Significant Figures Questions in English

(B) To subtract numbers in scientific notation,we first express them with the same power of $10$.
$3.9 \times 10^5 = 39.0 \times 10^4$.
Now,perform the subtraction: $(39.0 \times 10^4) - (2.5 \times 10^4) = (39.0 - 2.5) \times 10^4 = 36.5 \times 10^4$.
Convert this back to standard scientific notation: $3.65 \times 10^5$.
According to the rules of significant figures for subtraction,the result should be reported to the same number of decimal places as the measurement with the fewest decimal places. Here,$3.9 \times 10^5$ has one decimal place (when expressed as $39.0 \times 10^4$) and $2.5 \times 10^4$ has one decimal place. Thus,the result should be rounded to one decimal place.
Rounding $3.65 \times 10^5$ to one decimal place gives $3.6 \times 10^5$ (following the rule of rounding to the nearest even number or standard rounding convention).

(B) First,perform the addition inside the parentheses: $3.20 + 4.80 = 8.00$.
According to the rules of significant figures for addition,the result should be reported to the same number of decimal places as the measurement with the fewest decimal places.
Both $3.20$ and $4.80$ have two decimal places,so the sum $8.00$ is correct to two decimal places.
Thus,the expression becomes $8.00 \times 10^5$.
The digits $8$,$0$,and $0$ are all significant.
Therefore,there are $3$ significant figures.

(A) To round off $96378$ to two significant figures,we look at the first two digits,which are $9$ and $6$.
Since the third digit is $3$ (which is less than $5$),we keep the second digit as it is and replace the remaining digits with zeros.
Thus,$96378$ rounded to two significant figures is $96000$ or $9.6 \times 10^4$.

(A) The mass of a pearl is $85 \, mg$. This measurement is taken with a precision of $1 \, mg$.
If we express this mass in grams, the mass of the pearl becomes $0.085 \, g$. Here, the zeros added to change the unit are not significant. Thus, there are only two significant figures, $8$ and $5$.
If we express this mass in kilograms, the mass of the pearl becomes $0.000085 \, kg$. Here, the leading zeros are not significant. Thus, there are still only two significant figures, $8$ and $5$.
Therefore, it can be concluded that changing the units of measurement does not change the number of significant figures.

(N/A) The rule that trailing zeros are not significant applies to numbers without a decimal point. However,when a measurement is made with a specific instrument,the trailing zeros indicate the precision of the measurement.
Consider a cylinder measured with a Vernier caliper having a least count of $0.01 \ cm$. If the measured length is exactly $3.50 \ cm$,the trailing zero is significant because it indicates that the measurement was taken with an instrument capable of measuring up to the second decimal place.
Thus,the trailing zero in $3.50 \ cm$ is a significant figure,demonstrating that the general rule is not always applicable in the context of physical measurements.

(N/A) The formula for the plane angle $\theta$ in radians is given by $\theta = \frac{l}{r}$,where $l$ is the arc length and $r$ is the radius of the circle.
Given,$\theta = \frac{\pi}{6}$ radians and $r = 31.0 \, cm$.
Substituting these values into the formula:
$\frac{\pi}{6} = \frac{l}{31.0}$
Solving for $l$:
$l = 31.0 \times \frac{\pi}{6} \, cm$
Using $\pi \approx 3.14159$:
$l = 31.0 \times \frac{3.14159}{6} \, cm \approx 16.231 \, cm$.
Rounding to three significant figures (as the radius $31.0$ has three significant figures),the arc length is $16.2 \, cm$.

(C) $1$. The given readings are $5.50\, mm, 5.55\, mm, 5.45\, mm,$ and $5.65\, mm$. All these readings have $3$ significant figures.
$2$. The average value is $5.5375\, mm$. According to the rules of significant figures,the result should be rounded off to the same number of significant figures as the measured values,which is $3$. Thus,$5.5375$ is rounded to $5.54\, mm$.
$3$. The standard deviation is $0.07395\, mm$. The uncertainty (error) should generally be expressed to one or two significant figures. Rounding $0.07395$ to one significant figure gives $0.07\, mm$.
$4$. Therefore,the diameter should be recorded as $(5.54 \pm 0.07)\, mm$.

