Units Questions in English

(A) We know that $1 \mathring{A} = 10^{-10} \text{ m}$.
We also know that $1 \text{ fm} = 10^{-15} \text{ m}$.
To express $1 \mathring{A}$ in terms of $\text{fm}$, we perform the following calculation:
$1 \mathring{A} = \frac{10^{-10} \text{ m}}{10^{-15} \text{ m}} \text{ fm} = 10^{-10 - (-15)} \text{ fm} = 10^5 \text{ fm}$.
Thus, $1 \mathring{A} = 10^5 \text{ fm}$.

(A) The unit kilowatt-hour $(kWh)$ is a unit of energy.
$1 \text{ kWh} = 1 \text{ kilowatt} \times 1 \text{ hour}$.
Since $1 \text{ kilowatt} = 1000 \text{ watts} = 1000 \text{ J/s}$ and $1 \text{ hour} = 3600 \text{ seconds}$.
Therefore,$1 \text{ kWh} = 1000 \text{ J/s} \times 3600 \text{ s}$.
$1 \text{ kWh} = 3,600,000 \text{ J} = 3.6 \times 10^6 \text{ J}$.

(D) The unit of force in the $CGS$ system is $Dyne$ and in the $SI$ system is $Newton$. The relationship is $1 \ Newton = 10^5 \ Dyne$. Therefore,the statement $1 \ Newton = 10^{-5} \ Dynes$ is incorrect.

(D) We are given the unit $1 \ gm \ cm \ s^{-1}$.
To convert this to $Ns$ (Newton-seconds),we express the units in terms of $kg$,$m$,and $s$.
We know that $1 \ gm = 10^{-3} \ kg$ and $1 \ cm = 10^{-2} \ m$.
Substituting these values into the expression:
$1 \ gm \ cm \ s^{-1} = (10^{-3} \ kg) \times (10^{-2} \ m) \times s^{-1}$
$= 10^{-5} \ kg \ m \ s^{-1}$.
Since $1 \ N = 1 \ kg \ m \ s^{-2}$,it follows that $1 \ Ns = 1 \ kg \ m \ s^{-1}$.
Therefore,$1 \ gm \ cm \ s^{-1} = 10^{-5} \ Ns$.
Comparing this with $x \ Ns$,we get $x = 10^{-5}$.

(D) The Solar constant is given as $2 \ cal / (cm^2 \cdot min)$.
To convert this to $SI$ units ($W/m^2$ or $J / (m^2 \cdot s)$):
$1 \ cal = 4.18 \ J$
$1 \ min = 60 \ s$
$1 \ m^2 = 10^4 \ cm^2$
Solar constant $= 2 \times \frac{4.18 \ J}{10^{-4} \ m^2 \times 60 \ s}$
Solar constant $= 2 \times 4.18 \times \frac{10^4}{60} \ J/(m^2 \cdot s)$
Solar constant $= \frac{8.36 \times 10000}{60} \ W/m^2 = \frac{83600}{60} \ W/m^2 = 1393.33 \ W/m^2$.

(D) physical quantity is a property of a material or system that can be quantified by measurement. Volume,time,and length are all physical quantities that can be measured. The mole is a unit used to measure the amount of substance,not a physical quantity itself. Therefore,the correct answer is $D$.

(N/A) The statement is true because the terms 'large' and 'small' are relative. $A$ physical quantity can only be described as large or small when compared to a standard reference value.
$(a)$ Atoms are very small objects compared to a grain of sand.
$(b)$ $A$ jet plane moves with a speed much greater than that of a car.
$(c)$ The mass of Jupiter is very large compared to the mass of the Earth.
$(d)$ The air inside this room contains a large number of molecules compared to the number of molecules in a small test tube.
$(e)$ $A$ proton is much more massive than an electron (this statement is already meaningful as it compares two specific physical entities).
$(f)$ The speed of sound is much smaller than the speed of light (this statement is already meaningful as it compares two specific physical entities).

(B) The distance between the Sun and the Earth is given by the product of the speed of light and the time taken by light to travel this distance.
Given that in the new unit system,the speed of light $c = 1 \; \text{unit/s}$.
The time taken by light to travel from the Sun to the Earth is $t = 8 \; \text{min} \; 20 \; \text{s}$.
Converting the time into seconds: $t = (8 \times 60) \; \text{s} + 20 \; \text{s} = 480 \; \text{s} + 20 \; \text{s} = 500 \; \text{s}$.
Therefore,the distance $d = c \times t = 1 \times 500 = 500 \; \text{units}$.

