Question

The length of a rubber cord is $ {{l}_{1}} $ metre when the tension is $ 4\, N $ and $ {{l}_{2}} $ metre when the tension is $ 6\, N $ . The length when the tension is $9\,N$, is

• A. $ (2.5{{l}_{2}}-1.5{{l}_{1}})m $
• B. $ (6{{l}_{2}}-1.5{{l}_{1}})m $
• C. $ (3{{l}_{2}}-2{{l}_{1}})m $
• D. $ (3.5{{l}_{2}}-2.5{{l}_{1}})m $
• E. $ (2.5{{l}_{2}}+1.5{{l}_{1}})m $

Answer 1

Answer: (A)

Let the original unstretched length be $ l. $
$ Y=\frac{Stress}{Strain}=\frac{T/A}{\Delta l/l}=\frac{T}{A}\times \frac{l}{\Delta l} $
Now $ Y=\frac{4}{A}\frac{l}{({{l}_{1}}-l)}=\frac{6}{A}\frac{l}{({{l}_{2}}-l)} $
$ =\frac{9}{A}\frac{l}{({{l}_{3}}-l)} $
$ \therefore $ $ 4({{l}_{3}}-l)=9({{l}_{1}}-l) $
$ \Rightarrow $ $ 4{{l}_{3}}+5l=9{{l}_{1}} $ .... (i)
Again, $ 6({{l}_{3}}-l)=9({{l}_{2}}-l) $
$ \Rightarrow $ $ 2{{l}_{3}}+l=3{{l}_{2}} $ ...(ii)
Solve Eqs. (i) and (ii), we obtain
$ {{l}_{3}}=(2.5{{l}_{2}}-1.5{{l}_{1}}) $