question_answer

The actual value of resistance R, shown in the figure is \[30\,\Omega \]. This is measured in an experiment as shown using the standard formula \[R=\frac{V}{I}\] where V and I are the readings of the voltmeter and ammeter, respectively. If the measured value of R is \[5%\] less, then the internal resistance of the voltmeter is- [JEE Main Online Paper (Held On 10-Jan-2019 Evening]

A) 570 \[\Omega \]                         

B) 600 \[\Omega \]

C) 350 \[\Omega \]

D)                  35 \[\Omega \]

Correct Answer: A

Solution :

\[i=\frac{\varepsilon }{{{R}_{A}}+\frac{{{R}_{V}}R}{{{R}_{V}}+R}}\,\,=\,\,\frac{\varepsilon \,({{R}_{V}}+R)}{{{R}_{A}}{{R}_{V}}+{{R}_{A}}R+{{R}_{V}}R}\] \[V=\frac{{{R}_{V}}R\varepsilon }{({{R}_{V}}+R)\left[ {{R}_{A}}+\frac{{{R}_{V}}R}{{{R}_{V}}+R} \right]}\] \[V=\frac{{{R}_{V}}R\varepsilon }{{{R}_{A}}{{R}_{V}}+{{R}_{A}}R+{{R}_{V}}R}\] \[{{R}_{res}}=\,\frac{V}{i}\] \[{{R}_{res}}=\,30-30\times \frac{5}{100}\] \[=\,\,30\left[ 1-\frac{5}{100} \right]\,=\,28.5\] \[28.5\,=\,\frac{{{R}_{V}}R}{{{R}_{V}}+R}\] \[28.5\text{ }{{R}_{V}}+28.5\times 30={{R}_{V}}\times 30\] \[30\times 28.5={{R}_{V}}\times 1.5\] \[{{R}_{V}}=\frac{28.5}{1.5}\,\times 30=570\,\Omega \]