Mix Examples-Units, Dimensions and Measurement Questions in English

(C) The formula for Young's modulus is $Y = \frac{F \cdot L}{A \cdot \Delta L}$.
The dimensional formula for Young's modulus is $[Y] = [M L^{-1} T^{-2}]$.
In the $CGS$ system,the unit is $\text{dyne/cm}^2$,and in the $MKS$ $(SI)$ system,the unit is $\text{N/m}^2$.
$1 \text{ N/m}^2 = 1 \frac{\text{kg} \cdot \text{m/s}^2}{\text{m}^2} = 1 \text{ kg} \cdot \text{m}^{-1} \cdot \text{s}^{-2}$.
$1 \text{ dyne/cm}^2 = 1 \frac{\text{g} \cdot \text{cm/s}^2}{\text{cm}^2} = 1 \text{ g} \cdot \text{cm}^{-1} \cdot \text{s}^{-2}$.
Converting $1 \text{ dyne/cm}^2$ to $MKS$:
$1 \text{ g} = 10^{-3} \text{ kg}$
$1 \text{ cm}^{-1} = (10^{-2} \text{ m})^{-1} = 10^2 \text{ m}^{-1}$.
Therefore,$1 \text{ dyne/cm}^2 = 10^{-3} \text{ kg} \cdot 10^2 \text{ m}^{-1} \cdot \text{s}^{-2} = 10^{-1} \text{ N/m}^2 = 0.1 \text{ N/m}^2$.
Thus,the conversion factor from $CGS$ to $MKS$ is $0.1$.

(D) scalar quantity is defined as a physical quantity that has magnitude but no direction. Because it lacks direction,its value remains invariant under a rotation of the coordinate system. Therefore,a scalar quantity has the same value for all observers,regardless of the orientation of their axes.

(D) The $SI$ unit of pressure is $N/m^2$ (Pascal) and the $CGS$ unit is $dyne/cm^2$.
We know that $1\, N = 10^5\, dyne$ and $1\, m = 10^2\, cm$.
Therefore,$1\, N/m^2 = \frac{10^5\, dyne}{(10^2\, cm)^2} = \frac{10^5}{10^4}\, dyne/cm^2 = 10\, dyne/cm^2$.
Given pressure $P = 50\, N/m^2$.
Converting to $CGS$: $P = 50 \times 10\, dyne/cm^2 = 500\, dyne/cm^2$.

(D) $1$. Angular momentum $(L)$ is given by $L = mvr$. The dimensions are $[M] \times [LT^{-1}] \times [L] = [ML^2T^{-1}]$.
$2$. Latent heat $(L_h)$ is given by $L_h = Q/m$. The dimensions are $[ML^2T^{-2}] / [M] = [L^2T^{-2}]$.
$3$. Capacitance $(C)$ is given by $C = Q/V$. Since $V = W/Q$,$C = Q^2/W$. The dimensions are $[AT]^2 / [ML^2T^{-2}] = [M^{-1}L^{-2}T^4A^2]$.
Thus,the correct sequence is $[ML^2T^{-1}], [L^2T^{-2}], [M^{-1}L^{-2}T^4A^2]$.

(B) The volume $V$ of a cylinder is given by $V = \pi R^2 h = \frac{\pi}{4} D^2 h$.
Given $D = 12.6\, cm$,$\Delta D = 0.1\, cm$,$h = 34.2\, cm$,and $\Delta h = 0.1\, cm$.
First,calculate the volume: $V = \frac{3.14159}{4} \times (12.6)^2 \times 34.2 \approx 4264.4\, cm^3$.
Rounding to appropriate significant figures (three significant figures based on the given data),$V = 4260\, cm^3$.
Now,calculate the relative error: $\frac{\Delta V}{V} = 2\frac{\Delta D}{D} + \frac{\Delta h}{h}$.
$\Delta V = V \times (2 \times \frac{0.1}{12.6} + \frac{0.1}{34.2}) = 4264.4 \times (0.01587 + 0.00292) \approx 4264.4 \times 0.01879 \approx 80.13$.
Rounding the absolute error to one significant figure,we get $\Delta V = 80\, cm^3$.
Thus,the volume is $V = 4260 \pm 80\, cm^3$.

(B) We know that the time constant of an $LR$ circuit is $\tau = \frac{L}{R}$,which has the dimension of time $[T]$.
Also,the time constant of an $RC$ circuit is $\tau = RC$,which has the dimension of time $[T]$.
Therefore,the expression $\frac{L}{RCV}$ can be rewritten as $\frac{L/R}{CV} = \frac{\tau}{CV}$.
Since $Q = CV$,where $Q$ is charge,we have $\frac{L}{RCV} = \frac{L/R}{Q} = \frac{[T]}{[AT]} = [A^{-1}]$.
Thus,the dimension is $[M^0 L^0 T^0 A^{-1}]$.