(C) In subtraction,the number of decimal places in the result should be equal to the number of decimal places of the term in the operation that contains the fewest decimal places.
Given values:
$9.99\, m$ (has $2$ decimal places)
$0.0099\, m$ (has $4$ decimal places)
Performing the subtraction:
$9.99 - 0.0099 = 9.9801$
According to the rule of significant figures for subtraction,the result must be rounded to the same number of decimal places as the term with the fewest decimal places,which is $2$ decimal places.
Rounding $9.9801$ to $2$ decimal places gives $9.98$.
Thus,the final result is $9.98\, m$.

(C) The area of a rectangle is given by the formula: $Area = \text{Length} \times \text{Breadth}$.
Given,Length $= 55.3 \ m$ (which has $3$ significant figures) and Breadth $= 25 \ m$ (which has $2$ significant figures).
Calculating the product: $55.3 \times 25 = 1382.5 \ m^{2}$.
According to the rules of significant figures,the result of multiplication should have the same number of significant figures as the measurement with the least number of significant figures.
Here,the least number of significant figures is $2$ (from $25 \ m$).
Rounding $1382.5$ to $2$ significant figures,we get $1400 \ m^{2}$,which can be written as $14 \times 10^{2} \ m^{2}$.

(A) Accuracy in a measurement is determined by the number of significant figures. $A$ higher number of significant figures indicates a more precise and accurate measurement.
$A$. $400 \times 10^{-4} \,m = 0.0400 \,m$ (Significant figures: $3$)
$B$. $0.04 \,m$ (Significant figures: $1$)
$C$. $40 \,m$ (Significant figures: $2$)
$D$. $4 \times 10^1 \,m = 40 \,m$ (Significant figures: $1$)
Comparing the significant figures: $3 > 2 > 1$. Therefore,the measurement $400 \times 10^{-4} \,m$ has the highest number of significant figures,making it the most accurate.

(A) When adding measurements,the final result should be reported to the same number of decimal places as the measurement with the fewest decimal places.
Given measurements are $18.425 \, cm$ (three decimal places),$7.21 \, cm$ (two decimal places),and $5.0 \, cm$ (one decimal place).
The sum is $18.425 + 7.21 + 5.0 = 30.635 \, cm$.
Since the measurement with the fewest decimal places is $5.0 \, cm$ (one decimal place),the final result must be rounded to one decimal place.
Rounding $30.635$ to one decimal place gives $30.6 \, cm$.

(A) The given measurement is $0.0205 \times 10^{-4} \,m$.
According to the rules for significant figures:
$1$. Leading zeros in a decimal number are not significant.
$2$. The digits $2, 0, 5$ are significant.
$3$. The power of $10$ (i.e.,$10^{-4}$) does not contribute to the number of significant figures.
Therefore,the significant figures are $2, 0, 5$,which gives a total of $3$ significant figures.

(A) The area of a circle is given by the formula $A = \pi r^2$.
Given,radius $r = 2.12 \ m$.
Here,$r$ has $3$ significant figures.
$A = 3.14159 \times (2.12)^2$.
$A = 3.14159 \times 4.4944 = 14.11965... \ m^2$.
According to the rules of significant figures,the result of multiplication should have the same number of significant figures as the measurement with the least number of significant figures.
Since $2.12$ has $3$ significant figures,the final result must be rounded off to $3$ significant figures.
Rounding $14.11965$ to $3$ significant figures gives $14.1 \ m^2$.

(A) According to Ohm's law,the potential difference $V$ is given by $V = I \times R$.
Given,$R = 10.845 \, \Omega$ (which has $5$ significant figures) and $I = 3.23 \, \text{A}$ (which has $3$ significant figures).
Calculating the product: $V = 10.845 \times 3.23 = 35.02935 \, V$.
According to the rules of significant figures in multiplication,the final result should have the same number of significant figures as the measurement with the least number of significant figures.
Here,the least number of significant figures is $3$ (from $3.23$).
Rounding $35.02935$ to $3$ significant figures,we get $35.0 \, V$.

