Question
Question: In a meter bridge, the balance length from left end (standard resistance of \(1Ω\) is in the right g...
In a meter bridge, the balance length from left end (standard resistance of 1Ω is in the right gap) is found to be 20 cm, the length of resistance wire in left gap) is found to be 20cm the length of resistance wire in left gap is 21 m and radius is 2mm its specific resistance is
A. π×10−6ohmm
B. 2π×10−6ohmm
C. 2π×10−6ohmm
D. 3π×10−6ohmm
Solution
In order to solve the above question, we need to understand the concept of Meter Bridge. We will use the formula of balancing length to calculate the specific resistance.
Complete step by step answer:
From given data:
R=1Ω, X=20cm and the remaining length is 80cm.Let s be the value of specific resistance.
X=Asl…….. (1)
\dfrac{X}{R} = \dfrac{{20}}{{80}} \\\
Substituting the value of (1) in above equation
RX=41 ⇒41=RAsl ⇒41=π×(2×10−3)2s×21 ⇒41=2π×(2×10−3)2s ∴s=2π×10−6Ωm
Hence the correct option is B.
Additional information:
Meter Bridge instrument consists of a one meter standard resistance wire of uniform cross-sectional area fixed between two ends of metallic strip. There are two gaps, one on the right side and the other on the left side, we have two resistors, one of which is known resistance and the other is unknown resistance which are attached. When the jockey is moved on the wire until the galvanometer shows null reading. Certain positions on the wire show null deflection then reading is obtained.
Note: The instrument meter bridge also called by another name that is sliding Wire Bridge and it is a device or an instrument which works on the principle of a Wheatstone bridge. Usually we find the unknown resistance of a conductor by using the Wheatstone bridge.