Measuring Devices ( Galvanometer, Voltmeter and Ammeter & Meter Bridge ): Practice Problem & Solution
A resistance wire connected in the left gap of a metre bridge balances a $10\Omega$ resistance in the right gap at a point which divides the bridge wire in the ratio $3 : 2$. If the length of the resistance wire is $1.5\text{ m}$, then the length of $1\Omega$ of the resistance wire is: (2020)
Solution Explained:
To solve this problem, we apply the core principles of Measuring Devices ( Galvanometer, Voltmeter and Ammeter & Meter Bridge ). Understanding the underlying formula is key to arriving at the correct answer below:
Let the resistance of the left gap be $R$. For a balanced metre bridge, $\frac{R}{10} = \frac{3}{2} \implies R = 15\Omega$.nThe length of the $15\Omega$ resistance wire is $1.5\text{ m}$.nTherefore, the length of $1\Omega$ of the wire is $\frac{1.5}{15} = 0.1\text{ m} = 1.0 \times 10^{-1}\text{ m}$.
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