Supplementary Units – NTA TRAP! – NCERT Concepts + MCQs

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Listen up, NEET aspirants. Today, we are tackling a classic trap that the NTA loves to set in the Units and Measurements chapter. Usually, if a physical quantity has a unit, it also has a dimensional formula. But there is a massive exception to this standard rule: The Supplementary Physical Quantities. Let us break them down.

The Dimensional Exception Principle

In the SI system, there are two distinct supplementary units you must memorize.

Plane Angle ($d\theta$)

• This is the 2D angle subtended at the center of a circle by an arc.
• Formula: It is mathematically defined as the ratio of the arc length ($ds$) to the radius ($r$), giving us $d\theta = \frac{ds}{r}$.
• Unit: Radian ($\text{rad}$).
• Pro-Tip: To convert degrees to radians, remember that $1^\circ = \frac{\pi}{180} \text{ radians}$. If you ever see a trig function without a degree symbol (e.g., $\sin 15$), you must assume that the value is in radians.

Solid Angle ($d\Omega$)

• This is the 3D equivalent, representing the angle subtended at the center of a sphere by a patch of surface area.
• Formula: It is defined as the ratio of the intercepted area ($dA$) to the square of the radius ($r^2$), giving us $d\Omega = \frac{dA}{r^2}$.
• Unit: Steradian ($\text{sr}$).

The NEET High-Yield Secret: Look closely at those formulas. Plane angle is length divided by length, and solid angle is area divided by area. Because they are ratios of similar physical dimensions, the dimensions completely cancel out! This makes their dimensional formula exactly $[M^0L^0T^0]$.

Memorize this fundamental rule right now: A physical quantity can have a unit without having a dimension, but a dimensionless quantity can only have a unit if it is a supplementary angle.

The Golden Rule for Trigonometric Arguments

Another high-yield mock paper concept tied to dimensions is the constraint of arguments. The golden rule is this: the argument of any trigonometric function must be completely dimensionless.

For example, consider a wave equation represented by $y = A \sin(aEt – bVx)$. Because the argument of the sine function must be dimensionless, both the term $(aEt)$ and the term $(bVx)$ must have no dimensions (equaling $[M^0L^0T^0]$). If you know the dimensions of energy ($E$), time ($t$), velocity ($V$), and distance ($x$), you can easily calculate the unknown dimensions of $a$ and $b$ by equating them to one over the variables’ dimensions.

Advanced Application: Solid Angle Flux in Electrostatics

This supplementary concept perfectly blends with Electrostatics—a very common profile in top mock tests.

Imagine a point charge $q$ is placed on the axis of a disc, such that the disc subtends a cone of half-angle $\theta = 45^\circ$ at the position of the charge. What is the electric flux ($\phi$) passing through the disc?

• First, the solid angle ($\Omega$) subtended by a cone of half-angle $\theta$ is mathematically defined as:$$\Omega = 2\pi(1 – \cos\theta)\text{ steradians}$$.
• Applying Gauss’s law, a charge $q$ radiates a total flux of $\frac{q}{\epsilon_0}$ across a full 3D sphere of $4\pi$ steradians.
• To find the flux through just our disc, we multiply that total flux by the ratio of our solid angle to $4\pi$.
• Therefore, $\phi = \frac{q}{\epsilon_0}\left(\frac{\Omega}{4\pi}\right)$.

Test Your Knowledge: NEET MCQs on Supplementary Units

Question 1: The Dimensional Exception Principle (NEET 2022)

Plane angle and solid angle have:

1. Both units and dimensions
2. Units but no dimensions
3. Dimensions but no units
4. No units and no dimensions

Question 2: Dimension Matching and Ratios (NEET 2024)

The quantities which have the same dimensions as those of solid angle are:

1. Stress and angle
2. Strain and arc
3. Angular speed and stress
4. Strain and angle

Question 3: Solid Angle Flux Application (Aakash Test Series / NEET Mock Test)

A point charge $q$ is placed on the axis of a disc such that the disc subtends a cone of half-angle $\theta = 45^\circ$ at the position of the charge. The electric flux ($\phi$) passing through the disc is:

1. $\frac{q}{2\epsilon_0}$
2. $\frac{q}{\epsilon_0}$
3. $\frac{\sqrt{2}(\sqrt{2}-1)q}{4\epsilon_0}$
4. $\frac{q}{4\epsilon_0}$

Question 4: Argument Dimensionality Constraint (Mock Paper Concept)

In a wave equation represented by $y = A \sin(aEt – bVx)$, where $y$ is displacement, $E$ is energy, $V$ is velocity, $t$ is time, and $x$ is distance, the dimensional formula of the ratio $\frac{a}{b}$ is:

1. $[M^1L^1T^1]$
2. $[M^1L^2T^0]$
3. $[M^{-1}L^0T^0]$
4. $[MLT^{-2}]$

Question 5: Classification in the SI System (NCERT Core Concept)

Which of the following is correct regarding the classification of Plane Angle and Solid Angle in the SI system?

1. They are classified as Base Units because they are fundamental to geometry.
2. They are classified as Derived Units because they are formed as ratios of other physical quantities.
3. They are defined as Supplementary Units, which are dimensionless but possess unique SI units.
4. They are classified as Improper Units because they lack dimensions.

Solutions

Solution 1: Option (2) is correct. While physical quantities with units generally possess dimensions, supplementary quantities are the ultimate exception. They possess units (radian and steradian) but are completely dimensionless ($[M^0L^0T^0]$).

Solution 2: Option (4) is correct. Solid angle is dimensionless ($[M^0L^0T^0]$). Any other dimensionless quantities will share its dimensions. Strain (change in dimension over original dimension) and plane angle (arc over radius) are both perfectly dimensionless ratios.

Solution 3: Option (3) is correct.

• For $\theta = 45^\circ$, the solid angle is $\Omega = 2\pi(1 – \cos 45^\circ) = 2\pi\left(1 – \frac{1}{\sqrt{2}}\right) = \sqrt{2}\pi(\sqrt{2}-1)\text{ sr}$.
• Substituting this into the flux ratio gives: $\phi = \frac{q}{\epsilon_0}\left(\frac{\Omega}{4\pi}\right) = \frac{\sqrt{2}(\sqrt{2}-1)q}{4\epsilon_0}$.

Solution 4: Option (3) is correct.

• The argument of the sine function must be dimensionless.
• $[aEt] = [M^0L^0T^0] \implies [a] = [E^{-1}t^{-1}] = [M^{-1}L^{-2}T^1]$.
• $[bVx] = [M^0L^0T^0] \implies [b] = [V^{-1}x^{-1}] = [L^{-2}T^1]$.
• The ratio $\left[\frac{a}{b}\right] = \frac{[M^{-1}L^{-2}T^1]}{[L^{-2}T^1]} = [M^{-1}L^0T^0]$.

Solution 5: Option (3) is correct. NCERT explicitly notes that besides the seven base units, there are two supplementary units defined: plane angle (radian) and solid angle (steradian). They are dimensionless quantities that possess unique, official SI units.

Want a deeper dive into Units and Dimensions? Watch our full visual breakdown here!

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