Units Questions in English

(A) The torsional constant $(k)$ is defined by the restoring torque $(\tau)$ per unit angular twist $(\theta)$, given by the equation $\tau = k\theta$.
Rearranging for $k$, we get $k = \tau / \theta$.
The unit of torque $(\tau)$ is Newton-meter $(N \cdot m)$ and the unit of angular twist $(\theta)$ is radian $(rad)$.
Therefore, the unit of the torsional constant is $N \cdot m/rad$.

(A) The statement is true.
The potential energy $U$ of a spring is given by $U = \frac{1}{2} k x^2$,where $k$ is the spring constant and $x$ is the displacement.
Rearranging for $k$,we get $k = \frac{2U}{x^2}$.
The unit of energy $U$ is Joule $(J)$ and the unit of displacement $x$ is meter $(m)$.
Therefore,the unit of $k$ is $\frac{J}{m^2} = J\,m^{-2}$.

(A) We know that $1 \text{ MW} = 10^6 \text{ W}$.
Since $1 \text{ W} = 1 \text{ J/s}$ and $1 \text{ J} = 10^7 \text{ erg}$,
$1 \text{ W} = 10^7 \text{ erg/s}$.
Therefore,$1 \text{ MW} = 10^6 \times 10^7 \text{ erg/s} = 10^{13} \text{ erg/s}$.

The pressure $1 \text{ torr}$ is defined as the pressure exerted by a column of mercury $1 \text{ mm}$ high,which is approximately $133.322 \text{ Pa}$ $(N/m^2)$.
Since $1 \text{ bar} = 10^5 \text{ Pa}$,we have $1 \text{ Pa} = 10^{-5} \text{ bar}$.
Therefore,$1 \text{ torr} = 133.322 \times 10^{-5} \text{ bar} = 1.33322 \times 10^{-3} \text{ bar}$.
Since $1 \text{ bar} = 1000 \text{ millibar}$,we have $1 \text{ torr} = 1.33322 \times 10^{-3} \times 10^3 \text{ millibar} = 1.33322 \text{ millibar}$.

(N/A) The standard atmospheric pressure is defined as the pressure exerted by a mercury column of $760 \,mm$ height at $0 \,^{\circ}C$ under standard gravity.
$1 \,atm = 1.01325 \times 10^{5} \,N/m^{2}$ (or $Pa$).
Since $1 \,bar = 10^{5} \,Pa$, we can write:
$1 \,atm = 1.01325 \,bar$.

(D) To convert energy from Joules $(J)$ to electron-volts $(eV)$,we divide the energy value by the charge of an electron,which is $1.6 \times 10^{-19} \ C$.
Given energy $E = 10^{-20} \ J$.
$E \text{ (in } eV) = \frac{10^{-20} \ J}{1.6 \times 10^{-19} \ J/eV}$
$E = \frac{1}{1.6} \times 10^{-1} \ eV$
$E = 0.625 \times 0.1 \ eV$
$E = 0.0625 \ eV$.

(D) light-year is the distance that light travels in a vacuum in one Julian year ($365.25$ days).
Speed of light $(c)$ $\approx 3 \times 10^8 \text{ m/s}$.
Time in one year $(t)$ $= 365.25 \times 24 \times 60 \times 60 \text{ seconds} \approx 3.15576 \times 10^7 \text{ s}$.
Distance $(d)$ $= c \times t = (3 \times 10^8 \text{ m/s}) \times (3.15576 \times 10^7 \text{ s}) \approx 9.461 \times 10^{15} \text{ m}$.
Converting to kilometers: $9.461 \times 10^{15} \text{ m} = 9.461 \times 10^{12} \text{ km}$.
Therefore,the correct option is $D$.

