Physical World Units And Measurements – NEET UG PhysicsPractice Questions & PYQs

The expression $\left( \mu_0 \varepsilon_0 \right)^{-1/2}$ is fundamentally identified with the velocity of electromagnetic waves, such as light, propagating through a vacuum. Maxwell's equations demonstrate that the speed of light $c$ relates to the permeability and permittivity of free space via the formula $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$. Since the given expression is mathematically identical to this velocity, its dimensional formula must correspond to that of speed, which is $[LT^{-1}]$.

The damping force acting on an oscillator is described by the relation $F = kv$, where $k$ is the proportionality constant and $v$ is the velocity. By rearranging this expression to $k = F/v$, we can determine the unit of $k$ by dividing the standard SI unit of force, which is kg m s^-2, by the unit of velocity, m s^-1. This calculation simplifies to kg s^-1, which represents the dimensions of the constant of proportionality.

Density represents the mass contained in a unit volume, so changing the scale of the base units for mass and length naturally changes the numerical value associated with that density. In the new system, 1 g is equivalent to 1/100 of the specified mass unit, and 1 cm is equivalent to 1/10 of the specified length unit. The new density value is calculated by substituting these ratios into the equation $d_{new} = \frac{4/100}{(1/10)^3}$, which simplifies to 40.

The expression $\frac{1}{2} \varepsilon _0 E^2$ represents the energy density of an electric field, which is defined as the energy stored within a unit volume of space. Dimensional analysis follows directly from this definition by taking the ratio of the dimensions of energy, $[ML^2T^{-2}]$, to the dimensions of volume, $[L^3]$. Simplifying this fraction leads to the resulting dimensional formula of $[ML^{-1}T^{-2}]$.

The distance $h$ traversed by a body in free fall starting from rest is defined by the relation $h = \frac{1}{2}gt^2$. Rearranging this formula to isolate the acceleration due to gravity yields the equation $g = \frac{2h}{t^2}$.

According to the principles of error propagation, when calculating a value derived from division or multiplication, the maximum relative error is the sum of the relative errors of the independent variables, with each relative error multiplied by the magnitude of its exponent. Applying this rule to our expression, the maximum relative error in $g$ is the sum of the relative error in $h$ and twice the relative error in $t$, expressed as $\frac{\Delta g}{g} = \frac{\Delta h}{h} + 2\frac{\Delta t}{t}$. Given that the percentage errors in distance and time are $e_1$ and $e_2$ respectively, the total maximum percentage error in the estimation of $g$ becomes $e_1 + 2e_2$.

Dimensional analysis involves breaking down a physical quantity into its fundamental components of mass, length, and time to confirm the specific exponents given. Pressure is defined as force per unit area, which is expressed dimensionally as [P] = \frac{[MLT^{-2}]}}{[L^2]${}$. Simplifying this expression results in$[M^1L^{-1}T^{-2}]$, which corresponds exactly to the exponents$a = 1, b = -1, c = -2$ for pressure.

The volume of a sphere is defined by the formula $V = \frac{4}{3}\pi r^3$. Calculating the relative error involves differentiating this expression, which leads to the relationship $\frac{dV}{V} = 3 \frac{dR}{R}$. This ratio indicates that the percentage error in the volume is equal to three times the percentage error in the radius. By multiplying the provided 2% radius error by this factor of three, we find the resulting error in the volume is 6%.

Energy density, defined as energy per unit volume, has dimensions derived from work divided by volume, which simplifies to $[M L^{-1} T^{-2}]$. Young's modulus is the ratio of stress to strain, and since strain is a dimensionless ratio, the modulus shares the dimensions of stress, or force per unit area, also resulting in $[M L^{-1} T^{-2}]$. Because refractive index and dielectric constant are both dimensionless quantities, only energy density and Young's modulus share this identical dimensional formula.

Ohm's law establishes the foundational relationship for electrical resistance, defining it as the ratio of potential difference to current, which can be expressed as:
$R = \frac{V}{I}$
Since voltage represents the work done per unit charge, its dimensions are determined by dividing the dimensions of energy, $[ML^{2}T^{-2}]$, by the dimensions of charge, which is current multiplied by time, or $[IT]$. This calculation results in voltage having dimensions of $[ML^{2}T^{-3}I^{-1}]$. By substituting this back into the resistance formula and dividing by the dimension of current, $[I]$, we arrive at the final dimensional representation for resistance as $[ML^{2}T^{-3}I^{-2}]$.

Dimensional analysis relies on the principle of homogeneity, which dictates that any physical equation is only valid if the dimensions of all terms being added or subtracted are identical. In the expression $v=a t+\;\frac{b}{t+c}$, the constant $c$ is added directly to time $t$, meaning $c$ must represent a time value and carry the dimension $[T]$. Because the term $at$ must equate to the dimensions of velocity $[LT^{-1}]$, dividing these dimensions by the dimension of time reveals that $a$ corresponds to $[LT^{-2}]$. Finally, since the entire fractional term $\frac{b}{t+c}$ must also possess the dimensions of velocity, the variable $b$ divided by the time dimension $[T]$ must result in $[LT^{-1}]$, confirming that $b$ holds the dimension $[L]$.

Planck's constant $h$ is defined by the dimensions $[ML^{2}T^{-1}]$, while the moment of inertia $I$ corresponds to $[ML^{2}]$. Dividing these terms cancels the mass and squared length, leaving the inverse of time:

\frac{[ML^{2}T^{-1}]}$}{[ML^{2}]\${} = [T^{-1}]

This resultant dimension, $[T^{-1}]$, is characteristic of frequency.

Coulomb's Law establishes that the electrostatic force between two charges is expressed by the formula $F= \frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2}}{r^{2}}$ To determine the units for the permittivity of free space, we rearrange the equation to isolate $\varepsilon_{0}$ as $\varepsilon_{0}= \frac{1}{4 \pi} \frac{q^{2}}{F r^{2}}$ Substituting the standard SI units of charge \in coulombs, force \in newtons, and distance \in metres, we find that the resulting unit for $\varepsilon_{0}$ is coulomb squared per newton-metre squared.

Newton's law of universal gravitation establishes that the force between two masses is defined by $F = G \frac{m_1 m_2}{r^2}$, which allows us to isolate the gravitational constant as $G = \frac{F r^2}{m_1 m_2}$. By substituting the dimensions for force as $[MLT^{-2}]$, distance as $[L]$, and mass as $[M]$ into this rearranged formula, the expression becomes \frac{[MLT^{-2}]} × [L]^2}{[M] × [M]}$. Combining the terms through algebraic simplification results \in the final dimensional formula$[M^{-1} L^3 T^{-2}]$.

The fundamental quantum relation $E=hv$ reveals that Planck's constant is defined by the ratio of energy to frequency, yielding a dimensional formula of $[ML^{2}T^{-1}]$. Angular momentum is calculated as the product of the moment of inertia $[ML^{2}]$ and angular velocity $[T^{-1}]$, which simplifies to the identical representation of $[ML^{2}T^{-1}]$. This shared dimensional structure confirms that Planck's constant and angular momentum are physically equivalent in their units.

Dimensional analysis allows us to determine if two quantities share the same physical nature by comparing their dependence on mass, length, and time. Force acts as the product of mass and acceleration, characterized by the dimension $[MLT^{-2}]$, whereas impulse is calculated as the product of force and time, resulting in the dimension $[MLT^{-1}]$. Because these two expressions result in different powers for time, this specific pair fails to possess equivalent dimensions.