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In a meter bridge experiment resistances are connected as shown in the figure. Initially resistance $P\, = 4\,\Omega $ and the neutral point $N$ is at $60\,cm$ from $A$. Now an unknown resistance $R$ is connected in series to $P$ and the new position of the neutral point is at $80\,cm$ from $A$ . The value of unknown resistance $R$ is
$\frac{33}{5}\, \Omega $
$ 6\,\Omega $
$ 7\,\Omega $
$\frac{20}{3}\, \Omega $
Solution
In balance position of bridge, $\frac{P}{Q}=\frac{l}{(100-l)}$
Initially neutral position is $60\, \mathrm{cm} .$ from $\mathrm{A}$,
so
$\frac{4}{60}=\frac{Q}{40} \Rightarrow Q=\frac{16}{6}=\frac{8}{3}\, \Omega$
Now, when unknown resistance $\mathrm{R}$ is connected in series to $P$, neutral point is $80 \,cm$ from Athen,
$\frac{4+R}{80}=\frac{Q}{20}$
$\frac{4+R}{80}=\frac{8}{60}$
$R=\frac{64}{6}-4=\frac{64-24}{6}=\frac{40}{6} \,\Omega$
Hence, the value of unknown resistance $\mathrm{R}$
is $=\frac{20}{3} \,\Omega$