(A) According to Newton's law of universal gravitation,the force $F$ between two masses $m_1$ and $m_2$ separated by a distance $r$ is given by $F = G \frac{m_1 m_2}{r^2}$.
Rearranging for $G$,we get $G = \frac{F \cdot r^2}{m_1 \cdot m_2}$.
In the $CGS$ system,the unit of force $F$ is $dyne$ $(dyn)$,the unit of distance $r$ is $centimeter$ $(cm)$,and the unit of mass $m$ is $gram$ $(g)$.
Substituting these units into the formula for $G$,we get the unit of $G = \frac{dyn \cdot cm^2}{g \cdot g} = dyn \cdot cm^2 \cdot g^{-2}$.

(N/A) Heat is a form of energy that flows from a body at a higher temperature to a body at a lower temperature due to the temperature difference between them.
The $SI$ unit of heat is the Joule $(J)$.
The $CGS$ unit of heat is the calorie $(cal)$.
The relation between the $SI$ and $CGS$ units is: $1 \ cal = 4.184 \ J$ (often approximated as $4.2 \ J$).

(N/A) The mass of an atom is very small compared to a kilogram $(kg)$,so the kilogram is not a convenient unit to measure such small quantities.
In nuclear physics,the unit used for atoms is the atomic mass unit $(amu)$,which is also denoted by the unified mass unit $(u)$.
Definition: The twelfth part of the mass of an unexcited carbon $(C^{12})$ atom is called $1$ $amu$,which is $1$ $u$.
The mass of an unexcited carbon atom is $1.992647 \times 10^{-26} \ kg$.
$1$ $u = \frac{\text{Mass of unexcited carbon atom}}{12} = \frac{1.992647 \times 10^{-26}}{12} \ kg = 0.166 \times 10^{-26} \ kg = 1.66 \times 10^{-27} \ kg$.
The atomic masses of various elements expressed in atomic mass unit $(u)$ are close to being integral multiples of the mass of a hydrogen atom. However,there are many striking exceptions to this rule.
Accurate measurement of atomic masses is carried out with mass spectrometers.

(N/A) physical quantity is a property of a material or system that can be quantified by measurement. It is expressed as the product of a numerical value and a unit.
There are two main types of physical quantities:
$(1)$ Fundamental quantities: These are the basic physical quantities that are independent of each other and cannot be defined in terms of other physical quantities. Examples include length,mass,time,electric current,temperature,amount of substance,and luminous intensity.
$(2)$ Derived quantities: These are physical quantities that are expressed in terms of fundamental quantities. They are obtained by multiplying or dividing fundamental quantities. Examples include velocity,acceleration,force,work,and power.

(N/A) unit is defined as the comparison of a physical quantity with a certain basic,internationally accepted reference standard.
- The measurement of any physical quantity is expressed by a numerical value accompanied by a unit.
- Although the number of physical quantities is very large,we only need a limited number of units to express all of them,as they are interrelated.
- The units used for fundamental or base quantities are called fundamental or base units (e.g.,$meter$,$kilogram$,$second$).
- Physical quantities that are expressed as a combination of fundamental quantities are called derived quantities,and their units are called derived units (e.g.,$speed = distance/time$,so the unit is $m/s$).

(N/A) $(1)$ Fundamental quantities are physical quantities that do not depend on any other physical quantities for their measurement. Examples include length,mass,and time.
$(2)$ Derived quantities are physical quantities that are expressed as combinations of fundamental quantities. Examples include velocity,force,and work.

(N/A) system of units is a complete set of units,both fundamental and derived,for all kinds of physical quantities.
There are mainly four types of systems of units:
$(1)$ $CGS$ system: In this system,length is measured in centimeters $(cm)$,mass in grams $(g)$,and time in seconds $(s)$.
$(2)$ $FPS$ system: In this system,length is measured in feet $(ft)$,mass in pounds $(lb)$,and time in seconds $(s)$.
$(3)$ $MKS$ system: In this system,length is measured in meters $(m)$,mass in kilograms $(kg)$,and time in seconds $(s)$.
$(4)$ $SI$ system: This is the International System of Units,which is an extended version of the $MKS$ system. It is the globally accepted standard system of units.