(B) The length of the pendulum $(L)$ is given by the sum of the length of the thread $(l)$ and the radius of the bob $(r)$.
$L = l + r$
Given,$l = 63.2 \,cm$ and diameter $d = 2.256 \,cm$.
Radius $r = \frac{d}{2} = \frac{2.256}{2} = 1.128 \,cm$.
$L = 63.2 + 1.128 = 64.328 \,cm$.
According to the rules of significant figures in addition/subtraction,the result should be reported to the same number of decimal places as the measurement with the fewest decimal places.
Here,$63.2$ has one decimal place and $1.128$ has three decimal places.
Therefore,the result must be rounded to one decimal place.
$L = 64.3 \,cm$.

(B) The side length of the cube is $l = 1.2 \,m$.
Converting the side length to centimeters: $l = 1.2 \times 10^2 \,cm = 120 \,cm$.
The volume of the cube is given by $V = l^3 = (120 \,cm)^3 = 1,728,000 \,cm^3$.
This can be written as $V = 1.728 \times 10^6 \,cm^3$.
According to the rules of significant figures,the result of a calculation should have the same number of significant figures as the measurement with the least number of significant figures.
The given side length $1.2 \,m$ has $2$ significant figures.
Therefore,the volume should be rounded to $2$ significant figures.
Rounding $1.728$ to $2$ significant figures gives $1.7$.
Thus,the volume is $1.7 \times 10^6 \,cm^3$.

(B) Given masses are $m_1 = 1.6 \,g$,$m_2 = 7.32 \,g$,and $m_3 = 4.238 \,g$.
Summing these values: $m = 1.6 + 7.32 + 4.238 = 13.158 \,g$.
According to the rules of significant figures for addition,the final result should be reported to the same number of decimal places as the measurement with the fewest decimal places.
Here,$1.6 \,g$ has the fewest decimal places (one decimal place).
Therefore,we round $13.158 \,g$ to one decimal place,which gives $13.2 \,g$.

(C) Given:
Length $(l)$ = $10.2 \,cm$ (has $3$ significant figures).
Width $(w)$ = $6.8 \,cm$ (has $2$ significant figures).
Area = $l \times w = 10.2 \times 6.8 = 69.36 \,cm^2$.
According to the rules of significant figures in multiplication,the final result should have the same number of significant figures as the measurement with the least number of significant figures.
Here,the least number of significant figures is $2$ (from $6.8 \,cm$).
Rounding $69.36$ to $2$ significant figures gives $69 \,cm^2$.

(D) The correct option is $D$.
In any physical measurement,the digits that are reliably known plus the first uncertain digit are called significant figures.
For a measurement reported as $41.68 \,cm$,the digits $4, 1,$ and $6$ are certain,while the last digit $8$ is the uncertain digit.
Therefore,the uncertain digit is $8$.

(D) The given value is $4.700 \,m$.
According to the rules for significant figures,trailing zeros in a number with a decimal point are significant.
Therefore,$4.700$ has $4$ significant figures.
Now,let us check the options:
$(a)$ $4700$: Trailing zeros without a decimal point are not significant. It has $2$ significant figures.
$(b)$ $0.047$: Leading zeros are not significant. It has $2$ significant figures.
$(c)$ $4070$: Trailing zeros without a decimal point are not significant. It has $3$ significant figures.
$(d)$ $470.0$: Trailing zeros in a number with a decimal point are significant. It has $4$ significant figures.
Thus,the value $470.0 \,m$ has the same number of significant figures as $4.700 \,m$.

(C) The given value is $2.7465 \,g$.
Significant figures are the digits that carry meaningful information about the precision of a measurement.
If the value is restricted to three significant figures,we retain the first three digits $(2, 7, 4)$ and discard the remaining digits.
The digits to be discarded are the ones that are considered insignificant.
Therefore,the two insignificant digits are $6$ and $5$.

