Hint: Recall the concept of a balanced Wheatstone Bridge.

Explanation: The bridge is an arrangement of four resistances that can be used to measure one of them in terms of rest. Here arms \(AB\) and \(BC\) are called ratio arms and arms \(AC\) and \(BD\) are called conjugate arms.
   
Balanced bridge: The bridge is said to be balanced when deflection in the galvanometer is zero, i.e. no current flows through the galvanometer or in other words \({V}_{{B}}={V}_{{D}}.\)
In the balanced condition \(\frac{P}{Q}=\frac{R}{S},\) on mutually changing the position of the cell and galvanometer this condition will not change.
Unbalanced bridge: If the bridge is not balanced current will flow from \(D\) to \(B\) if \(V_D>V_B,\) i.e. \(\left(V_A-V_D\right)<\left(V_A-V_B\right)\) which gives \(P S>R Q.\)
According to the problem for first students, \({R}_2=10 \Omega, {R}_1=5 \Omega, {R}_3=5 \Omega\)
For the second student, \(R_1=500 \Omega, R_2=1000 \Omega, R_3=5 \Omega\)
Let us take \(R_4=R.\)
Now, according to the Wheatstone Bridge rule,
\(\frac{R_2}{R_1}=\frac{R_4}{R_3} \Rightarrow R_4=R_3 \times \frac{R_2}{R_1}\)
Now putting all the values in the above equation, we get \(R=10~ \Omega\) for both students. Thus, we can analyse that the Wheatstone bridge is most sensitive and accurate if resistances are of the same value.
Thus, the errors of measurement of the two students depend on the accuracy and sensitivity of the bridge,
which in turn depends on the accuracy with which \(R_2\) and \(R_1\) can be measured. The currents through the arms of the bridge are very weak when \(R_2\) and \(R_1\) are larger. This can make the determination of the null point accurately more difficult.
   
Hence, option (3) is the correct answer.