The current in a conductor and the potential difference across its ends are measured by an ammeter and a voltmeter. The meters draw negligible currents. The ammeter is accurate but the voltmeter has zero error(that different conditions are 1.75A, 14.4V and 2.75A, 22.4V)
a) 0.4volt
b) 0.8volt
c)-0.4volt
d)-0.8volt

Hint: In the question it is given that the instrument has a zero error i.e. its initial value on the meter is not zero. It is also given that the meters draw negligible currents. Hence the error is solely the calibration error. Hence by determining the actual value of the resistance and then obtaining the potential difference and then comparing with the answers on the meter will enable us to determine the error.
Formula used:
$R=\dfrac{{{V}_{1}}-{{V}_{2}}}{{{i}_{1}}-{{i}_{2}}}$
$V=iR$

Complete answer:
Let us say we connect a resistance of ‘R’ across a battery. If the current in the circuit is ‘i’, than from ohm's law, the potential difference (V) across the resistor is given by,
$V=iR$
Let us say we change the potential difference across the resistor. If for potential difference ${{V}_{1}}$ the current across it is ${{i}_{1}}$ and if for potential difference ${{V}_{2}}$ the current across it is ${{i}_{2}}$ than using the relation of ohm's law which is linear variation of voltage and current for a given resistance, taking the slope of the above two points the resistance ‘R’ equal to,
$R=\dfrac{{{V}_{1}}-{{V}_{2}}}{{{i}_{1}}-{{i}_{2}}}$
Let us say the voltmeter has a zero error of $\pm x$ . hence the above equation becomes,
$R=\dfrac{({{V}_{1}}\pm x)-({{V}_{2}}\pm x)}{{{i}_{1}}-{{i}_{2}}}$
If we observe the above equation, the error in both the readings will be the same as we have taken it from the same voltmeter. Hence we can imply the error of one will get cancelled with the error of the other. Hence the above equation now becomes,
$R=\dfrac{{{V}_{1}}-{{V}_{2}}}{{{i}_{1}}-{{i}_{2}}}$
The different conditions are 1.75A, 14.4V and 2.75A, 22.4V. hence from the above equation the value of resistance in the circuit is,
$\begin{align}
  & R=\dfrac{{{V}_{1}}-{{V}_{2}}}{{{i}_{1}}-{{i}_{2}}} \\
 & \Rightarrow R=\dfrac{22.4V-14.4V}{2.75A-1.75A} \\
 & \Rightarrow R=\dfrac{8V}{1A}=8\Omega \\
\end{align}$
From ohm's law for current equal to 1.75A, the actual potential difference ‘V’ across R is,
$\begin{align}
  & V=iR \\
 & \Rightarrow V=1.75A\times 8\Omega \\
 & \therefore V=14V \\
\end{align}$
Therefore the zero error ‘y’ i.e. the difference between the reading on the voltmeter and the actual voltage is equal to,
$\begin{align}
  & y={{V}_{1}}-V \\
 & \Rightarrow y=14.4V-14V \\
 & \therefore y=0.4V \\
\end{align}$

Hence the correct answer of the above question is option a.

Note:
The actual reading is less than the reading on the voltmeter. Therefore the error is positive. The above problem can also be solved by trial and error method but will be time consuming i.e. verifying for every option given. We have only obtained the error in ${{V}_{1}}$ but similarly it can be obtained for ${{V}_{2}}$ which will be the same.