Significant figures and Units for measurement Questions in English

(A) One fermi is defined as $1 \ fm = 10^{-15} \ m$.
Since $1 \ m = 10^2 \ cm$,
$1 \ fm = 10^{-15} \times 10^2 \ cm = 10^{-13} \ cm$.

(D) picometre $(pm)$ is a unit of length in the metric system.
It is defined as $1 \times 10^{-12} \ m$.
This is one trillionth of a metre, which is the $SI$ base unit of length.

(C) Pressure is defined as force per unit area: $P = \frac{F}{A} = \frac{[MLT^{-2}]}{[L^2]} = [ML^{-1}T^{-2}]$.
Energy per unit volume is defined as energy divided by volume: $\frac{E}{V} = \frac{[ML^2T^{-2}]}{[L^3]} = [ML^{-1}T^{-2}]$.
Since both have the same dimensional formula $[ML^{-1}T^{-2}]$,the dimensions of pressure are the same as that of energy per unit volume.

(D) The prefix $10^{18}$ is $\text{Exa}$. Its symbol is $E$.
$\text{Giga} (G)$ represents $10^9$.
$\text{Nano} (n)$ represents $10^{-9}$.
$\text{Mega} (M)$ represents $10^6$.
$\text{Kilo} (k)$ represents $10^3$.

(B) Rules for significant figures state that all non-zero digits are significant. Zeros preceding the first non-zero digit are not significant.
$1$. For $161 \ cm$,all digits are non-zero,so there are $3$ significant figures.
$2$. For $0.161 \ cm$,the leading zero is not significant,so there are $3$ significant figures.
$3$. For $0.0161 \ cm$,the leading zeros are not significant,so there are $3$ significant figures.
Thus,the numbers have $3, 3$ and $3$ significant figures respectively.

(C) In the given number $0.00051$,the zeros to the left of the non-zero digit are not considered as significant figures.
Thus,the given number has only $2$ significant figures.

(B) Step $1$: Perform the subtraction in the numerator: $(29.2 - 20.2) = 9.0$.
Since the subtraction rule states that the result should have the same number of decimal places as the term with the fewest decimal places,$9.0$ has $1$ decimal place (two significant figures).
Step $2$: The expression becomes $\frac{9.0 \times 1.79 \times 10^5}{1.37}$.
Step $3$: According to the multiplication and division rule,the final result should have the same number of significant figures as the term with the fewest significant figures.
Here,$9.0$ has $2$ significant figures,$1.79$ has $3$ significant figures,and $1.37$ has $3$ significant figures.
Therefore,the final answer must have $2$ significant figures.

(A) The mass of pure ethyl alcohol is calculated by subtracting the mass of water from the total mass of the sample.
Mass of pure ethyl alcohol $= 81.4 \ g - 0.002 \ g = 81.398 \ g$.
According to the rules of significant figures for addition and subtraction,the result should be reported to the same number of decimal places as the measurement with the fewest decimal places.
Here,$81.4 \ g$ has one decimal place,while $0.002 \ g$ has three decimal places.
Therefore,the result must be rounded to one decimal place.
Rounding $81.398 \ g$ to one decimal place gives $81.4 \ g$.

(A) The unit of energy (work) is Joule $(J)$ and the unit of pressure is Pascal $(Pa)$.
Dimensional formula of Joule $(J)$ is $[ML^2T^{-2}]$.
Dimensional formula of Pascal $(Pa)$ is $[ML^{-1}T^{-2}]$.
Therefore,the unit $J \, Pa^{-1}$ has dimensions:
$\frac{[ML^2T^{-2}]}{[ML^{-1}T^{-2}]} = [L^3]$.
Since the dimension $[L^3]$ represents volume,the $SI$ unit is $m^3$.

(C) The precision of a measurement is determined by the number of decimal places.
$1 \ g = 1000 \ mg$,which means $0.001 \ g = 1 \ mg$.
To express a mass nearest to the milligram,the value must be measured up to three decimal places (i.e.,the thousandths place in grams).
Comparing the options:
$16 \ g$ has $0$ decimal places.
$16.4 \ g$ has $1$ decimal place.
$16.428 \ g$ has $3$ decimal places,which corresponds to the milligram precision $(16428 \ mg)$.
$16.4284 \ g$ has $4$ decimal places,which is more precise than the milligram level.
Therefore,$16.428 \ g$ is the value expressed nearest to the milligram.

