Units Questions in English

(C) 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 option is $C$.

(C) The unit of force is $Newton$ $(N)$ and the unit of time is $second$ $(s)$.
Impulse is defined as the product of force and time,given by $J = F \times \Delta t$.
The $SI$ unit of impulse is $N \cdot s$.
According to the impulse-momentum theorem,impulse is equal to the change in linear momentum $(\Delta p = F \times \Delta t)$.
Therefore,the unit of momentum is also $Newton-second$ $(N \cdot s)$.

(D) According to the international standard definition,one second is defined as the duration of $9192631770$ periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the cesium-$133$ atom.

(C) By definition,$1 \, nm = 10^{-9} \, m$.
Since $1 \, m = 100 \, cm = 10^2 \, cm$,we can substitute this into the equation:
$1 \, nm = 10^{-9} \times 10^2 \, cm = 10^{-7} \, cm$.
Therefore,the correct option is $C$.

(D) By definition,$1 \, \text{micron}$ (also known as a micrometre) is equal to $10^{-6} \, \text{metres}$.
Since $1 \, \text{metre} = 100 \, \text{cm} = 10^2 \, \text{cm}$,we can substitute this into the expression:
$1 \, \text{micron} = 10^{-6} \times (10^2 \, \text{cm})$
$1 \, \text{micron} = 10^{-6+2} \, \text{cm}$
$1 \, \text{micron} = 10^{-4} \, \text{cm}$.
Therefore,the correct option is $D$.

(A) The $SI$ unit of pressure is the pascal (represented as $Pa$),which is defined as one newton per square metre ($N/m^2$ or $kg \cdot m^{-1} \cdot s^{-2}$).
This unit was officially adopted in $1971$.
Before this,pressure in the $SI$ system was expressed as $newtons$ $per$ $square$ $metre$.

(C) Angular acceleration $(\alpha)$ is defined as the rate of change of angular velocity $(\omega)$ with respect to time $(t)$.
Mathematically,$\alpha = \frac{d\omega}{dt}$.
The $SI$ unit of angular velocity $(\omega)$ is radians per second $(rad\,s^{-1})$.
Therefore,the $SI$ unit of angular acceleration is $\frac{rad\,s^{-1}}{s} = rad\,s^{-2}$.
Thus,the correct option is $C$.

(B) Energy is defined as the capacity to do work. The $SI$ unit of energy is the Joule $(J)$.
$1 \text{ Joule} = 1 \text{ Newton} \cdot \text{ meter} (N \cdot m) = 1 \text{ Watt} \cdot \text{ second} (W \cdot s)$.
Option $(b)$ $kg \cdot m/s$ represents the unit of linear momentum $(p = mv)$,not energy.
Therefore,the correct option is $(b)$.

(C) The $SI$ unit of energy is the $Joule$ $(J)$.
$1 \ J = 1 \ kg \cdot m^2/s^2$.
$Erg$ is the $CGS$ unit of energy.
$Calorie$ and $Electron \ volt$ are other units of energy,but they are not the $SI$ unit.
Therefore,the correct option is $C$.

(B) The given wavelength is $\lambda = 0.00006\,m = 6 \times 10^{-5}\,m$.
We know that $1\,micron = 1\,\mu m = 10^{-6}\,m$.
To convert the wavelength into microns,we divide by $10^{-6}\,m$:
$\lambda = \frac{6 \times 10^{-5}\,m}{10^{-6}\,m/micron} = 6 \times 10^{1}\,microns = 60\,microns$.
Therefore,the correct option is $B$.

(D) In the $SI$ system of units,there are seven fundamental physical quantities: length,mass,time,electric current,thermodynamic temperature,amount of substance,and luminous intensity.
Since temperature is one of the seven fundamental quantities,it cannot be expressed as a derived quantity in terms of length,mass,or time.
Therefore,the correct option is $D$.