(A) Statement $(A)$ is correct: $A$ physical quantity can be measured in different units (e.g.,length can be measured in $m, cm, ft, inch$).
Statement $(B)$ is correct: The dimensional formula of a physical quantity is unique,representing its fundamental nature.
Statement $(C)$ is correct: Different physical quantities can share the same dimensions (e.g.,work and energy both have dimensions $[ML^2T^{-2}]$).
Statement $(D)$ is correct: According to the principle of homogeneity of dimensions,only physical quantities with the same dimensions can be added or subtracted.
Since all four statements are correct,the total number of correct statements is $4$.

(C) The formula for energy is $E = F \times d$,where $F$ is force and $d$ is distance (length).
If the unit of force is increased by $4$ times,the new force $F' = 4F$.
If the unit of length is increased by $4$ times,the new distance $d' = 4d$.
The new energy $E'$ will be $E' = F' \times d' = (4F) \times (4d) = 16(F \times d) = 16E$.
Therefore,the unit of energy is increased by $16$ times.

(B) $1$. Pressure and stress: Both have dimensions $[M L^{-1} T^{-2}]$.
$2$. Tension and surface tension: Tension is a force with dimensions $[M L T^{-2}]$,while surface tension is force per unit length with dimensions $[M T^{-2}]$. Thus,they do not have the same dimensions.
$3$. Strain and angle: Both are dimensionless quantities $[M^0 L^0 T^0]$.
$4$. Energy and work: Both have dimensions $[M L^2 T^{-2}]$.
Therefore,the correct option is $B$.

(C) $1$. Potential gradient $(dV/dx)$ has dimensions $[M L T^{-3} A^{-1}]$,and electric field $(E = F/q)$ also has dimensions $[M L T^{-3} A^{-1}]$.
$2$. Torque $(\tau = r \times F)$ and kinetic energy $(K = 1/2 mv^2)$ both have dimensions $[M L^2 T^{-2}]$.
$3$. Light year is a unit of distance with dimensions $[L]$,while time period is a unit of time with dimensions $[T]$. Since $[L] \neq [T]$,this pair does not have the same dimensions.
$4$. Impedance $(Z)$ and reactance $(X)$ both have dimensions of resistance,$[M L^2 T^{-3} A^{-2}]$.
Therefore,the correct option is $C$.

(C) The product of charge $e$ and potential difference $V$ represents energy $(W = eV)$.
Thus,the expression becomes $\frac{eV}{T} = \frac{W}{T}$.
From the ideal gas equation $PV = nRT$,we know that energy $W$ is proportional to $RT$ (where $R$ is the universal gas constant).
Therefore,$\frac{W}{T}$ has the same dimensions as the gas constant $R$.
Since the Boltzmann constant $k_B$ is defined as $\frac{R}{N_A}$ (where $N_A$ is Avogadro's number),the units of $\frac{eV}{T}$ are equivalent to the units of the Boltzmann constant.

(B) The $Assertion$ is correct because physical quantities can be measured directly (e.g.,using a meter scale) or indirectly (e.g.,using parallax method for distance of stars).
The $Reason$ is also correct because the reliability of any measurement depends on the precision and accuracy of the instrument used,and the final result must account for the associated uncertainties or errors.
However,the $Reason$ does not explain why we use direct or indirect methods; it explains the criteria for reporting a measurement. Therefore,the $Reason$ is not the correct explanation of the $Assertion$.

(N/A) Since $1\; cm = 10^{-2}\; m$,the volume of a cube of side $1\; cm$ is $(10^{-2}\; m)^{3} = 10^{-6}\; m^{3}$.
$(b)$ The total surface area of a cylinder is $S = 2\pi r(r + h)$. Given $r = 2.0\; cm = 20\; mm$ and $h = 10.0\; cm = 100\; mm$. Thus,$S = 2 \times 3.14 \times 20\; mm \times (20\; mm + 100\; mm) = 2 \times 3.14 \times 20 \times 120\; mm^{2} = 1.5072 \times 10^{4}\; mm^{2}$.
$(c)$ Speed $= 18\; km/h = 18 \times (5/18)\; m/s = 5\; m/s$. Distance covered in $1\; s$ is $5\; m/s \times 1\; s = 5\; m$.
$(d)$ Relative density is the ratio of the density of a substance to the density of water $(1\; g/cm^{3})$. Thus,density of lead $= 11.3 \times 1\; g/cm^{3} = 11.3\; g/cm^{3}$. Since $1\; g/cm^{3} = 10^{3}\; kg/m^{3}$,the density is $11.3 \times 10^{3}\; kg/m^{3}$.