(B) When numbers are written in scientific notation,the number of digits in the coefficient (the part between $1$ and $10$) gives the number of significant figures.
In the number $6.02 \times 10^{23}$,the coefficient is $6.02$.
Since all non-zero digits and zeros between non-zero digits are significant,$6.02$ has $3$ significant figures.

(D) Zepto is a unit prefix in the metric system denoting a factor of $10^{-21}$.
$1 \ zepto = 10^{-21}$

(A) The correct option is $A$.
According to the rules for significant figures,trailing zeros in a number without a decimal point are not considered significant.
Therefore,only the digits $3$ and $4$ are significant.
The total number of significant figures is $2$.

(A) The number of significant figures in $6.0023$ is $5$.
According to the rules of significant figures,all zeros between two non-zero digits are significant.

(B) For $P = 0.0030 \ m$: Leading zeros are not significant. The digits $3$ and $0$ are significant. Thus,there are $2$ significant figures.
For $Q = 2.40 \ m$: Trailing zeros in a decimal number are significant. Thus,there are $3$ significant figures.
For $R = 3000 \ m$: In the absence of a decimal point,trailing zeros are generally not considered significant. Thus,there is $1$ significant figure.
Therefore,the significant figures are $2, 3, 1$.

(B) According to the rules for significant figures,all zeros between two non-zero digits are significant.
In the number $60.0001$,there are $4$ zeros between the digits $6$ and $1$.
Therefore,all $6$ digits are significant.
Thus,the number of significant figures is $6$.

(C) The precision of a result is limited by the least precise measurement.
In the given values,$3.929 \ g$ has $4$ significant figures and $4.0 \ g$ has $2$ significant figures.
The result should be reported to the same number of significant figures as the least precise measurement,which is $2$.
Rounding $3.929$ to $2$ significant figures gives $3.9 \ g$.

(D) We know that $1 \ m = 100 \ cm$.
Therefore,$1 \ m^3 = (100 \ cm)^3 = 100 \times 100 \times 100 \ cm^3$.
$1 \ m^3 = 1,000,000 \ cm^3 = 10^6 \ cm^3$.
Thus,the correct option is $D$.

(A) Newton is the unit of force,whereas $Torr$,$Pascal$,and $Bar$ are units of pressure.

(B) The relationship between temperature in Celsius $(C)$ and Fahrenheit $(F)$ is given by the formula: $\frac{C}{5} = \frac{F - 32}{9}$.
For a change in temperature ($\Delta C$ and $\Delta F$),the constant $32$ cancels out,resulting in: $\frac{\Delta C}{5} = \frac{\Delta F}{9}$.
Given $\Delta C = 1\,^{\circ}C$,we have: $\Delta F = \frac{9}{5} \times 1 = 1.8\,^{\circ}F$.
Therefore,a rise of $1\,^{\circ}C$ is equal to a rise of $1.8\,^{\circ}F$ or $\frac{9}{5}\,^{\circ}F$.

(C) The correct relations for volume are:
$1 \ L = 10^3 \ mL$
$1 \ L = 1 \ dm^3$
$1 \ L = 10^3 \ cm^3$
$1 \ L = 10^{-3} \ m^3$
Therefore,the relation $1 \ L = 10^3 \ m^3$ is incorrect.

(A) The standard atmospheric pressure is defined as $1 \ atm = 1.01325 \times 10^5 \ Pa$.
Since $1 \ Pa = 10 \ dynes \ cm^{-2}$,we have $1 \ atm = 1.01325 \times 10^5 \times 10 \ dynes \ cm^{-2} = 1.01325 \times 10^6 \ dynes \ cm^{-2}$.
Therefore,$1 \ atm$ is approximately equal to $10^6 \ dynes \ cm^{-2}$.

(C) The universal gas constant $R$ is a fundamental constant in the ideal gas equation $PV = nRT$.
Its numerical value is not dependent on the nature of the gas,the pressure,or the temperature.
However,its value changes depending on the units used for pressure,volume,and temperature.
Therefore,the value of $R$ depends only on the units of measurement.

