Q.(MCQ II — one or more options correct) Photon is quantum of radiation with energy $E = h\nu$ where $\nu$ is frequency and $h$ is Planck's constant. The dimensions of $h$ are the same as that of

The core idea here is Dimensional Analysis, which involves comparing the fundamental dimensions (Mass $M$, Length $L$, Time $T$) of different physical quantities.

First, we find the dimensions of Planck's constant $h$:

From $E = h\nu$, we have $h = \frac{E}{\nu}$.

The dimensions of energy $[E]$ are $[ML^2T^{-2}]$ (e.g., from $E = \frac{1}{2}mv^2$).

The dimensions of frequency $[\nu]$ are $[T^{-1}]$.

Therefore, the dimensions of $h$ are $[h] = \frac{[ML^2T^{-2}]}{[T^{-1}]} = [ML^2T^{-1}]$.

Next, we determine the dimensions for each given option:

• (A) Linear impulse: Impulse $J = F \Delta t$. Dimensions $[J] = [MLT^{-2}][T] = [MLT^{-1}]$.
• (B) Angular impulse: Angular impulse $J_{ang} = \tau \Delta t$. Torque $\tau = r \times F$. Dimensions $[\tau] = [L][MLT^{-2}] = [ML^2T^{-2}]$. So, $[J_{ang}] = [ML^2T^{-2}][T] = [ML^2T^{-1}]$. …

Understanding the dimensions of physical quantities is fundamental in physics. It allows us to check the consistency of equations, derive relationships between different quantities, and even predict the form of physical laws. The core idea is that for any valid physical equation, the dimensions on both sides must be identical. This principle is called the Principle of Homogeneity of Dimensions.

Here, we are asked to find quantities that share the same dimensions as Planck's constant ($h$). We will first determine the dimensions of $h$ from the given energy-frequency relation, and then systematically find the dimensions of each option to compare.

1. Determine the dimensions of Planck's constant ($h$).

   The given relation is $E = h\nu$, where $E$ is energy and $\nu$ is frequency.

   We need to recall the fundamental dimensions of energy and frequency.

   • Energy ($E$): Energy is defined as the capacity to do work. Work done is Force $\times$ Distance. Force ($F$) has dimensions of $[M][L][T^{-2}]$ (from $F=ma$). Distance ($d$) has dimensions of $[L]$. So, the dimensions of Energy are $[E] = [F][d] = ([M][L][T^{-2}])([L]) = [M][L^2][T^{-2}]$.
   • Frequency ($\nu$): Frequency is the number of cycles per unit time, so its dimension is the inverse of time. The dimensions of Frequency are $[\nu] = [T^{-1}]$.

   Now, we can find the dimensions of $h$ from $E = h\nu$:

   $[h] = \frac{[E]}{[\nu]}$

   $[h] = \frac{[M][L^2][T^{-2}]}{[T^{-1}]}$

   $[h] = [M][L^2][T^{-2}][T^1]$

   $[h] = [M][L^2][T^{-1}]$

   Important

   The dimensions of Planck's constant are $[M][L^2][T^{-1}]$. This is a frequently tested result.

2. Determine the dimensions of each given option.

   • (A) Linear impulse:

     Linear impulse ($J_{linear}$) is defined as the product of force and the time interval over which it acts, or equivalently, the change in linear momentum.

     $J_{linear} = F \Delta t$

     The dimensions of Force ($F$) are $[M][L][T^{-2}]$.

     The dimensions of Time ($\Delta t$) are $[T]$.

     So, the dimensions of Linear impulse are $[J_{linear}] = [F][\Delta t] = ([M][L][T^{-2}])([T]) = [M][L][T^{-1}]$.

     This does not match $[h]$.

   • (B) Angular impulse:

     Angular impulse ($J_{angular}$) is defined as the product of torque and the time interval over which it acts, or equivalently, the change in angular momentum.

     $J_{angular} = \tau \Delta t$

     First, let's find the dimensions of Torque ($\tau$). Torque is Force $\times$ Perpendicular distance.

     $[\tau] = [F][r] = ([M][L][T^{-2}])([L]) = [M][L^2][T^{-2}]$.

     The dimensions of Time ($\Delta t$) are $[T]$.

     So, the dimensions of Angular impulse are $[J_{angular}] = [\tau][\Delta t] = ([M][L^2][T^{-2}])([T]) = [M][L^2][T^{-1}]$.

     This matches $[h]$.

   • (C) Linear momentum:

     Linear momentum ($p$) is defined as the product of mass and velocity.

     $p = mv$

     The dimensions of Mass ($m$) are $[M]$.

     The dimensions of Velocity ($v$) are $[L][T^{-1}]$ (from $v = \text{distance}/\text{time}$). …

Concept: Dimensional Analysis of Planck's Constant

The core idea is to find the dimensions of $h$ from $E = h\nu$, then match them with the dimensions of the given physical quantities.

Method: Dimensional Matching

Step 1: Find dimensions of $h$

From $E = h\nu$:

• Energy $E$ has dimensions: $[E] = [ML^2T^{-2}]$
• Frequency $\nu$ has dimensions: $[\nu] = [T^{-1}]$

So:

Step 2: Find dimensions of each option

Step 3: Compare with $[h]$ …

Common Mistakes & How to Avoid Them

Mistake 1: Confusing “same dimensions” with “same units”

Many students think that if two quantities have the same units (like N·s and kg·m/s), they automatically have the same dimensions — which is true — but they then fail to check whether the given quantity actually has those dimensions.

How to avoid:

Always reduce the quantity to its base dimensions (mass $M$, length $L$, time $T$).

For $h$:

Energy $E$ has dimensions $[M L^2 T^{-2}]$ and frequency $\nu$ has $[T^{-1}]$.

So

Now check each option by writing its dimensional formula — don’t rely on memory alone.

Mistake 2: Thinking “impulse” always means linear impulse

Students see “impulse” and immediately write $[M L T^{-1}]$ (linear impulse), forgetting that angular impulse exists and has different dimensions.

How to avoid:

• Linear impulse = force × time = $[M L T^{-2}] \times [T] = [M L T^{-1}]$
• Angular impulse = torque × time = $[M L^2 T^{-2}] \times [T] = [M L^2 T^{-1}]$

Compare with $[h] = [M L^2 T^{-1}]$ — only angular impulse matches.

Mistake 3: Assuming “angular momentum” and “angular impulse” are the same

They do have the same dimensions ($[M L^2 T^{-1}]$), but students often pick only one and miss the other in a multi-correct question.

How to avoid:

In MCQ with one or more correct options, always check every option independently.

• Angular momentum $L = mvr$ → $[M L T^{-1}] \times [L] = [M L^2 T^{-1}]$ ✓
• Angular impulse = torque × time → same dimensions ✓ …