(N/A) Since $1\; kg = 10^{3}\; g$ and $1\; m^{2} = 10^{4}\; cm^{2}$,then $1\; kg\; m^{2}\; s^{-2} = 10^{3}\; g \times 10^{4}\; cm^{2}\; s^{-2} = 10^{7}\; g\; cm^{2}\; s^{-2}$.
$(b)$ $1\; ly$ (light year) is the distance light travels in one year. $1\; ly = (3 \times 10^{8}\; m/s) \times (365.25 \times 24 \times 3600\; s) \approx 9.46 \times 10^{15}\; m$. Thus,$1\; m = 1 / (9.46 \times 10^{15})\; ly \approx 1.057 \times 10^{-16}\; ly$.
$(c)$ $1\; m = 10^{-3}\; km$ and $1\; s = (1/3600)\; h$,so $1\; s^{-2} = (3600)^{2}\; h^{-2}$. Therefore,$3.0\; m\; s^{-2} = 3.0 \times 10^{-3}\; km \times (3600)^{2}\; h^{-2} = 3.0 \times 10^{-3} \times 12960000\; km\; h^{-2} = 3.888 \times 10^{4}\; km\; h^{-2}$.
$(d)$ $G = 6.67 \times 10^{-11}\; N\; m^{2}\; kg^{-2}$. Since $1\; N = 1\; kg\; m\; s^{-2}$,$G = 6.67 \times 10^{-11}\; (kg\; m\; s^{-2})\; m^{2}\; kg^{-2} = 6.67 \times 10^{-11}\; kg^{-1}\; m^{3}\; s^{-2}$. Converting units: $1\; kg^{-1} = (10^{3}\; g)^{-1} = 10^{-3}\; g^{-1}$ and $1\; m^{3} = (10^{2}\; cm)^{3} = 10^{6}\; cm^{3}$. Thus,$G = 6.67 \times 10^{-11} \times 10^{-3}\; g^{-1} \times 10^{6}\; cm^{3}\; s^{-2} = 6.67 \times 10^{-8}\; cm^{3}\; s^{-2}\; g^{-1}$.

(A) Radius of hydrogen atom,$r = 0.5 \mathring{A} = 0.5 \times 10^{-10} \, m$.
Volume of one hydrogen atom $= \frac{4}{3} \pi r^{3} = \frac{4}{3} \times 3.14159 \times (0.5 \times 10^{-10})^{3} \approx 0.5236 \times 10^{-30} \, m^{3}$.
$1$ mole of hydrogen contains $N_{A} = 6.022 \times 10^{23}$ atoms.
Total volume of $1$ mole of hydrogen atoms $= N_{A} \times \text{Volume of one atom} = 6.022 \times 10^{23} \times 0.5236 \times 10^{-30} \, m^{3} \approx 3.153 \times 10^{-7} \, m^{3}$.
Rounding to two decimal places,the result is $3.16 \times 10^{-7} \, m^{3}$.

(N/A) During monsoons,the average rainfall in India is about $215 \ cm$,i.e.,the height of the water column $h = 2.15 \ m$. The area of India is $A \approx 3.3 \times 10^{12} \ m^2$. The volume of rainwater $V = A \times h \approx 7.09 \times 10^{12} \ m^3$. Given the density of water $\rho = 10^3 \ kg/m^3$,the mass of rain water is $M = \rho V \approx 7.09 \times 10^{15} \ kg$.
$(b)$ Consider a ship of known base area $A$ floating in the sea. Measure its depth $d_1$. When an elephant is placed on the ship,the depth increases to $d_2$. The volume of water displaced by the elephant is $V = A(d_2 - d_1)$. The mass of the elephant is $M = \rho_{water} \times A(d_2 - d_1)$.
$(c)$ Wind speed during a storm can be estimated using an anemometer,which measures the rate of rotation of cups driven by the wind. Alternatively,one can observe the displacement of objects over a known time interval.
$(d)$ Measure the surface area of the head $A$ and the diameter of a single strand of hair $d$ using a screw gauge. The number of strands $N \approx A / (\pi (d/2)^2)$.
$(e)$ Let the volume of the room be $V$. At $NTP$,one mole of air occupies $22.4 \times 10^{-3} \ m^3$. The number of molecules in one mole is $6.023 \times 10^{23}$. Thus,the number of molecules in the room is $N = (V / 22.4 \times 10^{-3}) \times 6.023 \times 10^{23} \approx 2.7 \times 10^{25} \times V$.

(A) Diameter of sodium atom $= 2.5 \; \mathring{A} = 2.5 \times 10^{-10} \; m$.
Radius of sodium atom,$r = 1.25 \times 10^{-10} \; m$.
Volume of one sodium atom,$V = \frac{4}{3} \pi r^3 = \frac{4}{3} \times 3.14 \times (1.25 \times 10^{-10})^3 \approx 8.18 \times 10^{-30} \; m^3$.
Mass of one sodium atom,$m = \frac{\text{Atomic mass}}{\text{Avogadro's number}} = \frac{23 \times 10^{-3} \; kg}{6.023 \times 10^{23}} \approx 3.82 \times 10^{-26} \; kg$.
Density of sodium atom,$\rho = \frac{m}{V} = \frac{3.82 \times 10^{-26}}{8.18 \times 10^{-30}} \approx 4.67 \times 10^3 \; kg \; m^{-3}$.
Given density of crystalline sodium $= 970 \; kg \; m^{-3} \approx 10^3 \; kg \; m^{-3}$.
Both densities are of the order of $10^3 \; kg \; m^{-3}$. They are of the same order of magnitude because in the solid phase,atoms are closely packed,but the inter-atomic spacing is comparable to the atomic size.

