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)  \[(3{{l}_{2}}-2{{l}_{1}})\]       

D)  \[(3{{l}_{1}}-2{{l}_{2}})\]

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 \[=5{{l}_{2}}-5{{l}_{1}}+l\] \[=3{{l}_{2}}-2{{l}_{1}}\]