Tag: units and measurement: error analysis

Questions Related to units and measurement: error analysis

Given $P = 0.0030 \,m, Q = 2.40 \,m$ and $R = 3000 m$, then number of significant figure in $P, Q, R$ are respectively:

  1. $1, 2, 1$

  2. $2, 3, 4$

  3. $4, 2, 1$

  4. $4, 2, 4$


Correct Option: B
Explanation:

$P = 0.00 \,\underset{1}{\underset{\uparrow}{3}}\underset{2}{\underset{\uparrow}{0}} m$

No. of digits (significant) $= 2$
$Q = \underset{1}{\underset{\uparrow}{2}}.\underset{2}{\underset{\uparrow}{4}}\underset{3}{\underset{\uparrow}{0}} \,m$
Digit $= 3$

$R = \underset{1}{\underset{\uparrow}{3}}\underset{2}{\underset{\uparrow}{0}}\underset{3}{\underset{\uparrow}{0}}\underset{4}{\underset{\uparrow}{0}} \,m$
Digit $= 4$

The number of significant figure in the result of (5.0m+6.0m) is

  1. Two

  2. Three

  3. Four

  4. One


Correct Option: B

How many significant figures are there in 0.30100?

  1. 1

  2. 3

  3. 5

  4. none of these


Correct Option: C
Explanation:

As per the following rules of significant numbers:
 1) ALL non-zero numbers (1,2,3,4,5,6,7,8,9) are ALWAYS significant.
2) ALL zeroes between non-zero numbers are ALWAYS significant.
3) ALL zeroes which are SIMULTANEOUSLY to the right of the decimal point AND at the end of the number are ALWAYS significant.
4) ALL zeroes which are to the left of a written decimal point and are in a number are ALWAYS significant.
Hence in given number i.e. 0.30100
According to the rule 1 and rule 3, the total significant numbers are 5. 

How many significant figures are there in 0.0020 ?

  1. 1

  2. 2

  3. 5

  4. none of these


Correct Option: B
Explanation:

As we know zeroes only after a non zero digit and after decimal and zeroes between any two non zero digits are significant.

Therefore, option B is correct.

How many significant figures are there in 30100?

  1. 1

  2. 3

  3. 5

  4. none of these


Correct Option: B
Explanation:

As we know zeroes only after a non zero digit and after decimal and zeroes between any two non zero digits are significant.

Therefore, option (B) is correct.

With due regards to significant figures $ 5.4 \times 0.125$ is equal to:

  1. 0.7

  2. 0.68

  3. 0.667

  4. none of these


Correct Option: B
Explanation:

Least number of significant figures before multiplication is $2$ in one of the multiplying numbers, then after the multiplication also, the answer should be in same number of significant digits.
$5.4\times 0.125=0.675=0.68$

How many significant figures are there in $0.030100\times 10^6$?

  1. 1

  2. 3

  3. 5

  4. 7


Correct Option: C
Explanation:

As per the following rules of significant numbers: 

1) ALL non-zero numbers (1,2,3,4,5,6,7,8,9) are ALWAYS significant. 
2) ALL zeroes between non-zero numbers are ALWAYS significant. 
3) Zeroes placed after other digits but behind a decimal point are significant 4)Zeroes placed before other digits are not significant.
Hence in given number i.e. $0.030100 \times 10^6$ can be further simplified as $30100.00$ 
According to the rule 1, 3 and rule 4, the total significant numbers are 5.

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

  1. 0.001

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

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

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


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:

  1. 1

  2. 2

  3. 3

  4. 4


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 :

  1. n

  2. n-1

  3. 2n

  4. n/2


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.