(N/A) scalar quantity is defined by its magnitude only,without any associated direction. $A$ vector quantity is defined by both its magnitude and an associated direction.
$1$. Scalar quantities: Volume,mass,speed,density,number of moles,and angular frequency.
$2$. Vector quantities: Acceleration,velocity,displacement,and angular velocity.

(B) Work and current are scalar quantities.
Work done is defined as the dot product of force and displacement, $W = \vec{F} \cdot \vec{d}$. Since the dot product of two vectors is always a scalar, work is a scalar physical quantity.
Current is defined by its magnitude. Although it has a direction, it does not follow the laws of vector addition (it follows algebraic addition). Therefore, it is classified as a scalar quantity.

(E) False: Despite being a scalar quantity,energy is not conserved in inelastic collisions.
$(b)$ False: Despite being a scalar quantity,temperature can take negative values.
$(c)$ False: Total path length is a scalar quantity,yet it has the dimension of length.
$(d)$ False: $A$ scalar quantity such as gravitational potential can vary from one point to another in space.
$(e)$ True: The value of a scalar does not vary for observers with different orientations of axes.

(N/A) Physical quantities are classified based on whether they require both magnitude and direction (vector) or only magnitude (scalar) to be fully described.
$1$. Scalar quantities: These quantities have only magnitude.
- Speed
- Pressure
- Temperature
- Work
- Energy
$2$. Vector quantities: These quantities have both magnitude and direction.
- Position
- Velocity
- Acceleration
- Force

(N/A) The energy required to break one bond of $DNA$ is given by:
$\frac{10^{-20} \; J}{1.6 \times 10^{-19} \; J/eV} \simeq 0.06 \; eV$.
Note: $0.1 \; eV = 100 \; meV$ ($100$ millielectron volt).
$(b)$ The kinetic energy of an air molecule is given by:
$\frac{10^{-21} \; J}{1.6 \times 10^{-19} \; J/eV} \simeq 0.0062 \; eV$.
This is equivalent to $6.2 \; meV$.
$(c)$ The average daily energy intake of a human adult is approximately $10^7 \; J$. Converting this to kilocalories:
$\frac{10^7 \; J}{4.2 \times 10^3 \; J/kcal} \simeq 2400 \; kcal$.

(N/A) In the liquid phase,the molecules of water are quite closely packed. The density of a water molecule may therefore be regarded as roughly equal to the density of bulk water $= 1000 \; kg \; m^{-3}$.
To estimate the volume of a water molecule,we need to know the mass of a single water molecule.
We know that $1 \; mole$ of water has a mass approximately equal to $(2 + 16) \; g = 18 \; g = 0.018 \; kg$.
Since $1 \; mole$ contains about $6 \times 10^{23}$ molecules (Avogadro's number),the mass of a single molecule of water is $(0.018) / (6 \times 10^{23}) \; kg = 3 \times 10^{-26} \; kg$.
Therefore,a rough estimate of the volume of a water molecule is:
Volume of a water molecule $= (3 \times 10^{-26} \; kg) / (1000 \; kg \; m^{-3}) = 3 \times 10^{-29} \; m^3$.
Assuming the molecule is spherical,$V = (4/3) \pi r^3$,which gives a radius $r \approx 2 \times 10^{-10} \; m = 2 \; \mathring{A}$.

(N/A) The $MKS$ unit of work is the Joule $(J)$.
Definition of Joule: One Joule of work is said to be done when a force of $1 \text{ Newton}$ displaces a body by $1 \text{ metre}$ in the direction of the applied force.
$1 \text{ Joule} = 1 \text{ N} \cdot \text{m} = 1 \text{ kg} \cdot \text{m}^2 \cdot \text{s}^{-2}$.
The $CGS$ unit of work is the erg.
Definition of erg: One erg of work is said to be done when a force of $1 \text{ dyne}$ displaces a body by $1 \text{ cm}$ in the direction of the applied force.
$1 \text{ erg} = 1 \text{ dyne} \cdot \text{cm} = 1 \text{ g} \cdot \text{cm}^2 \cdot \text{s}^{-2}$.
Dimensional formula of work: $[M^1 L^2 T^{-2}]$.
Conversion: $1 \text{ Joule} = 10^7 \text{ erg}$.
Practical units of work/energy:

erg	$10^{-7} \text{ J}$
Electron volt (eV)	$1.6 \times 10^{-19} \text{ J}$
Calorie (cal)	$4.186 \text{ J}$
Kilowatt hour (kWh)	$3.6 \times 10^6 \text{ J}$

