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(D) Let the resistances of the two wires be $R_1$ and $R_2$.When connected in parallel,the equivalent resistance $R_p$ is given by $\frac{1}{R_p} = \f

Vedclass · Accepted Answer

$3$

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(D) Let the resistances of the two wires be $R_1$ and $R_2$.When connected in parallel,the equivalent resistance $R_p$ is given by $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}$,which implies $R_p = \frac{R_1 R_2}{R_1 + R_2}$.Given $R_p = \frac{6}{5} \ \Omega$,we have $\frac{R_1 R_2}{R_1 + R_2} = \frac{6}{5}$.If one wire breaks,the effective resistance becomes the resistance of the remaining wire. Let $R_1 = 2 \ \Omega$.Substituting $R_1 = 2$ into the parallel equation:$\frac{2 R_2}{2 + R_2} = \frac{6}{5}$$10 R_2 = 6(2 + R_2)$$10 R_2 = 12 + 6 R_2$$4 R_2 = 12$$R_2 = 3 \ \Omega$.Thus,the resistance of the broken wire is $3 \ \Omega$.

Two resistance wires on joining in parallel the resultant resistance is $\frac{6}{5} \ \Omega$. One of the wires breaks,the effective resistance is $2 \ \Omega$. The resistance of the broken wire is ............ $\Omega$.

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