(A) Given mass $m = 4.237 \, g$ (which has $4$ significant figures).
Given volume $V = 1.72 \, cm^3$ (which has $3$ significant figures).
Density $\rho = \frac{m}{V} = \frac{4.237 \, g}{1.72 \, cm^3} \approx 2.46337 \, g \, cm^{-3}$.
According to the rules of significant figures,the result of division should have the same number of significant figures as the measurement with the least number of significant figures.
Since $1.72$ has $3$ significant figures,the final answer must be rounded to $3$ significant figures.
Therefore,the density is $2.46 \, g \, cm^{-3}$.

(B) To round off a number to a specific number of significant figures,we look at the digit following the last significant figure.
Here,we need to round $2.845$ to $3$ significant figures.
The first $3$ digits are $2, 8, 4$. The next digit is $5$.
According to the rules of rounding off,if the digit to be dropped is $5$ followed by zeros or nothing,and the preceding digit is even,it remains unchanged.
Since the digit $4$ is even,$2.845$ rounds to $2.84$.

(A) To round off $5.997$ to three significant figures,we look at the fourth digit,which is $7$.
Since $7 > 5$,we increase the third digit $(9)$ by $1$.
Adding $1$ to $9$ results in $10$,so we carry over the $1$ to the next digit.
This process continues until we get $6.00$.
Thus,$5.997 \ m$ rounded to three significant figures is $6.00 \ m$.

(B) The speed of light in vacuum is approximately $c = 3 \times 10^8 \, m/s$.
To find the order of magnitude,we express the value in the form $a \times 10^n$,where $1 \le a < 10$.
Here,$a = 3$,which satisfies the condition $1 \le 3 < 10$.
Therefore,the order of magnitude is the exponent $n$,which is $8$.

(C) Step $1$: Perform the multiplication $1.53 \times 0.9995 = 1.529235$.
Step $2$: According to the rules of significant figures,the result of multiplication should have the same number of significant figures as the measurement with the fewest significant figures. Here,$1.53$ has $3$ significant figures and $0.9995$ has $4$. Thus,the product should be rounded to $3$ significant figures: $1.53$.
Step $3$: Perform the division $\frac{1.53}{1.592} = 0.961055...$.
Step $4$: The result of division should also be rounded to the least number of significant figures present in the operands. Here,$1.53$ has $3$ significant figures and $1.592$ has $4$. Therefore,the final result must be rounded to $3$ significant figures,which is $0.961$.

(C) To determine the number of significant figures,we follow these rules:
$1$. All non-zero digits are significant.
$2$. Zeros between non-zero digits are significant.
$3$. Leading zeros are not significant.
$4$. Trailing zeros in a number with a decimal point are significant.
Applying these rules:
- $A. 1001$: There are $4$ significant figures (all digits are significant).
- $B. 010.1$: The leading zero is not significant. The digits $1, 0, 1$ are significant. Total = $3$ significant figures.
- $C. 100.100$: All digits are significant because the zeros are between non-zero digits or are trailing zeros after a decimal point. Total = $6$ significant figures.
- $D. 0.0010010$: The leading zeros are not significant. The digits $1, 0, 0, 1, 0$ are significant. Total = $5$ significant figures.
Matching the results:
$A-II, B-I, C-IV, D-III$.

(C) The given readings are $4.62 \,s, 4.632 \,s, 4.6 \,s$,and $4.64 \,s$.
To find the arithmetic mean,we first calculate the sum: $4.62 + 4.632 + 4.6 + 4.64 = 18.492 \,s$.
According to the rules of significant figures for addition,the result should be reported to the same number of decimal places as the measurement with the fewest decimal places. Here,$4.6 \,s$ has only one decimal place.
Therefore,the sum should be rounded to one decimal place: $18.5 \,s$.
Now,calculate the arithmetic mean: $\text{Mean} = \frac{18.5}{4} = 4.625 \,s$.
Finally,applying the rules of significant figures for division,the result must have the same number of significant figures as the measurement with the fewest significant figures. The value $4.6 \,s$ has two significant figures.
Rounding $4.625 \,s$ to two significant figures gives $4.6 \,s$.

