Given l1 = 70 cm
l2 = 50 cm
E = 2
r = ?
For balancing length l1 = \((\frac{E}{R+r})\frac{R_p}{L}\times l_1\)
\((\frac2{2+r})\frac{R_p}{L}\times70\).......(i)
For balancing length l2
\((\frac{E}{R+r})\frac{R_p}{L}\times l_1\)
\(\left(\cfrac{2}{\frac23+r}\right)\frac{R_p}{L}\times50\).......(ii)
Comparing equation (i) and (ii)
\((\frac2{2+r})\frac{R_p}L\times70=\left(\cfrac{2}{\frac23+r}\right)\frac{R_p}L\times50\)
\(\frac{140}{2+r}=\cfrac{100}{\frac{(2+3i)}{3}}\)
\(\frac{140}{2+r}=\frac{300}{2+3r}\)
280 + 420 r = 600 + 300r
120r = 600 -280
120 r = 320
r = \(\frac{320}{120}\)
r = 2.6 Ω (d)