In the expression $P = E l^2 m^{-5} G^{-2}$, $E$, $m$, $l$ and $G$ denote energy, mass, angular momentum and gravitational constant, respectively. Show that $P$ is a dimensionless quantity.

In the expression P = E l² m⁻⁵ G⁻², E, m, l and G denote energy, mass, angular momentum and gravitational cons

NCERT Exemplar · Q39

P = E l^2 m^{-5} G^{-2}

Rajasthan RbseLong· 3mImportance★★★★★est

92% · 61/66 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Unit Conversion

Unit Conversion: Why 1 Metre and 100 Centimetres Are the Same Thing

Imagine you're measuring the length of your desk. You pull out a ruler marked in centimetres and find it's 120 cm long. Your friend, using a metre stick, says it's 1.2 m. You're both right — you've just used different units to describe the same physical length.

That's the core idea: unit conversion is the process of changing how you express a quantity without changing the quantity itself.

The Intuition: Same Quantity, Different Labels

Think of a pizza. Whether you call it "one pizza" or "8 slices," the amount of pizza hasn't changed. You've just used a different unit (pizza vs. slice) to describe it.

Similarly, 1 metre and 100 centimetres are the same length — just like 1 pizza and 8 slices are the same amount. The number changes (1 becomes 100, or 1 becomes 8), but the actual thing being measured stays identical.

Note

This is the most important idea to hold onto: conversion changes the number, not the quantity. If you ever feel like the quantity has changed, you've made a mistake.

The Precise Statement

Unit conversion is the multiplication of a quantity by a conversion factor — a fraction equal to 1 — that cancels the old unit and introduces the new one.

A conversion factor looks like this:

\frac{\text{new unit}}{\text{old unit}} = 1

For length: $\frac{1 \text{ m}}{100 \text{ cm}} = 1$ and $\frac{100 \text{ cm}}{1 \text{ m}} = 1$ .

Why are these fractions equal to 1? Because 1 metre is 100 centimetres. The numerator and denominator describe the same physical length, so their ratio is exactly 1.

How to Convert: The Only Rule You Need

Multiply by a conversion factor that cancels the unit you have and leaves the unit you want.

Let's convert 120 cm to metres:

Start with what you have: $120 \text{ cm}$
Choose the conversion factor that has "cm" in the denominator (to cancel it) and "m" in the numerator: $\frac{1 \text{ m}}{100 \text{ cm}}$
Multiply:

120 \text{ cm} \times \frac{1 \text{ m}}{100 \text{ cm}} = \frac{120}{100} \text{ m} = 1.2 \text{ m}

The "cm" units cancel just like numbers do: $\frac{\text{cm}}{\text{cm}} = 1$ .

Tip

Always write the units explicitly. If the units don't cancel correctly, you've used the wrong conversion factor. This catches 90% of conversion mistakes.

The Reverse: Metres to Centimetres

Now convert 1.2 m to cm. This time, you want "cm" to remain and "m" to cancel. Use $\frac{100 \text{ cm}}{1 \text{ m}}$ :

1.2 \text{ m} \times \frac{100 \text{ cm}}{1 \text{ m}} = 1.2 \times 100 \text{ cm} = 120 \text{ cm}

Notice: when going from a larger unit (m) to a smaller unit (cm), the number gets larger (1.2 → 120). When going from smaller to larger, the number gets smaller (120 → 1.2). This is a useful sanity check.

Common Conversion Factors You'll Use

Quantity	Relationship	Conversion Factors
Length	1 m = 100 cm	$\frac{1 \text{ m}}{100 \text{ cm}}$ , $\frac{100 \text{ cm}}{1 \text{ m}}$
Mass	1 kg = 1000 g	$\frac{1 \text{ kg}}{1000 \text{ g}}$ , $\frac{1000 \text{ g}}{1 \text{ kg}}$
Time	1 h = 60 min	$\frac{1 \text{ h}}{60 \text{ min}}$ , $\frac{60 \text{ min}}{1 \text{ h}}$
Speed	1 km/h = $\frac{1000}{3600}$ m/s	$\frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ h}}{3600 \text{ s}}$

Why this formula?

Dimensional Analysis: Why the Key Principles Hold

Dimensional Analysis is a powerful tool in physics and engineering that lets us check the consistency of equations, derive relationships, and convert units. But why does it work? Let's build the reasoning from the ground up.

