A) \[{{\text{w}}_{\text{A}}}>{{\text{W}}_{\text{B}}}\text{ and W}{{\text{ }\!\!\!\!\text{ }}_{A}}=\text{W}{{\text{ }\!\!\!\!\text{ }}_{B}}\]
B) \[{{\text{w}}_{\text{A}}}>{{\text{W}}_{\text{B}}}\text{ and W}{{\text{ }\!\!\!\!\text{ }}_{A}}\text{W}{{\text{ }\!\!\!\!\text{ }}_{B}}\]
C) \[{{\text{w}}_{\text{A}}}>{{\text{W}}_{\text{B}}}\text{ and W}{{\text{ }\!\!\!\!\text{ }}_{A}}\text{W}{{\text{ }\!\!\!\!\text{ }}_{B}}\]
D) \[{{\text{w}}_{\text{A}}}\text{}{{\text{W}}_{\text{B}}}\text{ and W}{{\text{ }\!\!\!\!\text{ }}_{A}}\text{W}{{\text{ }\!\!\!\!\text{ }}_{B}}\]
Correct Answer: B
Solution :
Work done in stretching a spring \[W=\frac{1}{2}k{{x}^{2}}\] Here, \[{{W}_{A}}=\frac{1}{2}{{k}_{A}}x_{A}^{2}\] and \[{{W}_{B}}=\frac{1}{2}{{k}_{B}}x_{B}^{2}\] As \[{{x}_{A}}={{x}_{B}}\] and \[{{k}_{A}}>{{k}_{B}}\] \[\therefore \] \[{{W}_{A}}>{{W}_{B}}\] Similarly, when forces are equal \[{{F}_{A}}={{F}_{B}}\] \[{{k}_{A}}{{x}_{A}}={{k}_{B}}{{x}_{B}}\] ?. (i) As \[{{k}_{A}}>{{k}_{B}}\] \[\therefore \] \[{{x}_{A}}<{{x}_{B}}\] ... (ii) Now, \[W_{A}^{}=\frac{1}{2}({{k}_{A}}{{x}_{A}}){{x}_{A}}\] and \[W_{B}^{}=\frac{1}{2}({{k}_{B}}{{x}_{B}}){{x}_{B}}\] From Eqs. (i) and (ii), we conclude that\[W_{A}^{}<W_{B}^{}\]
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