question_answer

A resistance wire connected in the left gap of a metre bridge balances a \[10\text{ }\Omega \] 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 \[1\text{ }\Omega \] of the resistance wire is:                                                                                   [NEET 2020]

A)  \[1.0\times {{10}^{1}}m\]       

B)       \[1.5\times {{10}^{1}}m\]

C)  \[1.5\times {{10}^{2}}m\]       

D)       \[1.0\times {{10}^{2}}m\]

Correct Answer: A

Solution :

[a]

Initially, \[\frac{P}{10}=\frac{{{l}_{1}}}{{{l}_{2}}}=\frac{3}{2}\]

\[\Rightarrow P=\frac{30}{2}=15\,\Omega \]

Now Resistance, \[R=\frac{{{p}^{l}}}{A}\]

\[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{{{l}_{1}}}{{{l}_{2}}}\]

\[\Rightarrow \frac{15}{1}=\frac{1.5}{{{l}_{2}}}\]

\[{{l}_{2}}=0.1\,\,m\]

\[=1.0\times {{10}^{-1}}m\]