In a meter bridge, the null point is found at a distance of 33.7 cm from terminal A (known resistance R is connected to gap near to A). If a resistance of 12 Ω is connected parallel to unknown resistance X (which is connected to gap near to terminal B), the null point is found at 51.9 cm. The value of X is 

Concept: 

Meter bridge:

• The Meter Bridge is an instrument based on the principle of the Wheatstone bridge and is used to measure an unknown resistance.
• In the case of a meter bridge, the resistance wire AC is 100 cm long. Varying the position of tapping point B, the bridge is balanced.
• If in the balanced position of bridge AB = l, BC (100 – l) so that

\(\frac{Q}{P} = \frac{{\left( {100 - l} \right)}}{l}\)

Also, \(\frac{P}{Q} = \frac{R}{S} ⇒ S = \frac{{\left( {100 - l} \right)}}{l}R\)

Explanation:

Given - Null point in 1st case (l1) = 33.7 cm, Null point in 2nd case (l1) = 51.9 cm

• The null point in 1​st case (l1)

\(⇒ \frac{R}{X} =\frac{l}{100-l}=\frac{33.7}{66.3}\)

• When X is connected in parallel to R = 12 Ω, the resistance across the gap is Req 

\(⇒ \frac{1}{R_{eq}}=\frac{1}{12}+\frac{1}{X}=\frac{12+X}{12X}\)

\(⇒ R_{eq}=\frac{12X}{12+X}\)

• The null point in 2nd case (l1)

\(⇒ \frac{R}{R_{eq}} =\frac{l}{100-l}=\frac{51.9}{48.1}\)

\(⇒ \frac{R}{\frac{12X}{12+X}}=\frac{R(12+X)}{12X}=\frac{51.9}{48.1}\)

Substitute \( \frac{R}{X} =\frac{33.7}{66.3}\) in the above equation, we get

\(⇒ \frac{33.7}{66.3}[\frac{12+X}{12}]=\frac{51.9}{48.1}=1.08\)

\(⇒ [\frac{12+X}{12}]=2.126\)

⇒ X = 13.5 Ω