(C) The dimension of pressure $(P)$ is defined as force per unit area,which is $[M L^{-1} T^{-2}]$.
Energy $(E)$ has dimensions $[M L^2 T^{-2}]$.
Volume $(V)$ has dimensions $[L^3]$.
Therefore,the dimension of energy per unit volume is $\frac{[M L^2 T^{-2}]}{[L^3]} = [M L^{-1} T^{-2}]$.
Since the dimensions of pressure and energy per unit volume are identical,the correct option is $C$.

(B) The calorie is a unit of energy.
By definition,$1 \ \text{calorie}$ is the amount of heat energy required to raise the temperature of $1 \ \text{g}$ of water by $1 \ ^\circ C$.
According to the conversion factor,$1 \ \text{calorie} = 4.184 \ J$.
Therefore,the correct option is $B$.

(C) We know the conversion factors for energy units:
$1 \ joule = 10^7 \ erg$
$1 \ calorie = 4.184 \ joule = 4.184 \times 10^7 \ erg$
Comparing these values:
$1 \ calorie (4.184 \times 10^7 \ erg) > 1 \ joule (10^7 \ erg) > 1 \ erg$
Therefore,the correct order is $1 \ calorie > 1 \ joule > 1 \ erg$.

(A) The conversion factors for energy units are as follows:
$1 \, cal = 4.184 \, J$
$1 \, J = 10^{7} \, erg$
$1 \, eV = 1.602 \times 10^{-19} \, J$
Comparing these values,$1 \, cal$ is equal to $4.184 \, J$,which is $4.184 \times 10^{7} \, erg$. Since $1 \, eV$ is an extremely small unit $(1.602 \times 10^{-19} \, J)$,the calorie represents the largest amount of energy among the given options.
Therefore,the correct option is $A$.

(A) The triple point of a substance is the temperature and pressure at which its three phases (solid,liquid,and gas) coexist in equilibrium.
For water,the triple point is $273.16 \ K$ at a pressure of $611.2 \ Pa$.
Therefore,the correct option is $A$.

(C) The density of the given metals are:
$1$. Iron $(Fe)$: $7.87 \ g/cm^3$
$2$. Copper $(Cu)$: $8.96 \ g/cm^3$
$3$. Silver $(Ag)$: $10.49 \ g/cm^3$
$4$. Gold $(Au)$: $19.32 \ g/cm^3$
Comparing these values,Gold $(Au)$ has the highest density and is the heaviest among the given options.
Therefore,the correct option is $(C)$.

(D) To compare the units of energy,we convert them into a common unit,the Joule $(J)$:
$1 \text{ electron volt (eV)} = 1.602 \times 10^{-19} \ J$
$1 \text{ erg} = 10^{-7} \ J$
$1 \text{ Joule (J)} = 1 \ J$
$1 \text{ calorie (cal)} = 4.184 \ J$
Comparing these values,$4.184 \ J$ is the largest value.
Therefore,the calorie is the largest unit of energy among the given options.

(A) The conversion factors for energy units are as follows:
$1 \, \text{calorie} = 4.184 \, \text{Joule}$
$1 \, \text{Joule} = 10^7 \, \text{erg}$
$1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{Joule}$
Comparing these,$1 \, \text{calorie}$ is equal to $4.184 \, \text{Joule}$,which is significantly larger than $1 \, \text{Joule}$,$1 \, \text{erg}$,or $1 \, \text{eV}$.
Therefore,the calorie has the highest energy value.

(B) The conversion factors for energy units are as follows:
$1 \text{ calorie} = 4.184 \text{ joules}$
$1 \text{ joule} = 10^7 \text{ ergs}$
Therefore,$1 \text{ calorie} \approx 4.184 \text{ joules} = 4.184 \times 10^7 \text{ ergs}$.
Comparing the values: $1 \text{ calorie} (4.184 \times 10^7 \text{ ergs}) > 1 \text{ joule} (10^7 \text{ ergs}) > 1 \text{ erg}$.
Thus,the correct order is $1 \text{ calorie} > 1 \text{ joule} > 1 \text{ erg}$.

