A) \[\frac{{{v}_{2}}-{{v}_{1}}}{k-1}\]
B) \[\frac{k{{v}_{1}}-{{v}_{2}}}{k-1}\]
C) \[\frac{k{{v}_{2}}-{{v}_{1}}}{k-1}\]
D) \[\frac{{{v}_{2}}-{{v}_{1}}}{k}\]
Correct Answer: B
Solution :
When frequency is \[{{v}_{1}}\], \[h{{v}_{1}}=h{{v}_{0}}+\frac{1}{2}\,mu_{1}^{2}\] When frequency is \[{{v}_{2}},\] \[h{{v}_{2}}=h{{v}_{0}}+\frac{1}{2}mu_{2}^{2}\] \[\because \] \[\frac{1}{2}\,mu_{1}^{2}=\frac{1}{k}\,\left( \frac{1}{2}\,mu_{2}^{2} \right)\] \[\therefore \] from Eq. (i) \[h{{v}_{1}}=h{{v}_{0}}+\frac{1}{2k}mu_{2}^{2}\] or \[\frac{1}{2}\,mu_{2}^{2}=kh{{v}_{1}}-kh{{v}_{0}}\] From Eqs. (ii) and (iv) \[h{{v}_{2}}=h{{v}_{0}}+kh{{v}_{1}}-kh{{v}_{0}}\] or \[{{v}_{0}}\,(1-k)={{v}_{2}}-k{{v}_{1}}\] or \[{{v}_{0}}=\frac{k{{v}_{1}}-{{v}_{2}}}{k-1}\]
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