(N/A) Since $1 \ \text{J} = 1 \ \text{kg} \cdot \text{m}^2/\text{s}^2$ and $1 \ \text{erg} = 1 \ \text{g} \cdot \text{cm}^2/\text{s}^2$,we have $1 \ \text{J} = (10^3 \ \text{g}) \times (10^2 \ \text{cm})^2 / \text{s}^2 = 10^7 \ \text{erg}$.
$(b)$ Since $1 \ \text{eV}$ is the energy gained by an electron accelerated through a potential difference of $1 \ \text{V}$,$1 \ \text{eV} = (1.602 \times 10^{-19} \ \text{C}) \times (1 \ \text{V}) = 1.602 \times 10^{-19} \ \text{J}$.
$(c)$ Since $1 \ \text{kWh} = (10^3 \ \text{W}) \times (3600 \ \text{s}) = 1000 \ \text{J/s} \times 3600 \ \text{s} = 3.6 \times 10^6 \ \text{J}$.

(N/A) The gravitational constant $G$ has the unit $N \cdot m^2 / kg^2$ and the dimensional formula $[M^{-1} L^3 T^{-2}]$.
The acceleration due to gravity $g$ has the unit $m / s^2$ and the dimensional formula $[M^0 L^1 T^{-2}]$.
Therefore,the ratio $\frac{G}{g}$ has the unit:
$\frac{N \cdot m^2 / kg^2}{m / s^2} = \frac{(kg \cdot m / s^2) \cdot m^2 / kg^2}{m / s^2} = \frac{m^2}{kg} = m^2 \cdot kg^{-1}$.
The dimensional formula is:
$\frac{[M^{-1} L^3 T^{-2}]}{[M^0 L^1 T^{-2}]} = [M^{-1} L^2 T^0]$.

(0.1, 1) $1$. We know that $1 \, Pa = 1 \, N/m^2$. Therefore,$0.1 \, Pa = 0.1 \, Nm^{-2}$.
$2$. To convert $Nm^{-2}$ to $\text{dyne} \, cm^{-2}$,we use the conversion factors: $1 \, N = 10^5 \, \text{dyne}$ and $1 \, m^2 = (10^2 \, cm)^2 = 10^4 \, cm^2$.
$3$. Thus,$1 \, Nm^{-2} = \frac{10^5 \, \text{dyne}}{10^4 \, cm^2} = 10 \, \text{dyne} \, cm^{-2}$.
$4$. Therefore,$0.1 \, Nm^{-2} = 0.1 \times 10 \, \text{dyne} \, cm^{-2} = 1 \, \text{dyne} \, cm^{-2}$.
$5$. The final result is $0.1 \, Pa = 0.1 \, Nm^{-2} = 1 \, \text{dyne} \, cm^{-2}$.

(D) The unit of force is $N$ (Newton),which is $kg\,m\,s^{-2}$.
Given unit is $N\,m^{-1}\,s^{-2}$.
Substituting the base units of force: $(kg\,m\,s^{-2}) \times m^{-1} \times s^{-2} = kg\,s^{-4}$.
Alternatively,let us check the dimensions: $[N] = [MLT^{-2}]$,$[m^{-1}] = [L^{-1}]$,$[s^{-2}] = [T^{-2}]$.
Multiplying these: $[MLT^{-2}] \times [L^{-1}] \times [T^{-2}] = [MT^{-4}]$.
This unit does not correspond to any standard fundamental physical quantity like pressure $(N\,m^{-2})$,surface tension $(N\,m^{-1})$,or force constant $(N\,m^{-1})$.
Therefore,the correct option is $D$.

(A) Let the side length of the cube be $l$.
Given that the volume $V$ is equal to the surface area $A$:
$V = A$
$l^3 = 6l^2$
Dividing both sides by $l^2$ (since $l \neq 0$):
$l = 6$
Now,the volume $V$ is:
$V = l^3 = (6)^3 = 216 \text{ cubic units}$.

(A) Given,distance of the galaxy $d = 10^{25} \ m$.
Speed of light $c = 3 \times 10^{8} \ m/s$.
The time taken $t$ is given by the formula $t = \frac{d}{c}$.
Substituting the values,$t = \frac{10^{25}}{3 \times 10^{8}} = \frac{1}{3} \times 10^{17} \ s$.
This can be written as $t = 0.333 \times 10^{17} \ s = 3.33 \times 10^{16} \ s$.
Since $3.33 < 5$,the order of magnitude is $10^{16} \ s$.

