 5x - 3y + (y-x)T`
`therefore 8(y-x) = (y-x)T rArr T = 8 N`