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In the experiment of calibration of voltmeter, a standard cell of e.m.f. $1.1$ volt is balanced against $440 \mathrm{~cm}$ of potential wire. The potential difference across the ends of resistance is found to balance against $220 \mathrm{~cm}$ of the wire. The corresponding reading of voltmeter is $0.5$ volt. The error in the reading of volmeter will be:
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Verified Answer
The correct answer is:
$-0.05$ volt
$-0.05$ volt
In a voltmeter
$$
\begin{gathered}
\mathrm{V} \propto l \\
\mathrm{~V}=\mathrm{k} l
\end{gathered}
$$
Now, it is given $\mathrm{E}=1.1$ volt for $l_1=440 \mathrm{~cm}$ and $\mathrm{V}=0.5$ volt for $l_2=220 \mathrm{~cm}$ Let the error in reading of voltmeter be $\Delta \mathrm{V}$ then,
$$
\begin{aligned}
&1.1=400 \mathrm{~K} \text { and }(0.5-\Delta \mathrm{V})=220 \mathrm{~K} . \\
&\Rightarrow \frac{1.1}{440}=\frac{0.5-\Delta \mathrm{V}}{220} \\
&\therefore \Delta \mathrm{V}=-0.05 \text { volt }
\end{aligned}
$$
$$
\begin{gathered}
\mathrm{V} \propto l \\
\mathrm{~V}=\mathrm{k} l
\end{gathered}
$$
Now, it is given $\mathrm{E}=1.1$ volt for $l_1=440 \mathrm{~cm}$ and $\mathrm{V}=0.5$ volt for $l_2=220 \mathrm{~cm}$ Let the error in reading of voltmeter be $\Delta \mathrm{V}$ then,
$$
\begin{aligned}
&1.1=400 \mathrm{~K} \text { and }(0.5-\Delta \mathrm{V})=220 \mathrm{~K} . \\
&\Rightarrow \frac{1.1}{440}=\frac{0.5-\Delta \mathrm{V}}{220} \\
&\therefore \Delta \mathrm{V}=-0.05 \text { volt }
\end{aligned}
$$
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