(D) Plane angle is defined as the ratio of arc length to radius $(d\theta = ds/r)$. Since both arc length and radius have the dimension of length $(L)$,the dimension of plane angle is $[L^1/L^1] = [M^0 L^0 T^0]$,which is dimensionless. However,it has a unit,which is radian $(rad)$.
Similarly,solid angle is defined as the ratio of the area of the spherical surface to the square of the radius $(d\Omega = dA/r^2)$. Since both area and the square of the radius have the dimension of length squared $(L^2)$,the dimension of solid angle is $[L^2/L^2] = [M^0 L^0 T^0]$,which is also dimensionless. However,it has a unit,which is steradian $(sr)$.
Therefore,plane angle and solid angle have units but no dimensions.

(C) According to Einstein's mass-energy equivalence, the energy difference $\Delta E$ between the excited state and the ground state corresponds to a mass difference $\Delta m$ given by $\Delta E = \Delta m c^2$.
Given the excitation frequency $\nu = 9.2 \times 10^9 \, Hz$, the energy difference is $\Delta E = h\nu$.
Therefore, the mass difference for one pair of atoms is $\Delta m = \frac{h\nu}{c^2}$.
Substituting the given values:
$\Delta m = \frac{6.63 \times 10^{-34} \times 9.2 \times 10^9}{(3 \times 10^8)^2} \, kg$
$\Delta m = \frac{6.63 \times 9.2 \times 10^{-25}}{9 \times 10^{16}} \, kg$
$\Delta m \approx 6.78 \times 10^{-41} \, kg$.
Since this mass difference is obtained using $2$ atoms, the number of atoms $N$ required to obtain a total mass of $1 \, kg$ is:
$N = \frac{2 \times 1 \, kg}{\Delta m} = \frac{2}{6.78 \times 10^{-41}} \approx 0.295 \times 10^{41} \approx 3 \times 10^{40}$.
Thus, the order of magnitude is $10^{40}$.

(B) The correct option is $B$.
One astronomical unit $(AU)$ is defined as the average distance between the Earth and the Sun.
By definition,$1 \text{ AU} = 1.496 \times 10^{11} \, m$.

(B) The correct option is $B$.
$1$. $Nm$ (Newton-meter) is the unit of torque or work.
$2$. $mN$ (Milli-newton) is a unit of force,where $1 \, mN = 10^{-3} \, N$.
$3$. $nm$ (Nanometer) is a unit of length.
$4$. $Ns$ (Newton-second) is the unit of impulse or change in momentum.
Therefore,$mN$ is the only unit of force listed.

(B) The plane angle $\theta$ subtended by an arc of length $s$ at the centre of a circle of radius $r$ is given by $\theta = \frac{s}{r}$.
For a complete circle,the arc length $s$ is equal to the circumference $2 \pi r$.
Therefore,the total plane angle subtended at the centre is $\theta = \frac{2 \pi r}{r} = 2 \pi \, rad$.
Thus,the correct option is $B$.

(D) The correct value for $1$ light year is the distance light travels in one year,which is approximately $9.46 \times 10^{15} \, m$.
Option $A$ is correct: $1$ fermi $= 10^{-15} \, m$.
Option $B$ is correct: $1$ astronomical unit $(AU)$ $= 1.496 \times 10^{11} \, m$.
Option $C$ is correct: $1$ parsec $\approx 3.26$ light years.
Option $D$ is incorrect because $1$ light year $= 9.46 \times 10^{15} \, m$,not $9.46 \times 10^{12} \, m$.

(B) The values of the given units in meters are:
$1 \text{ AU} = 1.496 \times 10^{11} \text{ m}$
$1 \text{ ly} = 9.46 \times 10^{15} \text{ m}$
$1 \text{ pc} = 3.08 \times 10^{16} \text{ m}$
Comparing these values,we get $1 \text{ AU} < 1 \text{ ly} < 1 \text{ pc}$.
Statement $I$ is correct as these are units of distance.
Statement $II$ is incorrect because the correct order is $AU < ly < pc$.

(A) The order of magnitude of a number expressed as $a \times 10^b$ is determined by the value of $a$.
If $1 \leq a \leq 5$,the order of magnitude is $b$.
If $5 < a < 10$,the order of magnitude is $b + 1$.
Therefore,$b$ is the order of magnitude when $a \leq 5$.