(D) The given expression is $y = \frac{32.3 \times 1125}{27.4}$.
Calculating the value: $y = 1326.18613...$
According to the rules of significant figures in multiplication and division,the result should be reported to the same number of significant figures as the operand with the least number of significant figures.
In the expression,$32.3$ has $3$ significant figures,$1125$ has $4$ significant figures,and $27.4$ has $3$ significant figures.
The least number of significant figures is $3$.
Rounding $1326.186$ to $3$ significant figures,we get $1330$.

(C) The given masses are $m_1 = 435.42 \ g$,$m_2 = 226.3 \ g$,and $m_3 = 0.125 \ g$.
To find the sum,we add them: $435.42 + 226.3 + 0.125 = 661.845 \ g$.
According to the rules for addition with significant figures,the final result should be reported to the same number of decimal places as the measurement with the fewest decimal places.
The measurements have $2$,$1$,and $3$ decimal places respectively. The least number of decimal places is $1$.
Therefore,rounding $661.845 \ g$ to one decimal place gives $661.8 \ g$.

(C) Statement-$1$ is true because the definition of significant figures includes all digits that are known with certainty plus the first digit that is uncertain.
Statement-$2$ is false because,according to the rules of significant figures,trailing zeros in a number containing a decimal point are always significant (e.g.,$3.500$ has $4$ significant figures).

(A) In addition and subtraction,the result should be reported to the same number of decimal places as the measurement with the fewest decimal places.
$(i)\ 11.3 \ cm$ (one decimal place) $+ 4 \ cm$ (zero decimal places) $= 15 \ cm$. Thus,$(i)$ is incorrect.
$(ii)\ 4.53 \ m$ (two decimal places) $- 1.2 \ m$ (one decimal place) $= 3.3 \ m$. Thus,$(ii)$ is correct.
$(iii)\ 5.45 \ kg$ (two decimal places) $- 3.2 \ kg$ (one decimal place) $= 2.3 \ kg$. Thus,$(iii)$ is incorrect.
$(iv)\ 84.8 \ cm$ (one decimal place) $+ 48.6 \ cm$ (one decimal place) $= 133.4 \ cm$. Thus,$(iv)$ is incorrect.
Therefore,only $(ii)$ is correct.

(B) Density $\rho$ is given by the formula $\rho = \frac{m}{V}$.
Given mass $m = 4.237 \ g$ (which has $4$ significant figures).
Given volume $V = 2.5 \ cm^3$ (which has $2$ significant figures).
Calculating the density: $\rho = \frac{4.237}{2.5} = 1.6948 \ g \ cm^{-3}$.
According to the rules of significant figures,the result of division should have the same number of significant figures as the measurement with the fewest significant figures.
Since $2.5$ has $2$ significant figures,the final answer must be rounded to $2$ significant figures.
Rounding $1.6948$ to $2$ significant figures gives $1.7 \ g \ cm^{-3}$.

(A) For $0.007\ m^2$,the leading zeros are not significant. Thus,there is $1$ significant figure. Matches $(q)$.
$(b)$ For $2.64 \times 10^{24}\ kg$,the power of $10$ is not significant. The digits $2, 6, 4$ are significant. Thus,there are $3$ significant figures. Matches $(s)$.
$(c)$ For $0.0021\ g\ cm^{-3}$,the leading zeros are not significant. The digits $2, 1$ are significant. Thus,there are $2$ significant figures. Matches $(p)$.
$(d)$ For $0.0006032\ J$,the leading zeros are not significant. The digits $6, 0, 3, 2$ are significant. Thus,there are $4$ significant figures. Matches $(r)$.
Therefore,the correct matching is $a-q, b-s, c-p, d-r$.

(B) Given dimensions are: $L = 10.5 \ cm$ ($3$ significant figures),$b = 0.05 \ mm = 0.005 \ cm$ ($1$ significant figure),and $t = 6.0 \ \mu m = 6.0 \times 10^{-4} \ cm$ ($2$ significant figures).
According to the rules of significant figures,the result of multiplication should have the same number of significant figures as the measurement with the least number of significant figures.
Here,the least number of significant figures is $1$ (from $b = 0.05 \ mm$).
Volume $V = L \times b \times t = 10.5 \ cm \times 0.005 \ cm \times 6.0 \times 10^{-4} \ cm$.
$V = 10.5 \times 5 \times 10^{-3} \times 6.0 \times 10^{-4} \ cm^3 = 315 \times 10^{-7} \ cm^3 = 3.15 \times 10^{-5} \ cm^3$.
Rounding to $1$ significant figure,we get $V = 3 \times 10^{-5} \ cm^3$.

