Tag: significant figures and rounding of digits

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Questions Related to significant figures and rounding of digits

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 $0.00060$ m is:

  1. 1

  2. 2

  3. 3

  4. 4
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

According to general rule for significant figures, zeros to the left of a significant figure and not bounded to the left by another significant figure are not significant. So for given number $0.00060$ only 6 and 0 after 6 will be the significant figures. Thus, it has 2 significant figures. 

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

If a measured quantity has $n$ significant figures, the reliable digits in it are :

  1. n

  2. n-1

  3. 2n

  4. n/2
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

A measured quantity containing $n$ significant figures has  $n-1$  reliable digits.

For example: The quantity $a = 1.2673$ has  $5$ significant figures but it contains only $4$ reliable digits i.e  $1.267$ only.
Because it can be a possiblity that  the quantity was acually  $a = 1.26728$ and after rounding off, it becomes $1.2673$. 
Hence the last digit i.e  $3$ is uncertain.

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 $6.023\times { 10 }^{ 23 }$ is

  1. 4

  2. 3

  3. 2

  4. 23
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The correct option is A.

Every experimental observation value is having some amount of uncertainty associated with it. Significant figures are the number of meaningful digits having certainty. All non – zero digits are significant. Thus in $6.023 \times 10^{23}$, we can see $4$ 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 $5$ cm. Its volume will be given by (according to the theory of significant figures) :

  1. $523.33\ { cm }^{ 3 }$

  2. $5.23\ { \times\  { 10 }^{ 2 }cm }^{ 3 }$

  3. $5.0\ { \times\  { 10 }^{ 2 }cm }^{ 3 }$

  4. $5\ { \times\  { 10 }^{ 2 }cm }^{ 3 }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Volume$=\dfrac{4}{3}{\pi r}^{3}=\dfrac{4}{3}\times \pi \left ( 5 \right )^{3}=523.33\ {cm}^{3}$$=5.2333\times 10^{2}\ cm^{3}$
                                                                                   $\downarrow $
                                                                          5 significant figures
Since radius has single significant figure, so, volume should also have single significant figure.

For single significant figure, we have to drop all after decimal.
$\Rightarrow$ Volume $= 5\times 10^{2}\ cm^{2}$ (if the digit to be dropped is less than 5, preceding digit is left unchanged)

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

When the number 0.046508 is reduced to 4 significant figures, then it becomes :

  1. $0.0465$

  2. $4650.8 \ \times $ ${ 10 }^{ -5 }$

  3. $4.651 \times  { 10 }^{ -2 }$

  4. $4.650 \times  { 10 }^{ -2 }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$0.046508\approx 4.6508\times 10^{-2}$
$\downarrow $
5 significant digits
To have 4 significant digits $\rightarrow 4.6508\approx 4.651$
(If the digit to be dropped is more than 5, then the preceding digit is raised by 1)
So, $0.046508\approx 4.651\times 10^{-2}$

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

When 13546 is rounded off to four significant figures, it becomes :

  1. 1355

  2. 13550

  3. $1355\times { 10 }^{ 1 }$

  4. 135.5
Reveal answer Fill a bubble to check yourself
B,C Correct answer
Explanation

$13546\approx 13550$
All zeros on the right non zero digit are not significant.
And, since last digit (6) is greater than 5,
So, preceding no. has to be raised by 1.

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

The diameter of a sphere is $4.24\ m$. Its surface area with due regard to significant figures is :

  1. 5.65 ${ m }^{ 2 }$

  2. 56.5 ${ m }^{ 2 }$

  3. 565 ${ m }^{ 2 }$

  4. 5650 ${ m }^{ 2 }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$d=4.24\;m$
$SA=$ Surface area $=4\pi r^{2}=4\pi \left ( \dfrac{d}{2} \right )^{2}=\pi d^{2}$
$=\pi \left ( 4.24 \right )^{2}$
$=56.47\ m^{2}$
Since significant figure in 4.24 is 3, so we express the answer in 3 significant figures.
$SA=56.47\ m^2 \approx 56.5\ m^2$
(Rounding off to 3 significant figures - if digit to be dropped is more than 5, the preceding digit is raised by 1)

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 the numbers $672.9$ and $2.520\times { 10 }^{ 7 }$ are :

  1. $4, 4$

  2. $3, 4$

  3. $4, 3$

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

$672.9 \rightarrow 6, 7, 2, 9 \rightarrow$ 4 significant digits since all non zero digits are significant.
$2.520\times 10^{7} \rightarrow 2, 5, 2, 0 \rightarrow 4$ significant digits, since all the zero on the right side of last non zero digit in the decimal part are significant.