Question

In a meter bridge experiment $S$ is a standard resistance, $R$ is a resistance wire. It is found that balancing length is $l=25cm$. If $R$ is replaced by a wire of half length and half diameter of that of $R$ of same material, then the balancing $l$ (in cm) will now be:

Answer 1

Step 1: Given

Standard resistance is $S$.

Resistance wire is $R$.

Balancing length is, $l=25cm$.

Diameter of wire is = $d$.

New resistance is = $R\text{'}$.

Step 2: Formula Used

The condition for balancing meter bridge is,

$\frac{S}{R}=\frac{100-l}{l}$.

Resistance is given by,

$R=\rho \frac{l}{A}$

Where $\rho$ is the resistivity of the material, $l$ is the length and $A$ is the area.

Step 3: Finding the ratio of $\frac{S}{R}$

$\begin{array}{rcl}\frac{S}{R}& =& \frac{100-l}{l}\\ & =& \frac{100-25}{25}\\ & =& \frac{75}{25}\\ & =& 3\end{array}$

Step 4: Calculating the new balancing length

Find an expression for resistance in terms of diameter

$\begin{array}{rcl}R& =& \rho \frac{l}{A}\\ & =& \rho \frac{l}{\pi r^2}\\ & =& \rho \frac{l}{\pi {\left({\displaystyle \frac{d}{2}}\right)}^2}\\ & =& \frac{4\rho l}{\pi d^2}\end{array}$

Finding the relation between new resistance and initial resistance by dividing the length and diameter by 2 in the formula for resistance.

$\begin{array}{rcl}R\text{'}& =& \frac{4\rho \left({\displaystyle \frac{l}{2}}\right)}{\pi {\left({\displaystyle \frac{d}{2}}\right)}^2}\\ & =& \frac{8\rho l}{\pi d^2}\\ & =& 2\times \frac{4\rho l}{\pi d^2}\\ & =& 2R\end{array}$

Find the new length by replacing $R$ by $R\text{'}$ in the condition for balancing meter bridge.

$$\begin{gathered} \frac{S}{2R}=\frac{100-l}{l} \\ \Rightarrow \frac{3}{2}=\frac{100-l}{l} \\ \Rightarrow 5l=200 \\ \Rightarrow l=40cm \end{gathered}$$

Hence, the new length is $40cm$.