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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 m, then the length of $1\ \Omega$ of the resistance wire is:
A
$1.0\times10^{-1}\ m$
B
$1.5\times10^{-1}\ m$
C
$1.5\times10^{-2}\ m$
D
$1.0\times10^{-2}\ m$
Detailed Solution
At balance: $\frac{P}{10} = \frac{l_1}{l_2} = \frac{3}{2} \Rightarrow P = 15\ \Omega$
The resistance wire of length 1.5 m has resistance 15 Ω, and $R \propto l$:
$\frac{15}{1} = \frac{1.5}{l_2} \Rightarrow l_2 = 0.1$ m $= 1.0 \times 10^{-1}$ m