(N/A) The $SI$ (Systeme Internationale d'Unites) is the internationally accepted system of units.
- It is the modern form of the metric system and is abbreviated as $SI$.
- The $SI$ system consists of $7$ fundamental physical quantities: length,mass,time,electric current,thermodynamic temperature,amount of substance,and luminous intensity.
- These fundamental quantities,along with their units and symbols,were established and recommended by the General Conference on Weights and Measures $(CGPM)$ in $1971$ for international use in scientific,technical,and commercial fields.
- The $SI$ system is based on the decimal system,which makes conversions between units simple and convenient.

(N/A) There are two supplementary quantities in the $SI$ system:
$(1)$ Plane angle $d \theta$
$(2)$ Solid angle $d \Omega$
$(1)$ Plane angle $d \theta$: The ratio of the arc length of a circle to its radius is called the plane angle $(d \theta)$.
From the figure,the plane angle $d \theta = \frac{\text{arc}}{\text{radius}} = \frac{ds}{r}$.
The plane angle subtended at the center by an arc of a circle having a length equal to the radius is called $1$ radian. It is represented as $rad$. The maximum value of the plane angle is $2 \pi \ rad$.
If $ds = r$,then $\theta = 1 \ rad$.
$[1^{\circ} = \frac{\pi}{180} \ rad]$ and $[1 \ rad = \frac{180}{\pi} \ \text{degree}]$.
$(2)$ Solid angle $d \Omega$: The angle subtended by an area $(\Delta A)$ on a spherical surface at the center of the sphere is called the solid angle $d \Omega$.
$d \Omega = \frac{dA}{r^2} \ \text{steradian}$.
From the figure,the solid angle $d \Omega = \frac{\text{area}}{(\text{radius})^2} = \frac{\Delta A}{r^2}$.
The maximum value of the solid angle is $4 \pi \ sr$.
The angle subtended by an area of $1 \ m^2$ on a sphere of $1 \ m$ radius at its center is called $1$ steradian. Its symbol is $sr$.
If $\Delta A = 1 \ m^2$ and $r = 1 \ m$,then $\Omega = 1 \ sr$.

Multiples are used to express very large values,while submultiples are used for very small values. Below are the standard $SI$ prefixes:
Multiples:

Value	Prefix / Symbol
$10^{18}$	Exa $(E)$
$10^{15}$	Peta $(P)$
$10^{12}$	Tera $(T)$
$10^{9}$	Giga $(G)$
$10^{6}$	Mega $(M)$
$10^{3}$	Kilo $(k)$
$10^{2}$	Hecto $(h)$
$10^{1}$	Deca $(da)$

Submultiples:

Factor	Prefix / Symbol
$10^{-1}$	deci $(d)$
$10^{-2}$	centi $(c)$
$10^{-3}$	milli $(m)$
$10^{-6}$	micro $(\mu)$
$10^{-9}$	nano $(n)$
$10^{-12}$	pico $(p)$
$10^{-15}$	femto $(f)$
$10^{-18}$	atto $(a)$

(N/A) The international system of units is known as the $SI$ (Système International d'Unités).
The supplementary quantities are:
$(1)$ Plane angle $(d \theta)$,measured in radians $(rad)$.
$(2)$ Solid angle $(d \Omega)$,measured in steradians $(sr)$.

(N/A) $1$ radian: The plane angle subtended at the center of a circle by an arc equal in length to the radius of the circle is defined as $1$ radian.
$1$ steradian: The solid angle subtended at the center of a sphere by a surface area equal to the square of the radius of the sphere is defined as $1$ steradian.

(A) The plane angle is defined as the ratio of the arc length to the radius of a circle. For a full circle,the arc length is $2 \pi r$,so the maximum plane angle is $\frac{2 \pi r}{r} = 2 \pi \; rad$.
The solid angle is defined as the ratio of the area of a spherical cap to the square of the radius of the sphere. For a full sphere,the surface area is $4 \pi r^2$,so the maximum solid angle is $\frac{4 \pi r^2}{r^2} = 4 \pi \; sr$.

