Understand the circuit and error source

In the usual arrangement for measuring resistance by voltmeter and ammeter:

• the voltmeter is connected across the resistor $R$,
• the ammeter measures the total current drawn from the source.

Since the voltmeter has finite internal resistance $R_v$, it also draws some current. Hence the ammeter reads $I = I_R + I_v$ where $I_R = \frac{V}{R}, \qquad I_v = \frac{V}{R_v}.$

Measured resistance using the standard formula

The experimenter calculates $R_{\text{meas}} = \frac{V}{I}.$

Substituting $I$: $I = \frac{V}{R} + \frac{V}{R_v} = V\left(\frac{1}{R} + \frac{1}{R_v}\right).$

Therefore,

Use the given data

Actual resistance: $R = 30\,\Omega.$

Measured value is $5\%$ less than actual: $R_{\text{meas}} = 0.95R = 0.95 \times 30 = 28.5\,\Omega.$

So, $\frac{30R_v}{30 + R_v} = 28.5.$

Solve for $R_v$

$30R_v = 28.5(30 + R_v)$ $30R_v = 855 + 28.5R_v$ $30R_v - 28.5R_v = 855$ $1.5R_v = 855$ $R_v = \frac{855}{1.5} = 570\,\Omega.$

Match with the options

$R_v = 570\,\Omega$

So the correct option is A.