Six equal resistances are connected between points $P, Q$ and $R$ as shown in figure. Then net resistance will be maximum between
JEE MAIN 2013, Diffcult
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Resistance between $\mathrm{P\,and} \,\mathrm{Q}$
$r_{P Q}=r_{11}\left(\frac{r}{3}+\frac{r}{2}\right)=\frac{r \times \frac{5}{6} r}{r+\frac{5}{6} r}=\frac{5}{11} r$
Resistance between $Q$ and $R$
$\mathrm{r}_{\mathrm{QR}}=\frac{\mathrm{r}}{2} 11\left(\mathrm{r}+\frac{\mathrm{r}}{3}\right)=\frac{\frac{\mathrm{r}}{2} \times \frac{4}{3} \mathrm{r}}{\frac{\mathrm{r}}{2}+\frac{4}{3} \mathrm{r}}=\frac{4}{11} \mathrm{r}$
Resistance between $\mathrm{P}$ and $\mathrm{R}$
${{\rm{r}}_{{\rm{PR}}}} = \frac{{\rm{r}}}{3}11\left( {\frac{{\rm{r}}}{2} + {\rm{r}}} \right) = \frac{{\frac{{\rm{r}}}{3} \times \frac{3}{2}{\rm{r}}}}{{\frac{{\rm{r}}}{3} + \frac{3}{2}{\rm{r}}}} = \frac{3}{{11}}{\rm{r}}$
Hence, it is clear that $\mathrm{r}_{\mathrm{PQ}}$ is maximum
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