Significant Figures and Error Analysis - Class XI

All non-zero numbers are always significant.All zeroes before a non zero number are insignificant. All zeroes which are simultaneously to the right of the decimal point and at the end of the number are significant. The only non significant digit here is the zero before the decimal point.
Hence, number of significant digits is 4.

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 $= 4$

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.

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
$l = 16.2 \pm 0.1 cm = 16.2 cm \pm 0.6$%

In the given number  $0.00210$,  only last digits $210$ are considered as significant figures. Thus the given number has three significant figures.

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.

Each measured number has only one uncertain digit ( the last digit written). It reflects the accuracy of the scale that was used in the measurement.

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 $1.728 \times 10^{-6}m^{3}$ 

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

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 $= 4$

When adding or subtracting, the result should have the same number of decimal places as the measurement with the fewest decimal places. 5.0m and 6.0m both have one decimal place, so the sum 11.0m has three significant figures.

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. 

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

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

As per the following rules of significant numbers: 

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

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. 

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

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.

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.

$0.046508\approx 4.6508\times 10^{-2}$
$\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, $0.046508\approx 4.651\times 10^{-2}$

$13546\approx 13550$
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.

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

$d=4.24;m$
$SA=$ 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)

$672.9 \rightarrow 6, 7, 2, 9 \rightarrow$ 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.

$\sqrt{79.62}=8.92300397$
Since 79.62 has 4 significant digits, so we express our answer in 4 significant digits.
$\Rightarrow \sqrt{79.62}=8.923 \rightarrow$ 4 significant figures.

Volume of 1 sphere $=1.76 cm^{3}$
Volume of 25 spheres $=25\times 1.76$
                                      $=44 {cm}^{3}$
                                      $=44.0 {cm}^{3}$
                                 (Since volume is reported in 3 significant figure)

$\left ( 2.0 \right )^{10}=1024$
But since 2.0 contains 2 significant figure.
So, $1024=1.0\times 10^{3}$

Momentum$=mv=3.513\times 5.00=17.565\approx 17.57$
$\approx 17.6 kg m/s$
(Since 5.00 contains  least no. of significant figures i.e. 3)

Since, the measurement $0.500 \ cm$ gives the reading up to 3 decimel, thus $0.500 \ cm$ is the most accurate measurement among the given three measurements.

The least significant digit is the lowest digit in a number, located at the far right of a string.

If there is a decimal point, the rightmost digit is the last or least significant figure. So 0.08076 is the least significant figure

Since, all the non zero digits and zeros between the non-zero digits are considered as significant figures. Thus the number $10.02$ has $4$ significant figures.

Zeros to the left of the non-zero digits are considered as significant figures. All non-zero digits are also considered as significant figures. Thus the number  $0.0632$ has $3$ significant figures.

In the number 6.0023, all non-zero digits are significant. Zeros between non-zero digits are also significant, so all five digits are counted.

When subtracting values, the absolute uncertainties are added. 20.23 - 5.75 = 14.48, and 0.01 + 0.01 = 0.02.

Trailing zeros in a whole number are not significant.Hence $3400$ has 2 significant digits only.

In numbers without a decimal point, trailing zeros are generally not considered significant unless specified otherwise. Thus, only 1 and 2 are significant.

Volume V = pi * r^2 * h. With diameter d = 12.6, r = 6.3. V = pi * (6.3)^2 * 34.2 = 4264.4. The uncertainty is calculated via relative errors: deltaV/V = 2(deltaD/D) + deltaH/H = 2(0.1/12.6) + 0.1/34.2 = 0.0158 + 0.0029 = 0.0187. deltaV = 4264.4 * 0.0187 = 79.7, which rounds to 81.

Leading zeros are never significant. The significant digits are 2, 0, 3, and 1.

The calculation for volume V = pi * r^2 * h with r = 0.8 and h = 5.0 yields approximately 10.05. The uncertainty calculation matches option D.

29.2 - 20.2 = 9.0 (two sig figs). 9.0 * 1.79 = 16.11. 16.11 / 1.37 = 11.759. The result should be limited by the precision of the subtraction (two sig figs), but the options suggest three.

To round 2.3456789 to 4 significant figures, look at the 5th digit (6). Since 6 is greater than 5, round up the 4th digit (5) to 6.

Significant figures upto three significant figures for the given question is 5.875 . Therefore, 5 is equal to 5 and we can round off the digit preceding 5 is increased by 1.

1.00 has three significant figures and 2.88 has three. The result of the multiplication should have three significant figures.

