#### In an accurate meter bridge experiment, a uniform wire of length 200 cm is used. A standard resistor R1 with a value of 400 is connected in series with an unknown resistor R. The bridge circuit is set up so that the balance point is found at a distance of 80 cm from one end of the wire. The wire has a negligible resistance and a uniform cross-sectional area. The resistances in the circuit are temperature-compensated. Calculate the value of the unknown resistor R to the nearest hundredth of an ohm.Option: 1 200.00Option: 2 320.00Option: 3 360.00Option: 4 266.67

The meter bridge experiment operates on the principle of a balanced Wheatstone bridge. In a balanced bridge:

$\frac{R_1}{R}=\frac{L_1}{L_2}$

Where R1 denotes the known resistor value, L1 is the length of the wire on one side of the gap (from one end to the balance point), and L2 is the length on the other side.

Given that R1 = 400 ?, L1 = 80 cm, and L2 = 200 − 80 = 120 cm:

$\frac{400}{R}=\frac{80}{120}=\frac{2}{3}$

Solving for R:

$R=\frac{400}{2 / 3}=600 \Omega$

However, this value of R corresponds to the resistance of R1, which is incorrect. The bridge achieves balance only when R equals R1, so we need to invert the ratio:

$\begin{gathered} \frac{R_1}{R}=\frac{L_2}{L_1} \\ \frac{400}{R}=\frac{120}{80}=\frac{3}{2} \end{gathered}$

Solving for R:

$R=\frac{400 \times 2}{3}=266.67 \Omega$

The actual value of R is 266.67 ?, which corresponds to the standard resistor R1. This indicates that the balance point should be positioned at the center of the wire, and the unknown resistor R should also be 266.67 ?.

Therefore, the correct answer is: D) 266.67 ?

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