Physics - Current Electricity Question with Solution | TestHub

Measured value of $R=5\%$ less than
actual value of $R$ i.e., $R^{\text{'}}=30-(5\%$ of $30$$)=30-\frac{5}{100}\times 30$

$R^{\text{'}}=28.5 \mathrm{\Omega }$ $...\left(\mathrm{i}\right)$

Now, let us assume that internal resistance of voltmeter $R_V$. Replacing voltmeter with its internal resistance, we get following circuit.

It is clear that the measured value, $R^{\text{'}}$ should be equal to parallel combination of $R$ and $R_V$. Mathematically,
$R^{\text{'}}=\frac{R\times R_V}{R+R_V}=28.5 \mathrm{\Omega }$

Given. $R=30 \mathrm{\Omega }$

$\Rightarrow \frac{30 R_V}{30+R_V}=28.5$

$\Rightarrow 30 R_V=\left(28.5\times 30\right)+28.5 R_V$

$\Rightarrow 1.5 R_V=28.5\times 30$

$\Rightarrow R_V=\frac{28.5\times 30}{1.5}=19\times 30$

or $R_V=570 \mathrm{\Omega }$