Here, $r _1=5 \Omega , r _2=10 \Omega, \Delta r _1=0.2 \Omega $ and $\Delta r _2=0.1 \Omega$

The error in two readings always adds up.
$l _1-l _2=(3.323\pm 0.001)-(3.321\pm 0.001)=(0.002\pm 0.002)cm$

$Area=(Length)^2$
$Length =(Area)^{1/2}$
             $=(100\pm 2)^{1/2}$
             $=(100)^{1/2}\pm \dfrac {1}{2}\times 2$
             $=(10\pm 1)m$

Relative density $=\dfrac {\text {Weight of body in air}}{\text {Loss of weight in water}}$$=\dfrac {5.00}{5.00-4.00}=\dfrac {5.00}{1.00}$

Since question contains atleast 1 significant number in one of the value, therefore, we should write the answer in such a form that it contains only one significant figure. $V=lbt=12\times 6\times 2.45=2\times 10^2\ cm^3$

Surface area, $\displaystyle S=4\pi r^{2}=\left ( 4 \right )\left ( \frac{22}{7} \right )\left ( 2.1 \right )^{2}$$\displaystyle =55.44=55.4cm^{2}$
Further, $\displaystyle \frac{\Delta S}{S}=2.\frac{\Delta r}{r}$
              $\displaystyle \Delta S=2\left ( \frac{\Delta r}{r} \right )\left ( S \right )=\frac{2\times 0.5\times 55.4}{2.1}$$\displaystyle =26.38=26.4 cm^{2}$

Mass $= 9.23\ kg$ .... 3 significant figures

l = 4.234 m 4 significant figures

Zeroes before, a non zero digit are not significant.

$2.00823 = 2.008$ because the digit 2 is less than 5.

According to general rule for determining of number of significant figures, zeroes placed after other digits but behind a decimal point are significant. Thus 12.00 has four significant figures. 

Zeroes before any non zero value are not significant.

Zeroes after decimal but after non zero value are significant.

For a number with $NO \ \ ZERO$ to the right, the number of significant figures is simply the total number of digits.

$1.23\times 2.345=2.88$

All the digits in 200.40 are significant since zeroes after non zero digit after decimal are significant . 

Here mass of the liquid , $m=m _2-m _1=17.3-10.1=7.2 gm$

Given :  $\Delta A = 0.5 \ m/s$        $\Delta B = 0.01 \ s$

S.F. = Five, trailing zeros after decimal point are significant.

S.F. = Three, as leading  zeros are not significant.

S.F. = Five, as trailing zeros after decimal place are significant.

S.F. = Two , the trailing zeros are not significant.

S.F. = Three ; 1, 6, 0; remaining 19 zeros are not significant.

S.F. = Seven, the trailing zeros after decimal place are significant.

S.F. = Four; 6, 0, 2, 0; remaining 23 zeros are not significant.

This question is intended to assess your knowledge of realistic values for physical quantities as well as an understanding of experimental error and significant figures. Only 60 kg and 64.3 kg are likely masses for an adult human. When recording measurements, it is assumed that there is some measurement error in the last digit recorded. For both possible values for the student's weight, 1/2 percent is about 0.3 kg. Therefore, a mass recorded to better than this accuracy will indicate tenths of a kilogram.

If $l$ be the length, $b$ be the breadth and $t$ be the thickness of the sheet,
then the area of sheet is $A=2(lb+bt+tl)=2[(4\times 2.5)+(2.5\times 1.56)+(1.56\times 4)]=40.28 cm^2$
As length has least significant figures i.e one so the answer must be expressed as one significant figures. 
Thus according rule of determining significant figures, the area will be $0.4\times 10^2 cm^2$

All non-zero digits and zeros between non-zero digits are considered as significant figures. Thus the number $500.06$ has $5$ significant digits.

We know that number of significant figures after addition will be equal to least number of significant figures present in any of the operand.
Therefore total weight after weight gain is $102.1+0.28=102.38 kg$
Now least number of significant figure present after the decimal in the operand is 1, present in $102.1\ kg$ .
Therefore, the number of significant figures present after the calculation should also be $1$ after the decimal. Hence, when we round the answer $102.38$ becomes $102.4\  kg$
Therefore correct answer is option (C)

$4.7\times { 10 }^{ -4 }-3.2\times { 10 }^{ -6 }$

$4.8\times { 10 }^{ 4 }-1.5\times { 10 }^{ 3 }$

$3.8\times { 10 }^{ -6 }+42\times { 10 }^{ -6 }=45.8\times { 10 }^{ -6 }=4.58\times { 10 }^{ -5 }$

Density = Mass / Volume. Volume = (4/3) * pi * r^3. Density = 5.975e24 / ((4/3) * pi * (6.37e6)^3) = 5518 kg/m^3, which is 5.52e3 kg/m^3.

No. of significant figures in $\underbrace { 5.69 } \times { 10 }^{ 15 }$ Kg is $3$.

$10\ cm$ of Main Scale $=200$ div of Main scale 

Let focal length  of the mirror be $f$  cm.

$Y=\dfrac{4MgL}{\pi D^2 l}$