Question

A resistance wire connected in the left gap of a meter bridge balances a $10\Omega$ resistance in the right gap at a point that divides the bridge wire in the ratio of $3:2$. If the length of the resistance wire is $1.5m$, then the length of $1\Omega$ of the resistance wire is _________.

Answer 1

Step 1. Given data

Balanced resistance = $10\Omega$

Bridge wire ratio ( Say $R:S$) = $3:2$

length of the resistance wire = $1.5m$,

Second resistance = $1\Omega$

Step 2. Formula used:

We know, Resistance $R=\frac{\rho l}{A}$, resistivity $\left(\rho \right)$ and area $\left(A\right)$ are constants in this case.

Step 3. Calculations:

As the meter bridge is balanced, so in this case, the ratio of resistance is equal to the ratio of length,

So, $$\begin{gathered} \frac{R}{10}=\frac{l_1}{l_2}=\frac{3}{2} \\ R=15\Omega \end{gathered}$$

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

Thus $\frac{R_1}{R_2}=\frac{l_1}{l_2}$

Putting the known values, we get

$\Rightarrow \frac{15}{1}=\frac{1.5}{l_2}$

$l_2=\frac{1.5}{15}=0.1mor1.0\times {10}^{1}m$

Thus, the length of $1\Omega$ of the resistance wire is $1.0\times {10}^{1}m$.