A) 2%
B) 4%
C) 8%
D) 1%
Correct Answer: B
Solution :
Volume of wire remains constant when the wire is stretched. Thus, \[\frac{1}{{{k}_{s}}}=\frac{1}{k}\,\left( \frac{1}{1-1/2} \right)\] area of cross-section \[\Rightarrow \] where, V = Volume of wire, \[\frac{1}{{{k}_{s}}}=\frac{2}{k}\Rightarrow \,{{k}_{s}}=\frac{k}{2}\] Differentiating, \[\Rightarrow \] Taking \[T=2\pi \sqrt{\frac{m}{{{k}_{s}}}}\,\Rightarrow \,T=2\pi \sqrt{\frac{2m}{k}}\] and \[{{f}_{approach}}=\left( \frac{v}{v-{{v}_{s}}} \right)\,{{f}_{0}}={{f}_{0}}={{f}_{1}}\] as fractional errors (or % cross) \[{{f}_{recede}}=\left( \frac{v}{v+{{v}_{s}}} \right){{f}_{0}}={{f}_{2}}\,(<{{f}_{1}})\]
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