The sizes of objects in the universe vary over a very wide range. The dimension of an atomic nucleus is approximately $10^{-14} \, m$, while the length of the observable universe is about $10^{26} \, m$.
We use special units for short and large lengths:
$1 \, fm$ (fermi) $= 10^{-15} \, m$
Size of nucleus $\approx 10^{-14} \, m$
$1 \, \mathring{A}$ ($\mathring{A}$) $= 10^{-10} \, m$
$1 \, nm$ (nanometer) $= 10^{-9} \, m$
$1 \, \mu m$ (micrometer) $= 10^{-6} \, m$
$1 \, mm$ (millimeter) $= 10^{-3} \, m$
$1 \, km$ (kilometer) $= 10^{3} \, m$
$1 \, Mm$ (megameter) $= 10^{6} \, m$
$1 \, AU$ (astronomical unit) $= 1.496 \times 10^{11} \, m$
$1 \, ly$ (light year) $= 9.46 \times 10^{15} \, m$
$1 \, pc$ (parsec) $= 3.08 \times 10^{16} \, m$
Size of galaxy $\approx 10^{21} \, m$

(N/A) An Astronomical Unit $(AU)$ is defined as the average distance between the Sun and the Earth.
It represents the physical quantity of length or distance.
The value of $1\,AU$ is approximately $1.496 \times 10^{11} \, m$.

(N/A) Light year is defined as the distance travelled by light in a vacuum in $1$ year.
Speed of light in vacuum,$c = 2.99 \times 10^{8} \text{ m/s}$.
Distance travelled in $1$ year $= c \times t$
$= 2.99 \times 10^{8} \text{ m/s} \times (365 \times 24 \times 3600 \text{ s})$
$= 9.46 \times 10^{15} \text{ m}$.
No,a light year is not a unit of time; it is a unit of distance used to measure the vast distances between celestial objects.

(A) parsec (parallax second) is a unit of distance used in astronomy,defined as the distance at which an arc of $1 \ \text{AU}$ (astronomical unit) subtends an angle of $1 \ \text{arcsecond}$ $(1'')$. It is approximately equal to $3.08 \times 10^{16} \ \text{m}$.
To convert $1 \ \text{fm}$ (femtometer) to $\mathring{A}$ $(\mathring{A}):$
$1 \ \text{fm} = 10^{-15} \ \text{m}$
$1 \ \mathring{A} = 10^{-10} \ \text{m}$
Therefore,$1 \ \text{fm} = \frac{10^{-15} \ \text{m}}{10^{-10} \ \text{m}} \ \mathring{A} = 10^{-5} \ \mathring{A}$.

(A) cesium clock is based on the vibrations of cesium-$133$ atoms.
Specifically,the second is defined as the duration of $9,192,631,770$ periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the cesium-$133$ atom.
Therefore,the time interval in a cesium clock depends on the atomic vibrations of cesium-$133$.

(A) We know that $1$ parsec is defined as the distance at which an arc of length $1 AU$ subtends an angle of $1$ arcsecond $(1'')$ at the center.
Using the formula for angle: $\theta = \frac{\text{arc}}{\text{radius}}$
Here,$\theta = 1'' = \frac{1}{3600} \text{ degrees} = \frac{1}{3600} \times \frac{\pi}{180} \text{ radians} \approx 4.848 \times 10^{-6} \text{ radians}$.
Arc length $= 1 AU$.
Radius $= 1 \text{ parsec}$.
Therefore,$1 \text{ parsec} = \frac{1 AU}{\theta} = \frac{1 AU}{4.848 \times 10^{-6}} \approx 2.06 \times 10^5 AU$.

(A) $nm$ stands for nanometer, which is a unit of length. $1 \, nm = 10^{-9} \, m$.
$mN$ stands for millinewton, which is a unit of force. $1 \, mN = 10^{-3} \, N$.
$Nm$ stands for newton-meter, which is a unit of torque or work/energy. $1 \, Nm = 1 \, N \cdot m$.

(C) The largest unit among the given options is the Parsec.
$1$ Parsec $= 3.08 \times 10^{16} \ m$
$1$ Light Year $= 9.46 \times 10^{15} \ m$
$1$ Astronomical Unit $= 1.50 \times 10^{11} \ m$
Comparing these values,we find that $3.08 \times 10^{16} \ m > 9.46 \times 10^{15} \ m > 1.50 \times 10^{11} \ m$. Therefore,the Parsec is the largest unit.