(C) The diameter of the Earth $(D_e)$ is $2 \times 6371 \ km = 12742 \ km = 1.2742 \times 10^4 \ km$.
The order of magnitude of the diameter of the Earth is $10^4$.
The diameter of the orbit $(D_o)$ is $2 \times 149 \times 10^6 \ km = 298 \times 10^6 \ km = 2.98 \times 10^8 \ km$.
The order of magnitude of the diameter of the orbit is $10^8$.
The difference in the order of magnitude is $8 - 4 = 4$.
Therefore,the order of magnitude of the diameter of the orbit is greater than that of the Earth by $10^4$.

(B) Density is defined as the ratio of mass to volume: $\text{Density} = \frac{\text{Mass}}{\text{Volume}}$.
Given: $\text{Mass} = 49.53 \ g$ (which has $4$ significant figures).
Given: $\text{Volume} = 1.5 \ cm^3$ (which has $2$ significant figures).
Calculation: $\text{Density} = \frac{49.53}{1.5} = 33.02 \ g \ cm^{-3}$.
According to the rules of significant figures,when dividing,the result should have the same number of significant figures as the measurement with the fewest significant figures.
Since $1.5$ has $2$ significant figures,the final answer must be rounded to $2$ significant figures.
Therefore,the density is $33 \ g \ cm^{-3}$.

(B) For the number $4.8000 \times 10^{4}$:
In scientific notation,the power of $10$ does not contribute to the number of significant figures.
The digits $4, 8, 0, 0, 0$ are all significant because trailing zeros after the decimal point are significant.
Thus,there are $5$ significant figures.
For the number $48000.50$:
All non-zero digits are significant.
Zeros between two non-zero digits are significant.
Trailing zeros after the decimal point are also significant.
Therefore,all digits $4, 8, 0, 0, 0, 5, 0$ are significant,totaling $7$ significant figures.

(B) According to the rules for significant figures,all non-zero digits are significant.
Additionally,trailing zeros in a number containing a decimal point are significant.
In the number $4.870$,the digits $4, 8, 7$ are non-zero,and the trailing zero after the decimal point is significant.
Therefore,the total number of significant figures is $4$.

(A) To determine the number of significant figures,we apply the following rules:
$(i)$ For $A = 0.001204 \ m$: Leading zeros are not significant. The digits $1, 2, 0, 4$ are significant. Thus,$N_A = 4$.
$(ii)$ For $B = 43120000 \ m$: Trailing zeros in a number without a decimal point are not significant. The digits $4, 3, 1, 2$ are significant. Thus,$N_B = 4$.
$(iii)$ For $C = 1.200 \ m$: Trailing zeros in a number with a decimal point are significant. The digits $1, 2, 0, 0$ are significant. Thus,$N_C = 4$.
Since $N_A = 4, N_B = 4$,and $N_C = 4$,we conclude that $N_A = N_B = N_C$.

(A) According to the rules of significant figures,for a number less than $1$,the zeros to the right of the decimal point and to the left of the first non-zero digit are not significant.
In the number $0.00764$,the zeros before the digit $7$ are not significant.
Therefore,the significant figures are $7, 6,$ and $4$,which makes a total of $3$ significant figures.
Thus,both Assertion $(A)$ and Reason $(R)$ are true,and $(R)$ is the correct explanation of $(A)$.

(A) For $74.083$,all non-zero digits and the zero between them are significant,so there are $5$ significant figures $(IV)$.
For $0.029$,leading zeros are not significant,so there are $2$ significant figures $(III)$.
For $0.002407$,leading zeros are not significant,but the zero between $2$ and $4$ is significant,so there are $4$ significant figures $(II)$.
For $2.74 \times 10^7$,the exponential term does not contribute to significant figures,so there are $3$ significant figures $(I)$.
Thus,the correct matching is $A-IV, B-III, C-II, D-I$.

