Maharashtra State BoardHSC Science (Electronics) 11th
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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.

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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?
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APPEARS IN

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