# Solve the numerical example. In a workshop a worker measures the length of a steel plate with a Vernier callipers having a least count 0.01 cm. Four such measurements of the - Physics

Numerical

Solve the numerical example.

In a workshop, a worker measures the length of a steel plate with Vernier calipers having a least count 0.01 cm. Four such measurements of the length yielded the following values: 3.11 cm, 3.13 cm, 3.14 cm, 3.14 cm. Find the mean length, the mean absolute error, and the percentage error in the measured value of the length.

#### Solution

Given:

a1 = 3.11 cm, a2 = 3.13 cm,

a3 = 3.14 cm, a4 = 3.14 cm

Least count L.C. = 0.01 cm.

To find:

i. Mean length ("a"_"mean")

ii. Mean absolute error (triangle "a"_"mean")

iii. Percentage error.

Formulae:

1. "a"_"mean" = ("a"_1 + "a"_2 + "a"_3 + "a"_4)/4

2. triangle "a"_"n" = |"a"_"mean" - "a"_"n"|

3. triangle "a"_"mean" = (triangle"a"_1 + triangle"a"_2 + triangle"a"_3 + triangle"a"_4)/4

4. Percentage error = (triangle "a"_"mean")/("a"_"mean") xx 100

Calculation:

From formula (i),

"a"_"mean" = (3.11 + 3.13 + 3.14 + 3.14)/4

= 3.13 cm

From formula (ii),

triangle"a"_1 = |3.13 - 3.11| = 0.02 cm

triangle"a"_2 = |3.13 - 3.13| = 0

triangle"a"_3 = |3.13 - 3.14| = 0.01 cm

triangle"a"_4 = |3.13 - 3.14| = 0.01 cm

From formula (iii),

triangle"a"_"mean" = (0.02 + 0 + 0.01 + 0.01)/4 = 0.01 cm

From formula (iii),

% error = 0.01/3.13 xx 100

= 1/3.13

= 0.3196  .....(using reciprocal table)

= 0.32 %

Ans:

1. Mean length is 3.13 cm.
2. Mean absolute error is 0.01 cm.
3. Percentage error is 0.32 %.
Is there an error in this question or solution?

#### APPEARS IN

Balbharati Physics 11th Standard Maharashtra State Board
Chapter 1 Units and Measurements
Exercises | Q 3. (vi) | Page 14