Tag: significant figures

Questions Related to significant figures

With due regards to significant figures 0.99-0.989 is equal to :

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

  1. 0.001

  2. $0.010\times 10^{-1}$

  3. $0.01\times 10^{-1}$

  4. $0.1\times 10^{-3}$

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Correct Option: C
Explanation:

$0.99-0.989=0.001=0.01\times 10^{-1}$

The number of significant figures in $0.00060$ m is:

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

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Correct Option: B
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. 

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

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

  1. n

  2. n-1

  3. 2n

  4. n/2

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Correct Option: B
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.

The number of significant figures in $6.023\times { 10 }^{ 23 }$ is

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

  1. 4

  2. 3

  3. 2

  4. 23

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Correct Option: A
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.

The radius of a sphere is $5$ cm. Its volume will be given by (according to the theory of significant figures) :

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

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

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

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

  4. $5\ { \times\  { 10 }^{ 2 }cm }^{ 3 }$

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Correct Option: D
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)

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

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

  1. $0.0465$

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

  3. $4.651 \times  { 10 }^{ -2 }$

  4. $4.650 \times  { 10 }^{ -2 }$

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Correct Option: C
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, $0.046508\approx 4.651\times 10^{-2}$

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

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

  1. 1355

  2. 13550

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

  4. 135.5

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Correct Option: B,C
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.

The value of $\sqrt { 2 } $ in correct significant digits is :

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

  1. 1.414

  2. 1.4

  3. 1.0

  4. 1

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Correct Option: D
Explanation:

$\sqrt{2}=1.41421356237$
Since $2$ has $1$ significant digit,
Hence the $\sqrt{2}$ has only one significant digit.

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

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

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

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

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

  4. 5650 ${ m }^{ 2 }$

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Correct Option: B
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)

The number of significant figures in the numbers $672.9$ and $2.520\times { 10 }^{ 7 }$ are :

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

  1. $4, 4$

  2. $3, 4$

  3. $4, 3$

  4. $3, 3$

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Correct Option: A
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.