(A) Power is defined as the rate of doing work or the rate of energy transfer. The $SI$ unit of power is the Watt $(W)$,which is defined as $1 \ J/s$. Among the given options,Kilowatt $(kW)$ is a unit of power,where $1 \ kW = 10^3 \ W$. Kilowatt-hour $(kWh)$ is a unit of energy,Dyne is a unit of force,and Joule $(J)$ is a unit of energy. Therefore,the correct option is $A$.

(A) To convert density from $CGS$ $(g/cm^3)$ to $SI$ $(kg/m^3)$:
$1 \, g/cm^3 = \frac{10^{-3} \, kg}{10^{-6} \, m^3} = 10^3 \, kg/m^3 = 1000 \, kg/m^3$.
Given density = $0.5 \, g/cm^3$.
Therefore,$0.5 \, g/cm^3 = 0.5 \times 1000 \, kg/m^3 = 500 \, kg/m^3$.

(B) Energy is defined as the product of power and time.
Since $\text{Power} = \text{Energy} / \text{Time}$,it follows that $\text{Energy} = \text{Power} \times \text{Time}$.
Given that the unit of power is $\text{Watt}$ and the unit of time is $\text{day}$,the product $\text{Watt} \cdot \text{day}$ represents a unit of energy.
Option $A$ $(J/\text{sec})$ is a unit of power.
Option $C$ $(\text{Kilowatt})$ is a unit of power.
Option $D$ $(\text{gm} \cdot \text{cm}/\text{sec}^2)$ is a unit of force (dyne).
Therefore,the correct option is $B$.

(D) The unit of force is $Newton$ $(N)$ and the unit of area is $meter^2$ $(m^2)$.
Pressure is defined as the force applied per unit area.
$\text{Pressure} = \frac{\text{Force}}{\text{Area}}$
Therefore,the unit of pressure is $\frac{N}{m^2}$ or $Newton/meter^2$ (also known as $Pascal$).

(B) Momentum $(p)$ is defined as the product of mass $(m)$ and velocity $(v)$.
Mathematically,$p = m \times v$.
The $SI$ unit of mass is $kg$ (kilogram).
The $SI$ unit of velocity is $m/s$ (meters per second).
Therefore,the $SI$ unit of momentum is $kg \cdot m/s$.

(B) Potential energy is given by the formula $U = mgh$.
Here,$m$ is mass in $gm$,$g$ is acceleration due to gravity in $cm/sec^2$,and $h$ is height in $cm$.
Substituting these units: $U = (gm) \cdot (cm/sec^2) \cdot (cm) = gm \cdot (cm^2/sec^2)$.
This can be rewritten as $gm \cdot (cm/sec)^2$.
Therefore,the correct option is $B$.

(A) $Parsec$ (parallax second) is a unit of length used to measure the large distances to astronomical objects outside the Solar System.
It is defined as the distance at which one astronomical unit $(AU)$ subtends an angle of one arcsecond.
$1 \ Parsec \approx 3.086 \times 10^{16} \ m$.
Therefore,the correct option is $A$.

(A) physical quantity $(Q)$ is expressed as the product of its numerical value $(n)$ and its unit $(u)$.
$Q = n \times u$
Since the physical quantity remains constant regardless of the system of measurement used,we have:
$Q = n_1u_1 = n_2u_2$
Therefore,the relationship between the two systems is ${n_1}{u_1} = {n_2}{u_2}$.

(C) The standard definition of the second in the International System of Units $(SI)$ is based on the atomic clock.
Specifically,one 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,universal time is based on the vibrations of the cesium atom.

(A) The number of cycles per unit time is known as frequency. The $SI$ unit of frequency is $Hertz (Hz)$.

(C) The prefix 'pico' represents a factor of $10^{-12}$.
Therefore,one pico Farad $(1 \ pF)$ is equal to $10^{-12} \ F$.

