When a voltmeter put in series, it still reads potential drop and when an ammeter is connected in parallel, it still shows current through it.

Let $I=$ current through cell, then potential drop read by voltmeter is $V=I \cdot R_V$ (this is reading of voltmeter)

Where, $R_V$ is the resistance of voltmeter In $\operatorname{loop} A B$,

$V_{A B}=I_1 \times 1=I_2 \times R_A$ and $I=I_1+I_2$

Where, $R_A$ is the resistance of ammeter We substitute for $I_1$ from above equation to get

$\Rightarrow \quad I=I_2 R_A+I_2=I_2\left(R_A+1\right)$

$\Rightarrow \quad I_2=\frac{I}{\left(R_A+1\right)}$

(this is reading of ammeter)

Now given, voltmeter reading ammeter reading $=1 \times 10^3=\frac{I R_V}{\left(\frac{I}{R_A+1}\right)}$

So, $R_V\left(R_A+1\right)=1000 \dots(i)$

Let $I=$ current through cell, then ammeter reading in this case is $I$.

Also, in loop $A B$,

$V_{A B}=I_1 \times 1=I_2 \times R_V$

$\text { As, } I=I_1+I_2=I_2 R_V+I_2$

$=I_2\left(R_V+1\right)$

So, $I_2=\frac{I}{\left(R_V+1\right)}$

Hence, voltmeter reading is $V=I_2 R_V$

$=\frac{I R_V}{\left(R_V+1\right)}$ (this is reading of voltmeter)

Now given, voltmeter reading $\div$ ammeter reading $=0.999 \,\Omega$.

So, $0.999=\left[\frac{I_V}{\left(R_V+1\right)}\right]$

$\Rightarrow 0.999=\frac{R_V}{R_V+1}$

So, $R_V=999 \,\Omega \dots(ii)$

$=10^3 \Omega$

Substituting $R_V$ in Eq $(i)$, we get

$R_A=\frac{1}{999}$

$\text { or } R_A=10^{-3} \,\Omega$