(A) Step $1$: Convert $1$ parsec to light-years.
$1 \text{ parsec} = 3.08 \times 10^{16} \text{ m}$
$1 \text{ light-year} = 9.46 \times 10^{15} \text{ m}$
$\frac{1 \text{ parsec}}{1 \text{ light-year}} = \frac{3.08 \times 10^{16}}{9.46 \times 10^{15}} \approx 3.26$
Therefore,$1 \text{ parsec} = 3.26 \text{ light-years}$.
Step $2$: Convert $1$ light-year to astronomical units $(AU)$.
$1 \text{ AU} = 1.496 \times 10^{11} \text{ m} \approx 1.5 \times 10^{11} \text{ m}$
$\frac{1 \text{ light-year}}{1 \text{ AU}} = \frac{9.46 \times 10^{15}}{1.5 \times 10^{11}} \approx 6.3 \times 10^{4}$
Therefore,$1 \text{ light-year} = 6.3 \times 10^{4} \text{ AU}$.

(B) No, they are not the same units of length.
$1 \text{ AU (Astronomical Unit)} = 1.496 \times 10^{11} \text{ m}$.
$1 \mathring{A} = 10^{-10} \text{ m}$.
Since their values differ by many orders of magnitude, they represent different scales of length measurement.

(C) To convert $1\,g\,cm^{-3}$ to $kg\,m^{-3}$,we use the conversion factors $1\,g = 10^{-3}\,kg$ and $1\,cm = 10^{-2}\,m$.
$\frac{1\,g}{1\,cm^3} = \frac{1 \times 10^{-3}\,kg}{(10^{-2}\,m)^3}$
$= \frac{10^{-3}\,kg}{10^{-6}\,m^3}$
$= 10^{-3 - (-6)}\,kg\,m^{-3}$
$= 10^3\,kg\,m^{-3}$

(N/A) $1$. The $SI$ unit of force is the newton $(N)$. One newton is defined as the force that,when applied to a body of mass $1 \ kg$,produces an acceleration of $1 \ m/s^2$ in the direction of the force. Mathematically,$1 \ N = 1 \ kg \cdot m/s^2$.
$2$. The $CGS$ unit of force is the dyne. One dyne is defined as the force that,when applied to a body of mass $1 \ g$,produces an acceleration of $1 \ cm/s^2$ in the direction of the force. Mathematically,$1 \ dyne = 1 \ g \cdot cm/s^2$.

(A) In the British unit system (also known as the Imperial system or $FPS$ system),the unit of power is the horsepower $(hp)$.
By definition,$1 \ hp = 550 \ ft-lb/s$,which is equivalent to approximately $746 \ W$ in the $SI$ system.

(N/A) Density is defined as the mass per unit volume of a substance. It is a measure of how much matter is packed into a given space.
Mathematically,$\text{Density} (\rho) = \frac{\text{Mass} (M)}{\text{Volume} (V)}$.
The $MKS$ $(SI)$ unit of density is $\text{kg/m}^3$.
The $CGS$ unit of density is $\text{g/cm}^3$.

(N/A) The unit $torr$ is named after Evangelista Torricelli. It is defined as $1 \ torr = 1/760$ of a standard atmosphere. Thus,$1 \ atm = 760 \ torr$. Since $1 \ atm = 1.01325 \times 10^5 \ Pa$,we have $1 \ torr \approx 133.322 \ Pa$.
The unit $bar$ is a metric unit of pressure,but not part of the $SI$ system. It is defined as $1 \ bar = 10^5 \ Pa$. This unit is commonly used in meteorology and industrial applications because it is very close to the standard atmospheric pressure $(1 \ atm = 1.01325 \ bar)$.

(0.239) The relationship between Joule $(J)$ and calorie $(\text{cal})$ is defined by the mechanical equivalent of heat.
$1 \text{ calorie} = 4.184 \text{ J}$.
Therefore, $1 \text{ J} = \frac{1}{4.184} \text{ calorie}$.
$1 \text{ J} \approx 0.239 \text{ calorie}$.

(N/A) The Boltzmann constant $(k_B)$ relates the average kinetic energy of particles in a gas to the thermodynamic temperature of the gas.
It is defined by the equation: $PV = N k_B T$,where $P$ is pressure,$V$ is volume,$N$ is the number of particles,and $T$ is temperature.
Rearranging for $k_B$: $k_B = \frac{PV}{NT}$.
The unit of pressure $(P)$ is $Pascal$ ($Pa$ or $J/m^3$),volume $(V)$ is $m^3$,and temperature $(T)$ is $Kelvin$ $(K)$.
Thus,the $SI$ unit of $k_B$ is $\frac{J}{K}$ (Joule per Kelvin).
To find the dimension,we use the formula for energy: $Energy = Force \times Displacement = [M L T^{-2}] \times [L] = [M L^2 T^{-2}]$.
Since $k_B = \frac{Energy}{Temperature}$,the dimension is $\frac{[M L^2 T^{-2}]}{[K]} = [M L^2 T^{-2} K^{-1}]$.