(D) The $Microsecond$ is a unit of time $(10^{-6} \ s)$.
The $Leap \ year$ is a unit of time used in calendars.
The $Lunar \ month$ is a unit of time based on the moon's cycle.
The $Parallactic \ second$ (or $parsec$) is a unit of distance used in astronomy to measure large distances to stars and galaxies.
Therefore,the $Parallactic \ second$ is not a unit of time.
Correct option is $D$.

(B) By definition,$1 \, \text{yard}$ is exactly $36 \, \text{inches}$.
Since $1 \, \text{inch} = 2.54 \, \text{cm} = 0.0254 \, \text{m}$,
$1 \, \text{yard} = 36 \times 0.0254 \, \text{m} = 0.9144 \, \text{m}$.
Therefore,the correct option is $B$.

(C) To determine the smallest unit, we compare their values in terms of metres:
$1 \text{ Millimetre } (mm) = 10^{-3} \, m$
$1 \text{ } \mathring{A} (\mathring{A}) = 10^{-10} \, m$
$1 \text{ Fermi } (fm) = 10^{-15} \, m$
$1 \text{ Metre } (m) = 1 \, m$
Comparing these values, $10^{-15} < 10^{-10} < 10^{-3} < 1$. Therefore, the Fermi is the smallest unit.
The correct option is $C$.

(B) The correct matches are:
$(A)$ Distance between earth and stars is measured in light years $(3)$.
$(B)$ Inter-atomic distance in a solid is typically in the order of $\mathring{A}$s $(2)$.
$(C)$ Size of the nucleus is measured in Fermi $(4)$.
$(D)$ Wavelength of an infrared laser is typically in the range of microns $(1)$.
Therefore, the correct sequence is $A-3, B-2, C-4, D-1$.

(A) $Torr$ is a non-$SI$ unit of pressure,named after the physicist $Evangelista$ $Torricelli$.
One $Torr$ is defined as exactly $1/760$ of an atmosphere $(1 \text{ atm} = 760 \text{ Torr})$.
Therefore,it is a unit used to measure pressure.

(D) Fundamental units are the base units of measurement that do not depend on any other physical quantities. These include mass $(kg)$,length $(m)$,and time $(s)$.
Derived units are physical units derived from the seven base units specified by the International System of Units $(SI)$.
Volume is defined as $length \times width \times height$. Since it is expressed as $m^3$,it is a derived unit.

(D) Energy is defined as the capacity to do work. The $SI$ unit of energy is the $Joule$ $(J)$. Other common units include $Calorie$ and $Electron \text{ } volt$ $(eV)$.
$Watt$ $(W)$ is the $SI$ unit of power,which is defined as the rate of doing work $(Power = \text{Energy} / \text{Time})$. Therefore,$Watt$ is not a unit of energy.

(B) The units $Fermi$ $(10^{-15} \ m)$,$Micron$ $(10^{-6} \ m)$,and $Light \ year$ $(9.46 \times 10^{15} \ m)$ are all units used to measure length or distance.
$Debye$ is a unit used to express the electric dipole moment of molecules,not length.
Therefore,length cannot be measured by $Debye$.

(D) The magnitude of a physical quantity $P$ remains constant regardless of the system of units used.
If a physical quantity is expressed as $P = n_1 u_1 = n_2 u_2$, it implies that the numerical value $n$ is inversely proportional to the size of the unit $u$.
Therefore, $n \propto \frac{1}{u}$.
Thus, the correct relation is $n \propto \frac{1}{u}$.

(B) The $SI$ unit of luminous intensity is candela $(cd)$.

(B) The International System of Units $(SI)$ defines $7$ base units for physical quantities.
These base units are: meter $(m)$ for length,kilogram $(kg)$ for mass,second $(s)$ for time,ampere $(A)$ for electric current,kelvin $(K)$ for thermodynamic temperature,mole $(mol)$ for amount of substance,and candela $(cd)$ for luminous intensity.

