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

Write the number of significant digits in 6.032 ?

  1. 4

  2. 3

  3. 2

  4. none of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The significant figures of a number are those digits that carry meaning contributing to its measurement resolution. This includes all digits except:
1. All leading zeros;
2. Trailing zeros when they are merely placeholders to indicate the scale of the number (exact rules are explained at identifying significant figures); 
3. Spurious digits introduced, for example, by calculations carried out to greater precision than that of the original data, or measurements reported to a greater precision than the equipment supports.

According to rules,
$6.032$ 
All the digits are significant. (no leading and trailing zero)
No. of significant digits $= 4$

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

The respective number of significant figures for the numbers 6.320, 6.032, 0.0006032 are then

  1. 3, 4, 8

  2. 4, 4, 8

  3. 4, 4, 4

  4. 4, 3, 4
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

According to the rules of significant figures $6.320$ has four significant figures.
$6.032$ has four significant figures.
$0.0006032$ has four significant figures.

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

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

  1. $7.30515\ m^{2}$
  2. $7.3051\ m^{2}$
  3. $7.305\ m^{2}$
  4. $7.31\ m^{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The area is calculated by multiplying length and breadth: 2.325 * 3.142 = 7.30515 m^2. Since both given measurements have four significant figures, the result must be rounded to four significant figures, giving 7.305 m^2.