b
At balance point

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

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

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

\({160-1.6 l_{1}=200-3.2 l_{1}} \)

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