A student records four measurements of a time interval as 2.3 s ,...

First, calculate the true value (mean of the observations):
$T_m = \dfrac{2.3 + 2.4 + 2.6 + 2.7}{4} = \dfrac{10.0}{4} = 2.5\text{ s}$.
So, (A) matches with (I).

Next, find the absolute errors for each measurement:
$\Delta T_1 = |2.3 - 2.5| = 0.2\text{ s}$
$\Delta T_2 = |2.4 - 2.5| = 0.1\text{ s}$
$\Delta T_3 = |2.6 - 2.5| = 0.1\text{ s}$
$\Delta T_4 = |2.7 - 2.5| = 0.2\text{ s}$

Calculate the mean absolute error:
$\Delta T_m = \dfrac{0.2 + 0.1 + 0.1 + 0.2}{4} = \dfrac{0.6}{4} = 0.15\text{ s}$.
So, (B) matches with (II).

Calculate the fractional error:
$\text{Fractional error} = \dfrac{\Delta T_m}{T_m} = \dfrac{0.15}{2.5} = 0.06$.
So, (C) matches with (III).

Calculate the percentage error:
$\text{Percentage error} = \text{Fractional error} \times 100 = 0.06 \times 100 = 6\%$.
So, (D) matches with (IV).

The correct matching is (A)-(I), (B)-(II), (C)-(III), (D)-(IV).

Answer: (A)-(I), (B)-(II), (C)-(III), (D)-(IV)