$\frac{5}{R}=\frac{l_{1}}{100-l_{1}}$  $\ldots(i)$

In the second case,

Substituting this value in eqn. $(i)$, we get

${\frac{5}{R}=\frac{25}{75}} $

${R=\frac{375}{25}\, \Omega=15\, \Omega}$

At balance point

$\frac{5}{(R / 2)}=\frac{1.6 l_{1}}{100-1.6 l_{1}}$         ....$(ii)$

Divide eqn. $(i)$ by eqn. $(ii),$ we get

${\frac{1}{2}=\frac{100-1.6 l_{1}}{1.6\left(100-l_{1}\right)}}$

${160-1.6 l_{1}=200-3.2 l_{1}} $

${1.6 l_{1}=40 \quad \text { or } \quad l_{1}=\frac{40}{1.6}=25\, \mathrm{cm}}$