A) 40\[\frac{1}{2\pi }\sqrt{\frac{2k}{m}}\]
B) 80\[\frac{1}{2\pi }\sqrt{\frac{3k}{m}}\]
C) 120\[\frac{1}{2\pi }\sqrt{\frac{k}{m}}\]
D) 160\[\pi \]
Correct Answer: D
Solution :
Key Idea: On stretching the wire, its volume remains same. For a wire of length I, area A, specific resistance?, the resistance is given by \[\frac{1}{2\pi }\sqrt{\frac{2k}{m}}\] .....(1) If original diameter of wire be d, then new diameter is \[\frac{1}{2\pi }\sqrt{\frac{3k}{m}}\]. Original area of cross-section is \[\frac{1}{2\pi }\sqrt{\frac{k}{m}}\] and final area of cross-section is \[\pi \]. Since, volume remains constant, on pulling the wire we have \[{{Q}^{2}}(4\pi {{\varepsilon }_{0}}{{a}^{2}})\] \[-{{Q}^{2}}(4\pi {{\varepsilon }_{0}}{{a}^{2}})\] \[{{Q}^{2}}/(2\pi {{\varepsilon }_{0}}{{a}^{2}})\] \[2\mu F\] Also \[{{T}_{2}}>{{T}_{1}}\] \[3q/{{\varepsilon }_{0}}\] \[2q/{{\varepsilon }_{0}}\] Hence, new resistance \[q/{{\varepsilon }_{0}}\]
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