Tag: units and measurement: error analysis

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Questions Related to units and measurement: error analysis

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

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

  1. 11.73 $cm^3$

  2. 11.736 $cm^3$

  3. 11.7 $cm^3$

  4. 117 $cm^3$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $= 11.7 cm^3$.
Multiple choice physics units and measurement: error analysis significant figures significant figures and rounding of digits units and measurements

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

  1. $5$ and $6.$

  2. $5$ and $ 7$.

  3. $2$ and $7.$

  4. $2$ and $6$.
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $(4,8,0,0,0,5, 0)$
Multiple choice physics units and measurement: error analysis significant figures significant figures and rounding of digits units and measurements

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:

  1. $164 \pm 3 cm^2$

  2. $163.62 \pm 2.6 cm^2$

  3. $163.6 \pm 2.6 cm^2$

  4. $163.62 \pm 3 cm^2$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $A = 164 \pm 3 cm^2$

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

The number of significant figures in 0.06900 is the

  1. 5

  2. 4

  3. 2

  4. 3
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $0.069.$
Multiple choice physics units and measurement: error analysis significant figures significant figures and rounding of digits units and measurements

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

  1. $3.282$

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

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

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $32.8\times10^2$
Multiple choice physics units and measurement: error analysis significant figures significant figures and rounding of digits units and measurements

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

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

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

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

  4. $210 ; \ 3.12 \times 10^{-4}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
By rounding off, 20.96 becomes 21.0 and 0.0003125 becomes $3.12\times 10^{-4} $


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

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

  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. $1.78\times 10^{-6} m^3$
Reveal answer Fill a bubble to check yourself
A Correct answer
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)$