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(C) Let the two resistances be $R_1$ and $R_2$. When connected in parallel,the equivalent resistance $R_{eq}$ is given by $\frac{R_1 R_2}{R_1 + R_2} =

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$6/5$

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(C) Let the two resistances be $R_1$ and $R_2$. When connected in parallel,the equivalent resistance $R_{eq}$ is given by $\frac{R_1 R_2}{R_1 + R_2} = \frac{6}{8} = 0.75 \, \Omega$ ..... $(i)$.When one wire is broken,the circuit consists of only the remaining resistance. Given the new effective resistance is $2 \, \Omega$,let $R_1 = 2 \, \Omega$ ..... $(ii)$.Substituting $(ii)$ into $(i)$:$\frac{2 R_2}{2 + R_2} = 0.75$$2 R_2 = 0.75(2 + R_2)$$2 R_2 = 1.5 + 0.75 R_2$$1.25 R_2 = 1.5$$R_2 = \frac{1.5}{1.25} = \frac{150}{125} = 1.2 \, \Omega$ or $6/5 \, \Omega$.

Two resistances are joined in parallel whose resultant is $6/8 \, \Omega$. One of the resistance wires is broken and the effective resistance becomes $2 \, \Omega$. Then the resistance in ohm of the wire that got broken was:

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