$(a)$ Plane angle $\theta = \frac{l}{r}$ has the unit radian, but its dimensional formula is $[M^{0} L^{0} T^{0}]$.
$(b)$ Strain $= \frac{\Delta l}{l} = \frac{\text{Change in length}}{\text{Original length}}$. It has neither a unit nor dimensions.
$(c)$ Gravitational constant $G = 6.67 \times 10^{-11} \text{ N} \cdot \text{m}^{2} \cdot \text{kg}^{-2}$. It is a constant with a unit.
$(d)$ Reynolds number $(Re)$ is a dimensionless constant.

(N/A) Since $1\,\mu m = 10^{-6}\,m$ and $1\,fm = 10^{-15}\,m$,then $\frac{10^{-6}}{10^{-15}} = 10^{9}$.
$(b)$ In the number $0.0060$,the leading zeros are not significant. The non-zero digit $6$ and the trailing zero after the decimal point are significant. Thus,there are $2$ significant figures.
$(c)$ The Scanning Tunneling Microscope $(STM)$ is used for the study of nanotechnology.

(A) Echo method or reflection method is used to measure the distance of a mountain.
$(b)$ Cesium atomic clock is the most accurate clock for the measurement of time.
$(c)$ To round off $15.753$ to three significant figures,we look at the fourth digit,which is $5$. Since the digit following the third significant figure $(7)$ is $5$ followed by non-zero digits $(3)$,we round up the third digit. Thus,$15.753$ becomes $15.8$.

(D) False. Weight is a derived quantity because it is defined as $W = mg$,where $m$ is mass (fundamental) and $g$ is acceleration due to gravity (derived).
$(b)$ False. An optical microscope or electron microscope is used to measure the size of an oleic acid molecule,not a micrometer (which is a mechanical measuring instrument).
$(c)$ False. The dimensions of a derived quantity can be zero for certain fundamental quantities. For example,velocity has dimensions $[M^0 L^1 T^{-1}]$,where the dimension of mass is zero.

(A) True. For example,plane angle is measured in radians but is dimensionless.
$(b)$ False. The unit of impulse is $N \cdot s$ (or $kg \cdot m/s$). The unit of the gradient of energy (force) is $N$ (or $kg \cdot m/s^2$). They are not the same.
$(c)$ True. By convention,the absolute error in a single measurement is taken as the least count of the measuring instrument.

(NONE) False. $A$ dimensionally incorrect equation must be physically incorrect.
$(b)$ False. $1 \ AU = 1.496 \times 10^{11} \ m$. The value $9.46 \times 10^{15} \ m$ represents $1 \ \text{light year}$.
$(c)$ False. The dimensional formula of force is $[M^1 L^1 T^{-2}]$, while the dimensional formula of stress (force per unit area) is $[M^1 L^{-1} T^{-2}]$.
$(d)$ False. The dimensions of a physical quantity are independent of the system of units used; only the numerical value changes.

(D) $1$. Wien's displacement law states that $\lambda_{max} T = b$,where $b$ is Wien's constant. The unit of wavelength $\lambda$ is $m$ and temperature $T$ is $K$. Thus,the unit of Wien's constant is $m K$,which corresponds to $(iii)$.
$2$. Stefan-Boltzmann's law states that the power radiated per unit area is $P/A = \sigma T^4$,where $\sigma$ is the Stefan-Boltzmann constant. The unit of power is $W$ (Watt),area is $m^2$,and temperature is $K$. Thus,the unit of $\sigma$ is $W m^{-2} K^{-4}$,which corresponds to $(i)$.
$3$. Therefore,the correct matching is $(a-iii), (b-i)$.

(N/A) The unit of both torque and work is $N \cdot m$ (Newton-meter) or $J$ (Joule). However,they are different physical quantities because:
$1$. Torque is a vector quantity,whereas work is a scalar quantity.
$2$. Torque represents the rotational effect of a force (turning effect),while work represents the energy transferred by a force acting over a displacement.

(A) The unit of energy (Joule) is $J = kg \cdot m^2/s^2$.
Multiplying this by $s^2$,we get:
$J \cdot s^2 = (kg \cdot m^2/s^2) \cdot s^2 = kg \cdot m^2$.
The physical quantity with the unit $kg \cdot m^2$ is the moment of inertia $(I = \sum mr^2)$.
Therefore,$J \cdot s^2$ is the unit of moment of inertia.

