Tag: units and measurement: error analysis

Questions Related to units and measurement: error analysis

The mass of a beaker is found to be ($10.1 \pm 0.1)gm $ when empty and $(17.3 \pm 0.1)gm$ when partially filled with a liquid. What is the best value for the mass of the liquid together with the possible limits of accuracy?

physics units and measurement: error analysis significant figures significant figures and rounding of digits units and measurements

  1. $(7.2  \pm 0.2)gm$

  2. $(7.2  \pm  0.1)gm$

  3. $(7.1  \pm  0.2)gm$

  4. $(7.2  \pm  0.3)gm$

Show answer
Correct Option: B
Explanation:

Here mass of the liquid , $m=m _2-m _1=17.3-10.1=7.2 gm$

So the relative error, $\dfrac{\Delta m}{m}=\pm\left[\dfrac{\Delta m _1}{m _1}+\dfrac{\Delta m _2}{m _2}\right]$
or $\Delta m=\pm\left[\dfrac{\Delta m _1}{m _1}+\dfrac{\Delta m _2}{m _2}\right]\times m=\pm[0.1/10.1+0.1/17.3]\times 7.2=\pm 0.1 gm$
Thus mass of liquid with possible accuracy $=(7.2\pm 0.1) gm$

Measure of two quantities along with the precision of respective measuring instrument is:
$A = 2.5 ms^{-1}\pm 0.5 ms^{-1}$
$B = 0.10s \pm 0.01 s$
The value of AB will be :

physics units and measurement: error analysis significant figures significant figures and rounding of digits units and measurements

  1. $(0.25\pm 0.08)m$

  2. $(0.25\pm 0.5)m$

  3. $(0.25\pm 0.05)m$

  4. $(0.25\pm 0.135)m$

Show answer
Correct Option: A
Explanation:

Given :  $\Delta A = 0.5 \ m/s$        $\Delta B = 0.01 \ s$

The absolute value of AB  $ = 2.5\times 0.10 = 0.25 \ m$

Relative error in AB,    $\dfrac{\Delta AB}{AB} = \dfrac{\Delta A}{A} +\dfrac{\Delta B}{B}$

$\therefore$   $\dfrac{\Delta AB}{0.25} = \dfrac{0.5}{2.5} +\dfrac{0.01}{0.10}$

$\implies \ \Delta AB = 0.075 = 0.08 \ m$
Thus value of AB is  $(0.25\pm0.08) \ m$

Count total number of S.F. in 300.00

physics units and measurement: error analysis significant figures significant figures and rounding of digits units and measurements

  1. 5

  2. 6

  3. 8

  4. 9

Show answer
Correct Option: A
Explanation:

S.F. = Five, trailing zeros after decimal point are significant.

Count total number of significant figures in 0.00418

physics units and measurement: error analysis significant figures significant figures and rounding of digits units and measurements

  1. 5

  2. 4

  3. 3

  4. 6

Show answer
Correct Option: C
Explanation:

S.F. = Three, as leading  zeros are not significant.

Count total number of Significant Figures in 3.0800

physics units and measurement: error analysis significant figures significant figures and rounding of digits units and measurements

  1. 3

  2. 5

  3. 1

  4. 2

Show answer
Correct Option: B
Explanation:

S.F. = Five, as trailing zeros after decimal place are significant.

Count total number of significant figures in 3500

physics units and measurement: error analysis significant figures significant figures and rounding of digits units and measurements

  1. 1

  2. 2

  3. 3

  4. 4

Show answer
Correct Option: B
Explanation:

S.F. = Two , the trailing zeros are not significant.

Count total number of S.F. in $1.60\, \times\, 10^{- 19}$ 

physics units and measurement: error analysis significant figures significant figures and rounding of digits units and measurements

  1. 3

  2. 5

  3. 2`

  4. 22

Show answer
Correct Option: A
Explanation:

S.F. = Three ; 1, 6, 0; remaining 19 zeros are not significant.

Count total number of S.F. in 5.003020

physics units and measurement: error analysis significant figures significant figures and rounding of digits units and measurements

  1. 3

  2. 20

  3. 7

  4. 9

Show answer
Correct Option: C
Explanation:

S.F. = Seven, the trailing zeros after decimal place are significant.

Count total number of S.F. in $6.020\, \times\, 10^{23}$

physics units and measurement: error analysis significant figures significant figures and rounding of digits units and measurements

  1. 4

  2. 3

  3. 6

  4. 7

Show answer
Correct Option: A
Explanation:

S.F. = Four; 6, 0, 2, 0; remaining 23 zeros are not significant.

If accuracy of weighing machine is better than 1/2 percent, a student record his own mass. Choose the option that explains the recording most accurately.

physics units and measurement: error analysis significant figures significant figures and rounding of digits units and measurements

  1. 6.43 kg

  2. 60 kg

  3. 64.3 kg

  4. 600kg

  5. 643 kg

Show answer
Correct Option: C
Explanation:

This question is intended to assess your knowledge of realistic values for physical quantities as well as an understanding of experimental error and significant figures. Only 60 kg and 64.3 kg are likely masses for an adult human. When recording measurements, it is assumed that there is some measurement error in the last digit recorded. For both possible values for the student's weight, 1/2 percent is about 0.3 kg. Therefore, a mass recorded to better than this accuracy will indicate tenths of a kilogram.