NEETPhysicsMathematics in Physics
A student records four measurements of a time interval as 2.3 s , 2.4 s , 2.6 s , and 2.7 s . Match the statistical terms in List-I with their corresponding calculated values in List-II. List-I List-II (A) True value (I) 2.5 s (B) Mean absolute error (II) 0.15 s (C) Fractional error (III) 0.06 (D) Percentage error (IV) 6 % Choose the correct answer from the options given below:
Options
- A(A)-(I), (B)-(II), (C)-(III), (D)-(IV)
- B(A)-(I), (B)-(II), (C)-(IV), (D)-(III)
- C(A)-(I), (B)-(III), (C)-(II), (D)-(IV)
- D(A)-(II), (B)-(I), (C)-(III), (D)-(IV)
Correct answer
A. (A)-(I), (B)-(II), (C)-(III), (D)-(IV)
Step-by-step solution
First, calculate the true value (mean of the observations): T_m = 2.3 + 2.4 + 2.6 + 2.7 4 = 10.0 4 = 2.5 s . So, (A) matches with (I). Next, find the absolute errors for each measurement: T₁ = |2.3 - 2.5| = 0.2 s T₂ = |2.4 - 2.5| = 0.1 s T₃ = |2.6 - 2.5| = 0.1 s T₄ = |2.7 - 2.5| = 0.2 s Calculate the mean absolute error: T_m = 0.2 + 0.1 + 0.1 + 0.2 4 = 0.6 4 = 0.15 s . So, (B) matches with (II). Calculate the fractional error: Fractional error = T_m T_m = 0.15 2.5 = 0.06 . So, (C) matches with (III). Calculate the