Current Electricity question for JEE Main (JEE Main 2022 (26 Jun, Shift 2)), with solution

Answer: (C) $5\ \Omega$

Reduce from the far end, always measuring resistance to $A$:

$P$–$A$: $5\ \Omega$. $Q$–$A$: $(5 + 5)\parallel10 = 5\ \Omega$.

$R$–$A$: $(5 + 5)\parallel10 = 5\ \Omega$. $S$–$A$: $(5 + 5)\parallel10 = 5\ \Omega$.

$B$–$A$: $(5 + 5)\parallel10 = 5\ \Omega$.

Solution by Sreeraj P, M.Sc Physics