Question

A resistance $2\Omega$ is connected across a gap of a meter bridge (the length of the wire is $100cm$ ) and an unknown resistance greater than $2\Omega$ , is connected across the other gap. When these resistances are interchanged, the balance point shifts by 20cm. Neglecting any corrections, the unknown resistance is:s
A. $3\Omega$
B. $4\Omega$
C. $5\Omega$
D. $6\Omega$

Answer 1

Meter bridge is an instrument that works on the principle of Wheatstone bridge. We find the null point, a point on the wire where galvanometer deflection is zero. The accuracy of the null point device is very high. Using a meter bridge we want to find the value of the unknown resistance of the circuit.

Complete step by step solution:
Let us first write the given information in the question.
Resistance across the gap $R = 2\Omega$ , the unknown resistance is greater than $2\Omega$, length of the wire is 100cm and when these resistances are interchanged the balance point shifts by 20cm.
We have to find the unknown resistance.
The balancing condition for a meter bridge is given below.
$\dfrac{R}{l} = \dfrac{S}{{100 - l}}$
Here, $S$is the unknown resistance, $R$ is the given resistance, and $l$ is the balancing point.
Let us substitute the values.
Initially, the balancing condition can be written as below.
$\dfrac{2}{l} = \dfrac{S}{{100 - l}}$……………………(1)
Now after interchanging the resistances, the balancing point shifts by 20cm. so, we can write the following.
$\dfrac{S}{{l + 20}} = \dfrac{2}{{80 - l}}$ …………………….(2)
Let us multiply equations (1) and (2) and write the following.
$\dfrac{2}{l} \times \dfrac{S}{{l + 20}} = \dfrac{S}{{100 - l}} \times \dfrac{2}{{80 - l}}$
$\Rightarrow \dfrac{1}{{l(l + 20)}} = \dfrac{1}{{(100 - l)(80 - l)}}$
$20l = - 180l + 8000$
$\Rightarrow \dfrac{{8000}}{{200}} = l = 40cm$
Now, let us substitute this value in equation (1).
$\dfrac{2}{{40}} = \dfrac{S}{{100 - 40}}$
$\Rightarrow S = 3\Omega$
Hence, the correct option is (C) $3\Omega$.

Note:
The name of the meter bridge is because it works on the principle of the Wheatstone bridge and the length of the wire is one meter.
We obtain the null point on the meter bridge by sliding a jockey on the wire till we get the point where the current is zero.