A wire of length ' $r$ ' and resistance $100 \Omega$ is divided into $10$ equal parts. The first $5$ parts are connected in series while the next $5$ parts are connected in parallel. The two combinations are again connected in series. The resistance of this final combination is:
NEET 2024, Medium
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$R =\frac{\rho l}{A}$
$R^{\prime} =\frac{\rho l}{10 A}=\frac{R}{10}$
$R_S =5 \times \frac{R}{10} \quad \text { [series] }$
$R_S =50$
$R_p =\frac{K}{50} \quad \text { [parallel] }$
$R_{e q} =R_S+R_p$
$=52 \Omega$
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