In the experiment of calibration of voltmeter, a standard cell of $e.m.f\,\,1.1\,volt$ is balanced against $440\,cm$ of potential wire. The potential difference across the ends of resistance is found to balance against $220\,cm$ of the wire. The corresponding reading of voltmeter is $0.5\,volt.$ The error in the reading of volmeter will be .................. $volt$
JEE MAIN 2014, Diffcult
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In a voltmeter
$V \propto l$
$\mathrm{V}=\mathrm{k} l$
Now, it is given $\mathrm{E}=1.1$ $volt$ for $l_{1}=440\, \mathrm{cm}$ and $V=0.5$ $volt$ for $l_{2}=220\, \mathrm{cm}$
Let the error in reading of voltmeter be $\Delta \mathrm{V}$ then, $1.1=400\, \mathrm{K}$ and $(0.5-\Delta \mathrm{V})=220\, \mathrm{K}$
$\Rightarrow \frac{1.1}{440}=\frac{0.5-\Delta V}{220}$
$\therefore $ $\Delta V=-0.05 \text { volt }$
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