(A) The $MKS$,$FPS$,and $CGS$ systems are based on the three fundamental physical quantities: mass,length,and time.
In the $SI$ system,there are seven fundamental physical quantities: length,mass,time,electric current,thermodynamic temperature,amount of substance,and luminous intensity.
Therefore,the $SI$ system is not based on units of mass,length,and time alone.

(B) The prefix 'femto' represents a factor of $10^{-15}$. Therefore,one femtometer $(1 \, fm)$ is defined as $10^{-15} \, m$.

(B) The definition of the metre was historically based on the wavelength of light emitted by the krypton-$86$ $(Kr^{86})$ atom.
One metre is defined as exactly $1650763.73$ wavelengths of the orange-red light emitted by the $Kr^{86}$ atom in a vacuum.
Therefore,the correct option is $B$.

(D) The unit of energy is $Joule$ $(J)$.
$Watt$ is the $SI$ unit of power.
$Horsepower$ $(hp)$ is a practical unit of power.
Therefore,none of the given options ($Watt$,$Horsepower$) are units of energy.

(A) We know that $1 \, kcal = 1000 \, cal = 4200 \, J$.
The conversion factor from Joules to kilowatt-hour $(kWh)$ is $1 \, kWh = 3.6 \times 10^6 \, J$.
Therefore,$1 \, kcal = \frac{4200}{3.6 \times 10^6} \, kWh = \frac{4.2 \times 10^3}{3.6 \times 10^6} \, kWh = \frac{4.2}{3.6} \times 10^{-3} \, kWh \approx 1.166 \times 10^{-3} \, kWh$.
For $700 \, kcal$,the energy in $kWh$ is:
$E = 700 \times \frac{4200}{3.6 \times 10^6} \, kWh$
$E = \frac{2,940,000}{3,600,000} \, kWh$
$E = \frac{294}{360} \, kWh = 0.8166... \, kWh$.
Rounding to two decimal places,we get $0.81 \, kWh$.

(A) Given,latent heat $L = 536 \, cal/g$.
We know that $1 \, cal = 4.2 \, J$ and $1 \, g = 10^{-3} \, kg$.
Substituting these values into the expression:
$L = 536 \times \frac{4.2 \, J}{10^{-3} \, kg}$
$L = 536 \times 4.2 \times 10^3 \, J/kg$
$L = 2251.2 \times 10^3 \, J/kg \approx 2.25 \times 10^6 \, J/kg$.
Therefore,the correct option is $A$.

(B) The unit of illuminance (or illumination) in the International System of Units $(SI)$ is the lux $(lx)$.
Illuminance is defined as the luminous flux per unit area incident on a surface.
Mathematically,$E = \frac{\Phi}{A}$,where $E$ is illuminance in lux,$\Phi$ is luminous flux in lumens $(lm)$,and $A$ is the area in square meters $(m^2)$.
Therefore,$1 \ lx = 1 \ lm/m^2$.
Thus,the correct option is $B$.

(D) In the $SI$ system of units,temperature is considered a fundamental physical quantity. Its base unit is the $kelvin$ $(K)$. It cannot be derived from or expressed in terms of mass $(M)$,length $(L)$,or time $(T)$. Therefore,none of the given options are correct.

(B) We know that $1 \, amu = 1.66 \times 10^{-27} \, kg$.
Since $1 \, kg = 10^3 \, g$,we can convert the mass to grams:
$1 \, amu = 1.66 \times 10^{-27} \times 10^3 \, g = 1.66 \times 10^{-24} \, g$.
To find the value of $1 \, g$ in $amu$,we take the reciprocal:
$1 \, g = \frac{1}{1.66 \times 10^{-24}} \, amu$.
Calculating the value:
$1 \, g \approx 0.6024 \times 10^{24} \, amu = 6.02 \times 10^{23} \, amu$.

