The length of an elastic string is ‘a’ meter when the tension is 4 N and ‘b’ meter when the tension is 5 N. The length, (in meter), when the tension is 9 N, is

Let l be the natural length and k=YAIbe the force constant of wire. Then,a=l+4k ⋯ (i) ∵F=kΔL or Δl=Fkand b=l+5kOn solving Eqs. (i) and (ii), we getor  1k=(b−a) and  l=(5a−4b)Now, when T = 9N L′=l+9k=(5a−4b)+9(b−a)=(5b−4a)