question_answer

An elastic spring has a length \[{{l}_{1}}\] when it is stretched with a force of 2 N and a length of \[{{l}_{2}}\] when it is stretched with a force of 3N. What will be the length of the spring if it is stretched with force of 5N?

A) \[(l{{  }_{1}}+{{l}_{2}})\]            

B) \[\frac{1}{2}(l{{  }_{1}}+{{l}_{2}})\]   

C) \[(3l{{  }_{2}}-2{{l}_{1}})\]        

D) \[(3l{{  }_{1}}-2{{l}_{2}})\]

Correct Answer: C

Solution :

In equilibrium position for a spring reaction.         \[k{{y}_{0}}=mg\] where \[{{y}_{0}}\]is the extension in spring.                           Let \[l\] be the length of the spring.      Then for \[{{l}_{1}}\]                       Extension \[=({{l}_{1}}-l)\] \[\therefore \]  \[k({{l}_{1}}-l)=2\]                           ?(i) Similarly for\[{{l}_{2}}\] \[k({{l}_{2}}-l)=3\] Eliminating fc from Eqs. (i) and (ii), we get \[l=3{{l}_{1}}-2{{l}_{2}}\] Moreover adding Eqs. (i) and (ii),we get \[k[{{l}_{1}}+{{l}_{2}}-2l]=5\] Substituting the value of I in above equation we get \[k[5{{l}_{2}}-5{{l}_{1}}]=5\] \[5{{l}_{2}}-5{{l}_{1}}\]is the extension for 5 N force \[\therefore \] Total length \[=5{{l}_{2}}-5{{l}_{1}}+1\]                                 \[=3{{l}_{2}}-2{{l}_{1}}\]