(C) Given: Density of $Hg = 13.6 \, g/cm^3$.
We know that $1 \, cm^3 = 1 \, mL$.
Therefore,density $= 13.6 \, g/mL$.
Volume $= 1 \, L = 1000 \, mL$.
Mass $= \text{Density} \times \text{Volume}$.
Mass $= 13.6 \, g/mL \times 1000 \, mL = 13600 \, g$.

(A) Volume of the block in cubic inches = $3.0 \times 4.0 \times 5.0 = 60.0 \, \text{inch}^3$.
Conversion factor: $1 \, \text{inch} = 2.54 \, \text{cm}$,so $1 \, \text{inch}^3 = (2.54)^3 \, \text{cm}^3 = 16.387 \, \text{cm}^3$.
Volume in $\text{cm}^3 = 60.0 \times 16.387 = 983.22 \, \text{cm}^3$.
Mass = $\text{Density} \times \text{Volume} = 2.7 \, \text{g/cm}^3 \times 983.22 \, \text{cm}^3 = 2654.694 \, \text{g}$.
Rounding to two significant figures (based on the given dimensions),the mass is approximately $2.7 \times 10^3 \, \text{g}$. However,calculating $60 \times (2.54)^3 \times 2.7$ gives $2654.69$,which is $2.65 \times 10^3 \, \text{g}$. Given the options,the calculation $2.65 \times 10^4$ appears to be a typo in the provided options' exponent,but $2.65$ is the correct numerical coefficient.

(C) Given that $1 \ kg = 2.205 \ lb$.
To convert the weight from $lb$ to $kg$,we use the conversion factor:
$\text{Weight in } kg = \frac{\text{Weight in } lb}{2.205 \ lb/kg}$
$\text{Weight in } kg = \frac{275 \ lb}{2.205 \ lb/kg} \approx 124.716 \ kg$.
Rounding to the appropriate significant figures,we get $124.70 \ kg$.

(B) The conversion factors are:
$1 = \frac{1000 \ m}{1 \ kg}$ and $1 = \frac{1.094 \ yards}{1 \ m}$.
To convert $3.00 \ kg$ to yards,we apply the conversion factors sequentially:
$3.00 \ kg \times \frac{1000 \ m}{1 \ kg} \times \frac{1.094 \ yards}{1 \ m} = 3.00 \times 1000 \times 1.094 \ yards$.
Calculation:
$3000 \times 1.094 = 3282 \ yards$.

(C) The conversion factors are as follows:
$1 \ \text{femtometer} (fm) = 10^{-15} \ m = 10^{-3} \ pm$ (Matches $R$)
$1 \ \text{picometer} (pm) = 10^{-12} \ m = 10^{-3} \ nm$ (Matches $W$)
$1 \ \text{nanometer} (nm) = 10^{-9} \ m = 10^{-3} \ \mu m$ (Matches $P$)
$1 \ \text{micrometer} (\mu m) = 10^{-6} \ m = 10^{-3} \ mm$ (Matches $Q$)
Therefore, the correct matching is $1-R, 2-W, 3-P, 4-Q$.

(A) Volume of water = $20 \ L = 20 \times 1000 \ cm^3 = 20000 \ cm^3$.
Volume of cuboid = $Length \times Width \times Height$.
$20000 \ cm^3 = 30 \ cm \times 25 \ cm \times Height$.
$Height = \frac{20000}{30 \times 25} \ cm = \frac{20000}{750} \ cm \approx 26.67 \ cm$.
Since $1 \ dm = 10 \ cm$,then $Height = \frac{26.67}{10} \ dm = 2.667 \ dm$.

(C) Given that $1 \ L = 1000 \ mL$.
Since $1000 \ mL$ contains $6.023 \times 10^{16}$ $H^{+}$ ions,
the number of $H^{+}$ ions in $1 \ mL$ is calculated as:
$\frac{6.023 \times 10^{16}}{1000} = 6.023 \times 10^{13}$.

(A) The triple point of water is the temperature and pressure at which the three phases of water (solid,liquid,and gas) coexist in thermodynamic equilibrium.
The triple point of water is exactly $273.16 \ K$.

