When a known standard resistor of 2.0 is connected to the 0.0cm end of a metre bridge, the balance point is found to be at 55.0cm
1.10Ω
1.64Ω
2.44Ω
27.50Ω
Correct answer is C
\(\frac{R_1}{R_2}\) = \(\frac{L_1}{(100 − L_1)}\)
\(\frac{R_1}{2}\) = \(\frac{55}{45}\)
R\(_1\) = 2 × \(\frac{55}{45}\)
= 2.44Ω