A resistance wire connected in the left gap of a meter bridge balances a $10 Ω$ 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 $1.5 m,$ then the length of $1 Ω$ of the resistance wire is :

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$1 .0 \times 10^{- 1} m$

$1 .0 \times 10^{- 2} m$

$1 .5 \times 10^{- 1} m$

$1 .5 \times 10^{- 2} m$

answer is A.

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Detailed Solution

Let the resistance of the wire be R.
According to the question, R balances 10 Ω resistance in the ratio 3 : 2.
$\begin{aligned} \Rightarrow \frac{R}{10} = \frac{3}{2} \\ \Rightarrow R = 15 Ω \end{aligned}$
Given length of R is 1.5 m.
$∴$ The length of 1 Ω of R will be,
$= (1 Ω) \times (\frac{1.5 m}{15 Ω}) = 1 \times 10^{- 1} m$ .