(D) Let us analyze each option:
$A$: $1 \text{ calorie} = 4.18 \text{ J}$ is correct.
$B$: $1 \text{ Å} = 10^{-10} \text{ m}$ is correct.
$C$: $1 \text{ MeV} = 10^6 \text{ eV} = 10^6 \times (1.6 \times 10^{-19} \text{ J}) = 1.6 \times 10^{-13} \text{ J}$. This is correct.
$D$: $1 \text{ Newton}$ is the $SI$ unit of force and $1 \text{ dyne}$ is the $CGS$ unit of force. Since $1 \text{ N} = 10^5 \text{ dyne}$,the statement $1 \text{ N} = 10^{-5} \text{ dyne}$ is incorrect.

(C) The atomic mass unit $(amu)$ is defined as one-twelfth of the mass of a carbon-$12$ atom.
$1 \; amu = 1.660539 \times 10^{-27} \; kg \approx 1.66 \times 10^{-27} \; kg$.
To find the value of $10 \; amu$ in kilograms,we multiply the value of $1 \; amu$ by $10$.
$10 \; amu = 10 \times (1.66 \times 10^{-27} \; kg)$.
$10 \; amu = 1.66 \times 10^{-26} \; kg$.
Therefore,the correct option is $C$.

(A) To convert $g/cm^3$ to $kg/m^3$,we use the conversion factors: $1 \, g = 10^{-3} \, kg$ and $1 \, cm = 10^{-2} \, m$.
Substituting these into the expression:
$10^3 \, g/cm^3 = 10^3 \times \frac{10^{-3} \, kg}{(10^{-2} \, m)^3}$
$= 10^3 \times \frac{10^{-3} \, kg}{10^{-6} \, m^3}$
$= \frac{10^0}{10^{-6}} \, kg/m^3$
$= 10^6 \, kg/m^3$.

(A) The density of wood is given as $0.5 \; g/cm^3$.
To convert this to $MKS$ units $(kg/m^3)$,we use the conversion factors:
$1 \; g = 10^{-3} \; kg$
$1 \; cm^3 = (10^{-2} \; m)^3 = 10^{-6} \; m^3$
Therefore,$0.5 \; g/cm^3 = 0.5 \times \frac{10^{-3} \; kg}{10^{-6} \; m^3} = 0.5 \times 10^3 \; kg/m^3 = 500 \; kg/m^3$.

(A) We know that $1 \mu m = 10^{-6} \text{ m}$ and $1 \text{ nm} = 10^{-9} \text{ m}$.
Thus,the ratio of $1 \text{ micron}$ to $1 \text{ nanometer}$ is given by:
$\frac{1 \mu m}{1 \text{ nm}} = \frac{10^{-6} \text{ m}}{10^{-9} \text{ m}} = 10^{-6 - (-9)} = 10^3$.

(C) The unit $second^{-1}$ (or $s^{-1}$) represents the number of cycles or events occurring per unit time.
Frequency is defined as the number of oscillations or cycles per unit time.
The $SI$ unit of frequency is the hertz $(Hz)$,which is equivalent to $1/second$ or $s^{-1}$.
Therefore,$second^{-1}$ is the unit of frequency.

(D) The micron $(10^{-6} \ m)$,light year $(9.46 \times 10^{15} \ m)$,and $\mathring{A}$ $(10^{-10} \ m)$ are all units used to measure length or distance.
Conversely,the radian is the $SI$ unit of plane angle,which is a dimensionless quantity representing the ratio of arc length to radius.

(C) We know that $1 \, m \approx 3.28084 \, ft$.
To convert the acceleration due to gravity from $m/s^2$ to $ft/s^2$,we multiply the value by the conversion factor:
$g = 9.8 \, m/s^2 \times 3.28084 \, ft/m$
$g \approx 32.14 \, ft/s^2$.
Rounding to the nearest whole number,we get $g \approx 32 \, ft/s^2$.