Current Electricity

JEE Main level, single correct, medium

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Question 371 of 380

In the experiment of calibration of voltmeter, a standard cell of e.m.f. 1.11.1 volt is balanced against 440 cm440 \mathrm{~cm} of potential wire. The potential difference across the ends of resistance is found to balance against 220 cm220 \mathrm{~cm} of the wire. The corresponding reading of voltmeter is 0.50.5 volt. The error in the reading of volmeter will be:

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Answer: (D) −0.05-0.05 volt

In a voltmeter
V∝l V=kl\begin{gathered} \mathrm{V} \propto l \\ \mathrm{~V}=\mathrm{k} l \end{gathered}
Now, it is given E=1.1\mathrm{E}=1.1 volt for l1=440 cml_1=440 \mathrm{~cm} and V=0.5\mathrm{V}=0.5 volt for l2=220 cml_2=220 \mathrm{~cm} Let the error in reading of voltmeter be ΔV\Delta \mathrm{V} then,
1.1=400 K and (0.5−ΔV)=220 K.⇒1.1440=0.5−ΔV220∴ΔV=−0.05 volt \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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