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 \[=5{{l}_{2}}-5{{l}_{1}}+l\] \[=3{{l}_{2}}-2{{l}_{1}}\]You need to login to perform this action.
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