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
The correct answer is: B
Explanation
\(\frac{R_1}{R_2}\) = \(\frac{(100- L_1)}{( L_1)}\)
\(\frac{R_1}{2}\) = \(\frac{45}{55}\)
R\(_1\) = 2 × \(\frac{45}{55}\)
= 1.64Ω
this is because the known resistor is connected to 0.00cm end and balances at 55cm
There is an explanation video available .