(A) Density of solution $= 3.12 \ g \ mL^{-1}$
Volume of solution $= 1.5 \ mL$
Mass of solution = Volume $\times$ Density
$= 1.5 \ mL \times 3.12 \ g \ mL^{-1} = 4.68 \ g$
Since the volume $1.5 \ mL$ has $2$ significant figures,the result should be rounded to $2$ significant figures.
Therefore,$4.68 \ g$ becomes $4.7 \ g$.

(C) To express a number with a specific number of significant figures,we use scientific notation in the form $a \times 10^n$.
For $5000$ to have $3$ significant figures,it must be written as $5.00 \times 10^3$.
The digits $5, 0, 0$ are significant here,totaling $3$ significant figures.

(C) We know that $1 \, m = 100 \, cm$ and $1 \, m = 10 \, dm$.
Therefore,$100 \, cm = 10 \, dm$.
This implies $1 \, cm = \frac{10}{100} \, dm = 0.1 \, dm = 10^{-1} \, dm$.
Now,convert $5 \times 10^2 \, cm$ to $dm$:
$5 \times 10^2 \, cm = 5 \times 10^2 \times 10^{-1} \, dm = 5 \times 10^1 \, dm$.

(D) $1 \, \mathring{A} = 10^{-10} \, \text{m}$
$1 \, \text{nm} = 10^{-9} \, \text{m}$
Therefore,$1 \, \mathring{A} = 10^{-1} \, \text{nm} = 0.1 \, \text{nm}$
$0.15 \, \mathring{A} = 0.15 \times 0.1 \, \text{nm} = 0.015 \, \text{nm}$

(B) $1 \, \mu g = 1 \times 10^{-6} \, g$
$1 \, mg = 1 \times 10^{-3} \, g$
Therefore,$1 \, \mu g = \frac{1 \times 10^{-6} \, g}{1 \times 10^{-3} \, g/mg} = 10^{-3} \, mg$.

(B) To convert $1 \, kg \, m^{-3}$ to $g \, cm^{-3}$:
$1 \, kg \, m^{-3} = \frac{1 \, kg}{1 \, m^3}$
Since $1 \, kg = 10^3 \, g$ and $1 \, m = 10^2 \, cm$,then $1 \, m^3 = (10^2 \, cm)^3 = 10^6 \, cm^3$.
Substituting these values:
$\frac{1 \times 10^3 \, g}{10^6 \, cm^3} = 10^{3-6} \, g \, cm^{-3} = 10^{-3} \, g \, cm^{-3}$.

(D) Mass $= 4 \ g = 4 \times 10^{-3} \ kg$
Volume $= 2 \ cm^3 = 2 \times 10^{-6} \ m^3$
Density $= \frac{\text{Mass}}{\text{Volume}} = \frac{4 \times 10^{-3} \ kg}{2 \times 10^{-6} \ m^3} = 2 \times 10^3 \ kg \ m^{-3}$

(A) The relationship between Celsius and Fahrenheit scales is given by the formula: $F = \frac{9}{5}C + 32$.
Since the melting point of ice is $0^{\circ}C$,we substitute $C = 0$ into the formula.
$F = \frac{9}{5}(0) + 32 = 32$.
Therefore,the melting point of ice is $32^{\circ}F$.

(C) The relationship between Fahrenheit and Celsius is given by: $^\circ F = \frac{9}{5} ( ^\circ C ) + 32$.
Substituting $68 \, ^\circ F$: $68 = \frac{9}{5} ( ^\circ C ) + 32$.
$68 - 32 = \frac{9}{5} ( ^\circ C ) \implies 36 = \frac{9}{5} ( ^\circ C )$.
$^\circ C = \frac{36 \times 5}{9} = 20 \, ^\circ C$.
To convert Celsius to Kelvin: $K = ^\circ C + 273.15$.
$K = 20 + 273 = 293 \, K$ (using $273$ as the standard conversion factor).

(C) The relationship between Celsius $(C)$ and Fahrenheit $(F)$ scales is given by the formula: $F = \frac{9}{5}C + 32$.
To find the temperature where $C = F$,let $x$ be the temperature.
Substituting $x$ into the equation: $x = \frac{9}{5}x + 32$.
Rearranging the terms: $x - \frac{9}{5}x = 32$.
$-\frac{4}{5}x = 32$.
$x = 32 \times (-\frac{5}{4}) = -40$.
Thus,at $-40^\circ$,the values on both scales are equal.