(A) The unit $fm$ stands for femtometer,which is also known as a fermi.
By definition,$1 \, fm = 10^{-15} \, m$.
This unit is commonly used in nuclear physics to express the size of atomic nuclei.

(C) In the $SI$ system,besides the seven base physical quantities,there are two additional quantities defined to describe geometric measurements,which are known as supplementary physical quantities.
These two quantities are plane angle (measured in $radians$) and solid angle (measured in $steradians$).

(B) The statement is incorrect. According to the $SI$ system of units,the fundamental unit of length is the meter $(m)$ and the fundamental unit of mass is the kilogram $(kg)$. Therefore,the given statement is false.

(B) light year is defined as the distance that light travels in a vacuum in one Julian year ($365.25$ days).
Since it measures the length of the path covered by light,it is a unit of distance.
Therefore,the correct answer is distance.

(A) The $SI$ system defines two supplementary units:
$(1)$ Radian $(rad)$: Used for measuring plane angles.
$(2)$ Steradian $(sr)$: Used for measuring solid angles.

(D) In the $SI$ system,prefixes are used to indicate powers of $10$. The prefix $M$ (Mega) stands for $10^6$.
If we write $M \, km$,it implies $10^6 \, km$.
However,the standard convention in physics is to avoid using two prefixes for a single unit. Since $k$ (kilo) is already a prefix for $10^3$,writing $M \, km$ would mean $10^6 \times 10^3 \, m = 10^9 \, m$.
Therefore,$10^6 \, km$ is equivalent to $10^9 \, m$,which is written as $1 \, Gm$ (Giga-meter),not $M \, km$.

(N/A) The value of any given physical quantity may vary over a wide range,so different units for the same physical quantity are required to express these values conveniently.
For example: The speed of a bicycle can be measured in $m/s$,the speed of a car can be measured in $km/h$,and the speed of a satellite can be measured in $km/s$.

(N/A) The reasons for choosing length,mass,and time as base quantities in mechanics are:
$(i)$ Length,mass,and time are independent physical quantities and cannot be derived from any other physical quantities.
$(ii)$ All other physical quantities in mechanics,such as velocity,acceleration,force,work,and energy,can be expressed in terms of these three base quantities.

(N/A) Since $\pi \, rad = 180^{\circ}$,therefore $1 \, rad = \frac{180}{\pi} \approx 57.3^{\circ}$.
$(b)$ $A$ light year is the distance that light travels in a vacuum in one year,so it is a unit of distance.
$(c)$ The solid angle subtended by a sphere at its center is $4\pi \, sr$. Since a hemisphere is half of a sphere,the solid angle subtended by a hemisphere at its center is $\frac{4\pi}{2} = 2\pi \, sr$.

(B) $(1)$ The distance between Earth and stars is extremely large, so it is measured in light-years $(ly)$. Thus, $(1-b)$.
$(2)$ The wavelength of infrared radiation is very small, typically in the range of micrometers to nanometers, which is conveniently expressed in $\mathring{A}$s. Thus, $(2-c)$.

(B) The values for the astronomical units are as follows:
$(1)$ Light year is the distance light travels in one year,which is approximately $9.46 \times 10^{15} \ m$. Thus,$(1-b)$.
$(2)$ Parsec is a unit of distance used in astronomy,equal to approximately $3.08 \times 10^{16} \ m$. Thus,$(2-a)$.
$(3)$ $1 \ AU$ (Astronomical Unit) is the average distance between the Earth and the Sun,which is approximately $1.496 \times 10^{11} \ m$. Thus,$(3-c)$.
Therefore,the correct matching is $(1-b, 2-a, 3-c)$.

(D) $1$. A light year is the distance light travels in one year, which is approximately $9.46 \times 10^{15} \, m$.
$2$. A parsec is a unit of distance used in astronomy, equal to about $3.26$ light years.
$3$. An $\mathring{A}$ is a unit of length equal to $10^{-10} \, m$.
$4$. A millisecond is a unit of time, equal to $10^{-3} \, s$.
Therefore, a millisecond is not a unit of distance.