(B) $1$. Rydberg constant $({R}_{H})$: The formula for the Rydberg constant is derived from the Rydberg equation,and its $SI$ unit is ${m}^{-1}$. Thus,$(a)-(iii)$.
$2$. Planck's constant $(h)$: From the relation $E = h\nu$,the unit of $h$ is $J \cdot s = (kg \cdot m^{2} \cdot s^{-2}) \cdot s = {kg} {m}^{2} {s}^{-1}$. Thus,$(b)-(ii)$.
$3$. Magnetic field energy density $(u_{B})$: Energy density is energy per unit volume,$u = \frac{E}{V}$. Its $SI$ unit is $J/m^{3} = (kg \cdot m^{2} \cdot s^{-2}) / m^{3} = {kg} {m}^{-1} {s}^{-2}$. Thus,$(c)-(iv)$.
$4$. Coefficient of viscosity $(\eta)$: From Stokes' law or the formula $F = \eta A \frac{dv}{dx}$,the unit of $\eta$ is $kg \cdot m^{-1} \cdot s^{-1}$. Thus,$(d)-(i)$.
Matching these,we get $(a)-(iii), (b)-(ii), (c)-(iv), (d)-(i)$.

(A) The dimensional formulas are calculated as follows:
$(a)$ Torque $(\tau) = \text{Force} \times \text{Distance} = [MLT^{-2}] \times [L] = [ML^2T^{-2}]$, which matches $(iii)$.
$(b)$ Impulse $(I) = \text{Force} \times \text{Time} = [MLT^{-2}] \times [T] = [MLT^{-1}]$, which matches $(i)$.
$(c)$ Tension is a force, so its dimension is $[MLT^{-2}]$, which matches $(iv)$.
$(d)$ Surface Tension $(S) = \frac{\text{Force}}{\text{Length}} = \frac{[MLT^{-2}]}{[L]} = [MT^{-2}]$, which matches $(ii)$.
Therefore, the correct matching is $(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)$.

(A) $1$. Capacitance $(C)$: From $q = CV$,we have $[C] = [q/V] = [q^2 / (Work)] = [A^2 T^2 / (M L^2 T^{-2})] = M^{-1} L^{-2} T^4 A^2$. Thus,$(a) \rightarrow (iii)$.
$2$. Permittivity of free space $(\varepsilon_0)$: From Coulomb's law $F = (q_1 q_2) / (4 \pi \varepsilon_0 r^2)$,we have $[\varepsilon_0] = [q^2 / (F L^2)] = [A^2 T^2 / (M L T^{-2} L^2)] = M^{-1} L^{-3} T^4 A^2$. Thus,$(b) \rightarrow (ii)$.
$3$. Permeability of free space $(\mu_0)$: Using $c = 1 / \sqrt{\mu_0 \varepsilon_0}$,we get $\mu_0 = 1 / (\varepsilon_0 c^2)$. Substituting dimensions,$[\mu_0] = [1 / (M^{-1} L^{-3} T^4 A^2 \cdot L^2 T^{-2})] = M^1 L^1 T^{-2} A^{-2}$. Thus,$(c) \rightarrow (iv)$.
$4$. Electric field $(E)$: From $F = qE$,we have $[E] = [F/q] = [M L T^{-2} / (A T)] = M^1 L^1 T^{-3} A^{-1}$. Thus,$(d) \rightarrow (i)$.
The correct matching is $(a) \rightarrow (iii), (b) \rightarrow (ii), (c) \rightarrow (iv), (d) \rightarrow (i)$.

(D) $1$. Wave number $(k = 1/\lambda)$ and Rydberg's constant $(R)$ both have dimensions of $[L^{-1}]$.
$2$. Stress and Coefficient of elasticity (Young's modulus,Bulk modulus,etc.) both have dimensions of $[M L^{-1} T^{-2}]$.
$3$. Coercivity $(H)$ and Magnetisation $(M)$ have different dimensions. Coercivity is the magnetic field strength $(H)$ with dimensions $[A L^{-1}]$,while Magnetisation is magnetic moment per unit volume with dimensions $[A L^{-1}]$. Wait,both are $[A L^{-1}]$. Let us re-evaluate.
$4$. Specific heat capacity $(S)$ has dimensions $[L^2 T^{-2} K^{-1}]$ (from $S = Q / (m \Delta T)$),while Latent heat $(L)$ has dimensions $[L^2 T^{-2}]$ (from $L = Q / m$).
Since $[L^2 T^{-2} K^{-1}] \neq [L^2 T^{-2}]$,the pair with different dimensions is Specific heat capacity and Latent heat.

(B) $1$. Torque is defined as the product of force and the perpendicular distance from the axis of rotation. Its $SI$ unit is $Nm$.
$2$. Stress is defined as the restoring force per unit area. Its $SI$ unit is $N/m^2$ or $Nm^{-2}$.
$3$. Latent heat is the energy required to change the state of a unit mass of a substance. Its $SI$ unit is $J/kg$ or $Jkg^{-1}$.
$4$. Power is defined as the rate of doing work. Since $Work = Force \times Displacement$,$Power = Force \times Velocity$. Its $SI$ unit is $N \times (m/s) = Nms^{-1}$.
Therefore,the correct matching is $(A)-(III), (B)-(IV), (C)-(II), (D)-(I)$.