(A) To determine the largest unit of length,we compare their values in meters:
$1$. $1 \text{ parsec} \approx 3.08 \times 10^{16} \text{ m}$
$2$. $1 \text{ light year} \approx 9.46 \times 10^{15} \text{ m}$
$3$. $1 \text{ astronomical unit (AU)} \approx 1.496 \times 10^{11} \text{ m}$
$4$. $1 \text{ fermi} = 10^{-15} \text{ m}$
Comparing these values,$3.08 \times 10^{16} \text{ m} > 9.46 \times 10^{15} \text{ m} > 1.496 \times 10^{11} \text{ m} > 10^{-15} \text{ m}$.
Therefore,the parsec is the largest unit of length among the given options.

(B) $\text{Density} = \frac{\text{Mass}}{\text{Volume}}$
$\text{The SI unit of mass is kilogram } (kg)$.
$\text{The SI unit of volume is cubic meter } (m^3)$.
$\text{Therefore, the SI unit of density is } \frac{kg}{m^3} \text{ or } kg \cdot m^{-3}$.

(D) The correct option is $D$.
Angular momentum is defined as $L = r \times p$.
The $SI$ unit of angular momentum is $kg \cdot m^2/s$,which is equivalent to $J \cdot s$.
According to Bohr's quantization condition,$L = \frac{n h}{2 \pi}$.
Since $n$ and $2 \pi$ are dimensionless,the unit of angular momentum $L$ is the same as the unit of Planck's constant $h$,which is $J \cdot s$.

(C) The values of the given units in meters are as follows:
$1 \text{ parsec} = 3.08 \times 10^{16} \text{ m}$
$1 \text{ nanometer} = 1 \times 10^{-9} \text{ m}$
$1 \text{ fermi} = 1 \times 10^{-15} \text{ m}$
$1 \text{ } \mathring{A} = 1 \times 10^{-10} \text{ m}$
Comparing these values, $1 \times 10^{-15} \text{ m}$ is the smallest magnitude.
Therefore, the least unit for length among the given options is fermi.

(B) By definition, $1 \text{ } \mathring{A} = 10^{-10} \text{ m}$. To convert this into millimeters $(\text{mm})$, we know that $1 \text{ m} = 10^3 \text{ mm}$. Therefore, $1 \text{ } \mathring{A} = 10^{-10} \times 10^3 \text{ mm}$. $1 \text{ } \mathring{A} = 10^{-7} \text{ mm}$. Thus, the correct option is $B$.

(D) The $SI$ unit of length is $1 \ m$.
Given that the new unit of length is $x \ m$.
Therefore,$1 \ m = \frac{1}{x}$ new units.
We need to express the area of $1 \ m^2$ in terms of the new unit.
Area $= 1 \ m^2 = 1 \ m \times 1 \ m$.
Substituting the value of $1 \ m$ in terms of the new unit:
Area $= (\frac{1}{x} \text{ new units}) \times (\frac{1}{x} \text{ new units}) = \frac{1}{x^2} \text{ (new units)}^2$.
Thus,the magnitude is $\frac{1}{x^2}$.

(B) light-year is a unit of astronomical distance,not time. It is defined as the distance that light travels in a vacuum in one Julian year,which is approximately $9.4607 \times 10^{12} \ km$.
Lunar month,leap year,and microsecond are all units used to measure time intervals.
Therefore,the correct option is $B$.

(B) Given energy $E = 13.6 eV$.
We know that $1 eV = 1.6 \times 10^{-19} J$.
Therefore,$E = 13.6 \times 1.6 \times 10^{-19} J = 21.76 \times 10^{-19} J$.
To convert Joules $(J)$ to kilowatt-hours $(kWh)$,we divide by $(1000 \times 3600)$ because $1 kWh = 1000 W \times 3600 s = 3.6 \times 10^6 J$.
$E = \frac{21.76 \times 10^{-19}}{3.6 \times 10^6} kWh$.
$E \approx 6.04 \times 10^{-25} kWh$.