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
A
1.10Ω
B
1.64Ω
C
2.44Ω
D
27.50Ω
Verified Correct Answer: B.
Physical Principles & Numerical Solution
\(\frac{R_1}{R_2}\) = \(\frac{(100- L_1)}{( L_1)}\)
\(\frac{R_1}{2}\) = \(\frac{45}{55}\)
R\(_1\) = 2 \times \(\frac{45}{55}\)
= 1.64\Omega
this is because the known resistor is connected to 0.00cm end and balances at 55cm
Source: Myschool Past Questions Archive (Record #47761)