(B) The surface area over which rain is received is $A = 600 \, km^2 = 600 \times (10^3)^2 \, m^2 = 6 \times 10^8 \, m^2$.
The average annual rainfall is $h = 2.4 \, m$.
The total volume of water received by rain is $V = A \times h = 6 \times 10^8 \times 2.4 = 14.4 \times 10^8 \, m^3 = 1.44 \times 10^9 \, m^3$.
Since $1 \, m^3 = 1000 \, L$,the total volume in liters is $1.44 \times 10^9 \times 10^3 = 1.44 \times 10^{12} \, L$.
The amount of water conserved is $10 \%$ of the total volume: $V_{cons} = 0.10 \times 1.44 \times 10^{12} \, L = 1.44 \times 10^{11} \, L$.
The percentage of Mumbai's annual water needs met by this conserved water is $\frac{1.44 \times 10^{11}}{1.4 \times 10^{12}} \times 100 \approx 10 \%$.

(D) The dimension of force is $[MLT^{-2}]$.
$A$. Weight is a force $(W = mg)$,so its dimension is $[MLT^{-2}]$.
$B$. According to Newton's second law,the rate of change of momentum is equal to force $(F = dp/dt)$,so its dimension is $[MLT^{-2}]$.
$C$. Work per unit length is $W/L$. Since $W = F \times d$,$W/L = (F \times d)/L = F$. Thus,its dimension is $[MLT^{-2}]$.
$D$. Work done per unit charge is defined as electric potential $(V = W/q)$. The dimension of potential is $[ML^2T^{-3}A^{-1}]$,which is not equal to the dimension of force.
Therefore,the correct option is $D$.

(C) Let the original units be $L$,$v$,and $F$. The new units are $L' = 2L$,$v' = 2v$,and $F' = 2F$.
We know that velocity $v = L/T$,so the unit of time $T = L/v$. The new unit of time $T' = L'/v' = (2L)/(2v) = L/v = T$. Thus,the unit of time remains unchanged.
We know that force $F = ma = m(v/T)$,so the unit of mass $m = FT/v$. The new unit of mass $m' = F'T'/v' = (2F \times T)/(2v) = FT/v = m$. Thus,the unit of mass remains unchanged.
We know that momentum $p = mv$. The new unit of momentum $p' = m'v' = m \times (2v) = 2mv = 2p$. Thus,the unit of momentum is doubled.
We know that energy $E = F \times L$. The new unit of energy $E' = F' \times L' = (2F) \times (2L) = 4FL = 4E$. Thus,the unit of energy becomes four times the original value.
Therefore,the correct statement is that the unit of momentum is doubled.

(A) The dimension of impulse is $J = F \times \Delta t = [MLT^{-2}] \times [T] = [MLT^{-1}]$.
Therefore,the dimension of "impulse per unit area" is $\frac{[MLT^{-1}]}{[L^2]} = [ML^{-1}T^{-1}]$.
Now,let us check the dimensions of the given options:
$A$. Viscosity (Coefficient of viscosity $\eta$): The force $F = \eta A \frac{dv}{dx}$,so $\eta = \frac{F}{A (dv/dx)} = \frac{[MLT^{-2}]}{[L^2] [LT^{-1}/L]} = [ML^{-1}T^{-1}]$.
$B$. Surface tension: $[MT^{-2}]$.
$C$. Bulk modulus: $[ML^{-1}T^{-2}]$.
$D$. Force: $[MLT^{-2}]$.
Thus,the unit of "impulse per unit area" is the same as that of viscosity.

(C) The correct answer is $C$.
$1$. Strain is the ratio of change in dimension to the original dimension,so it is dimensionless and unitless.
$2$. Reynold's number is a dimensionless quantity used in fluid mechanics.
$3$. Angular displacement is defined as the angle subtended by an arc at the center of a circle,given by $\theta = \frac{\text{arc}}{\text{radius}}$. It has units of radians $(rad)$ or degrees,but since it is a ratio of two lengths,it is dimensionless.
$4$. Poisson's ratio is the ratio of lateral strain to longitudinal strain,making it dimensionless and unitless.

$(A)$ Surface Tension $= \frac{F}{l} = \frac{MLT^{-2}}{L} = MT^{-2} = kg s^{-2}$ $(IV)$.
$(B)$ Pressure $= \frac{F}{A} = \frac{MLT^{-2}}{L^2} = ML^{-1}T^{-2} = kg m^{-1} s^{-2}$ $(III)$.
$(C)$ Viscosity $= \frac{F}{A(\frac{dv}{dz})} = \frac{MLT^{-2}}{L^2(\frac{LT^{-1}}{L})} = ML^{-1}T^{-1} = kg m^{-1} s^{-1}$ $(I)$.
$(D)$ Impulse $= F \times \Delta t = MLT^{-2} \times T = MLT^{-1} = kg m s^{-1}$ $(II)$.
Therefore, the correct matching is $(A)-(IV), (B)-(III), (C)-(I), (D)-(II)$.