Tag: units and measurement: error analysis

Questions Related to units and measurement: error analysis

The radius of a sphere is 1.41 cm. Its volume to an appropriate number of significant figures is then

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

  1. 11.73 $cm^3$

  2. 11.736 $cm^3$

  3. 11.7 $cm^3$

  4. 117 $cm^3$

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Correct Option: C
Explanation:
Radius of the sphere, $r = 1.41 cm$ ($3$ significant figures)
Volume of the sphere,
$\displaystyle V = \dfrac {4}{3} \pi r^3 = \dfrac{4}{3} \times 3.14 \times (1.41)^3\,cm^3 = 11.736\, cm^3$
Rounded off upto $3$ significant figures $= 11.7 cm^3$.

The number of significant figure In the numbers $4.8000 \times 10^4$ and $48000.50 $ are respectively then

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

  1. $5$ and $6.$

  2. $5$ and $ 7$.

  3. $2$ and $7.$

  4. $2$ and $6$.

Show answer
Correct Option: B
Explanation:
If any number has more than one digit after decimal place are significant. Power of 10 does not affect number of significant figures.
$4.8000 \times 10^{4} = 5$ significant figure $(4,8,0,0,0)$
$48000.50 = 7$ significant figure $(4,8,0,0,0,5, 0)$

The length and breadth of a rectangular sheet are 16.2 cm and 10.1 cm, respectively. The area of the sheet in appropriate significant figures and error is:

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

  1. $164 \pm 3 cm^2$

  2. $163.62 \pm 2.6 cm^2$

  3. $163.6 \pm 2.6 cm^2$

  4. $163.62 \pm 3 cm^2$

Show answer
Correct Option: A
Explanation:

Let the length and breadth of a rectangular sheet are measured by using a meter scale as 16.2 cm and 10.1 cm respectively. Each measurement has three significant figures.
$\therefore$ Length l can be written as
$l = 16.2 \pm 0.1 cm = 16.2 cm \pm 0.6$%

Similarly, the breadth b can be written as
$b = 10.1 \pm 0.1 cm = 10.1 cm \pm 1$%

Area of the sheet,
$A = l \times b = 163.62 cm^2 \pm 1.6% =163.62 \pm 2.6 cm^2$
Therefore, as per rule, the area will have only three significant figures and errors will have only one significant figure.
Rounding off, we get $A = 164 \pm 3 cm^2$

The number of significant figures in 0.06900 is the

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

  1. 5

  2. 4

  3. 2

  4. 3

Show answer
Correct Option: B
Explanation:
The significant figures of a number are those which adds the valuable importance to the number. The prefixing and suffixing of zeros in decimal number and normal number value have some significant figure value.
Therefore, given is the decimal number which has zero in front and at the last. The zero in front makes the decimal attain a value but the last end zeros are mere bulky figure without importance.
Therefore, the number of significant figures in $0.06900$ is $4$ because of actual valuable number to be $0.069.$

The value of 3.00 km in yards upto correct significant figures and the scientific notation is, (1m=1.094 yards)

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

  1. $3.282$

  2. $3.28\times { 10 }^{ 2 }$

  3. $3.28\times { 10 }^{ 3 }$

  4. none of these

Show answer
Correct Option: C
Explanation:
The right option is B
Explanation:
As we know
$1$km$=1093.61$yards
Therefore,
$3$km$=3\times1093.61$
$=3280.84$yards
$=32.8\times10^2$yards
Hence, In the scientific notation value of $3$km in yards is $32.8\times10^2$

The number of significant figures in $0.00210$ is

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

  1. five

  2. six

  3. four

  4. three

Show answer
Correct Option: D
Explanation:

In the given number  $0.00210$,  only last digits $210$ are considered as significant figures. Thus the given number has three significant figures.

the length and breath of a metal sheet are $2.325\ m$ and $3.142\ m$ respectively. What is the area of this sheet using proper number of significant figures

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

  1. $7.30515\ m^{2}$

  2. $7.3051\ m^{2}$

  3. $7.305\ m^{2}$

  4. $7.31\ m^{2}$

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Correct Option: A

The number of uncertain digit/digits in measured length reported as $ 41.68$ units is 

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

  1. One

  2. Two

  3. Four

  4. Eight

Show answer
Correct Option: A
Explanation:

Each measured number has only one uncertain digit ( the last digit written). It reflects the accuracy of the scale that was used in the measurement.

By rounding off,  (a) $20.96$ and (b) $0.0003125$ to three significant figures, we get 

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

  1. $21.0 ; \ 312 \times 10^{-4}$

  2. $21.0 ; \ 3.125 \times 10^{-4}$

  3. $2.10 ; \ 3.12 \times 10^{-4}$

  4. $210 ; \ 3.12 \times 10^{-4}$

Show answer
Correct Option: C
Explanation:
By rounding off, 20.96 becomes 21.0 and 0.0003125 becomes $3.12\times 10^{-4} $


The edge of a cube is a $=1.2\times 10^{-2}m$. Then its volume will be recorded as :

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

  1. $1.7\times 10^{-6} m^3$

  2. $1.70\times 10^{-6} m^3$

  3. $1.70\times 10^{-7} m^3$

  4. $1.78\times 10^{-6} m^3$

Show answer
Correct Option: A
Explanation:

Given that the edge of a cube is  $a=1.2 \times 10^{-2}m$.
The volume of a cube $= a^{3}$.
Volume $=[1.2 \times 10^{-2} ]^{3}$
The given volume of the cube is $1.728 \times 10^{-6}m^{3}$ 

$ \approx 1.7 \times 10^{-6}m^{3} \ \ \ (upto \ two \  significant  \ digits)$