(B) For any number less than $1$,the leading zeros before or after the decimal point are not significant.
In the given number $0.00005041$,the zeros to the left of the digit $5$ are leading zeros and are not significant.
The digits $5, 0, 4, 1$ are significant.
Therefore,the number of significant figures is $4$.

(B) According to the rules for significant figures:
$1$. Leading zeros (zeros to the left of the first non-zero digit) are not significant. In $0.03240$,the zeros before $3$ are not significant.
$2$. Non-zero digits are always significant. Here,$3, 2, 4$ are significant.
$3$. Trailing zeros in a decimal number are significant. The zero at the end of $0.03240$ is significant.
Therefore,the significant figures are $3, 2, 4, 0$,which gives a total of $4$ significant figures.

(A) Step $1$: Perform the subtraction inside the parentheses. $0.312 - 0.03 = 0.282$. According to the rules of significant figures for subtraction,the result should have the same number of decimal places as the measurement with the fewest decimal places. Here,$0.03$ has two decimal places,so the result $0.282$ is rounded to $0.28$ (two decimal places).
Step $2$: Perform the division. $\frac{0.501}{0.05} = 10.02$.
Step $3$: Multiply the results. $10.02 \times 0.28 = 2.8056$.
Step $4$: Apply the rules for multiplication. The result should have the same number of significant figures as the measurement with the fewest significant figures. $0.05$ has $1$ significant figure,and $0.28$ has $2$ significant figures. Therefore,the final result should be rounded to $1$ significant figure. Thus,the number of significant figures is $1$.

(D) The length of the side of the cube is $l = 1.2 \times 10^{-2} \ m$.
The volume of a cube is given by $V = l^3$.
$V = (1.2 \times 10^{-2} \ m)^3 = 1.728 \times 10^{-6} \ m^3$.
According to the rules of significant figures,the result of a multiplication or division should have the same number of significant figures as the measurement with the fewest significant figures.
The given length $1.2 \times 10^{-2} \ m$ has $2$ significant figures.
Therefore,the volume should be rounded to $2$ significant figures.
Rounding $1.728$ to $2$ significant figures gives $1.7$.
Thus,the volume is $1.7 \times 10^{-6} \ m^3$.

(C) In scientific notation,every number is expressed as $a \times 10^{b}$,where $a$ is a number between $1$ and $10$,and $b$ is any positive or negative integer exponent.
Significant figures are determined only by the digits in the coefficient $a$.
The power of $10$ (i.e.,$10^{22}$) is irrelevant to the determination of significant figures.
In the given value $3.78 \times 10^{22}$,the coefficient is $3.78$.
Since all non-zero digits are significant,the number of significant figures in $3.78$ is $3$.

(C) According to the rules for significant figures:
$1$. Leading zeros (zeros to the left of the first non-zero digit) are not significant. Here,the zeros before $7$ are not significant.
$2$. Trailing zeros in a decimal number are significant.
In the measurement $0.079000 \ m$,the digits $7, 9, 0, 0, 0$ are significant.
Therefore,the number of significant figures is $5$.

(A) The length of the original rod is $43.4 \ m$ (which has $3$ significant figures and is precise up to the first decimal place).
The length of the piece cut is $3.532 \ m$ (which has $4$ significant figures and is precise up to the third decimal place).
When subtracting,the result should be reported to the same number of decimal places as the measurement with the fewest decimal places.
Remaining length $= 43.4 \ m - 3.532 \ m = 39.868 \ m$.
Since $43.4$ has only one decimal place,we must round the result to one decimal place.
Rounding $39.868$ to one decimal place gives $39.9 \ m$.

(D) According to the rules for significant figures:
$1$. All non-zero digits are significant.
$2$. Trailing zeros in a number without a decimal point are generally not considered significant unless specified by measurement precision.
In the number $830600$,the digits $8, 3, 0, 6$ are significant.
The two trailing zeros are not significant because there is no decimal point.
Therefore,the significant figures are $8, 3, 0, 6$,which gives a total of $4$ significant figures.