1. The Core Idea: Physical Quantities Have Dimensions

Every physical quantity (like length, time, mass) can be expressed in terms of fundamental dimensions. The most common set in mechanics is:

L = Length
M = Mass
T = Time

For example:

Speed has dimensions $[L T^{-1}]$
Force has dimensions $[M L T^{-2}]$
Energy has dimensions $[M L^2 T^{-2}]$

Why this matters: Two quantities can only be meaningfully compared or equated if they have the same dimensions. You cannot add apples to oranges — and you cannot add length to time.

2. The Principle of Dimensional Homogeneity

The key formula that underpins everything is:

Every valid physical equation must be dimensionally homogeneous.

This means: the dimensions on the left-hand side must equal the dimensions on the right-hand side.

Why must this hold?

Consider an equation like:

v = u + at

Left side: $[v] = L T^{-1}$
Right side: $[u] = L T^{-1}$ , $[at] = (L T^{-2})(T) = L T^{-1}$

Both sides have dimensions $L T^{-1}$ . If they didn't match, the equation would be physically meaningless — you'd be comparing quantities that cannot be equal in any real experiment.

Reasoning: Physical laws describe relationships between measurable quantities. If the dimensions don't match, the equation cannot represent a real physical relationship, because the numerical value would depend on the arbitrary choice of units.

3. The Buckingham Pi Theorem: Why We Can Derive Relationships

This is the deeper mathematical reason. The Buckingham Pi Theorem states:

If a physical problem involves $n$ variables and $k$ fundamental dimensions, then it can be reduced to $n-k$ independent dimensionless groups (called $\pi$ groups).

Why does this work?

Imagine you have a relationship:

f(Q_1, Q_2, \dots, Q_n) = 0

where each $Q_i$ has dimensions. Because the equation must be dimensionally homogeneous, we can rearrange it into a function of dimensionless products only:

F(\pi_1, \pi_2, \dots, \pi_{n-k}) = 0

The reasoning: Dimensions act as constraints. Each fundamental dimension (M, L, T) gives one constraint. So if you have $n$ variables and $k$ constraints, you only have $n-k$ independent dimensionless combinations.

Example: For a simple pendulum, the period $T$ depends on length $L$ , mass $m$ , and gravity $g$ . That's 4 variables with 3 dimensions (M, L, T). So $4-3 = 1$ dimensionless group: $\pi = \frac{T^2 g}{L}$ . This tells us $T \propto \sqrt{L/g}$ without solving any differential equation.

4. Why We Can Convert Units Using Dimensional Analysis

The conversion factor formula:

\text{Value in new unit} = \text{Value in old unit} \times \left(\frac{\text{old unit}}{\text{new unit}}\right)^{\text{dimension exponent}}

Why this works: …

Start your 14-day free trial to unlock the full solution →

To show that $P$ is dimensionless, we determine the fundamental dimensions (Mass, Length, Time) for each variable ( $E, l, m, G$ ) and substitute them into the expression. After combining the powers of each fundamental dimension, we find that all powers become zero, proving that $P$ is indeed dimensionless.

Understanding the dimensions of physical quantities is crucial in physics. A quantity is said to be dimensionless if it has no physical units associated with it, meaning its dimensions in terms of fundamental quantities like Mass (M), Length (L), and Time (T) are $M^0L^0T^0$ . Dimensional analysis is a powerful tool to check the consistency of equations, derive relationships between physical quantities, and understand the nature of physical constants.

In this problem, we are given an expression $P = E l^2 m^{-5} G^{-2}$ and asked to show it is dimensionless. This means we need to find the dimensions of each term ( $E, l, m, G$ ) and then combine them according to the given formula. If the final expression has all fundamental dimensions raised to the power of zero, then $P$ is dimensionless.

Here's how we approach this:

Identify the fundamental dimensions:

The fundamental physical quantities we typically use for dimensional analysis are Mass (M), Length (L), and Time (T). All other physical quantities can be expressed in terms of these fundamental dimensions.
Determine the dimensions of each variable:

We need to find the dimensional formula for energy ( $E$ ), angular momentum ( $l$ ), mass ( $m$ ), and the gravitational constant ( $G$ ).
- Mass ( $m$ ):
  
  Mass is a fundamental quantity.
  
  Dimensions of $m = [M]$
- Energy ( $E$ ):
  
  Energy can be defined as the capacity to do work. Work done is Force $\times$ distance.
  
  Force ( $F$ ) = mass $\times$ acceleration = $[M] \times [LT^{-2}] = [MLT^{-2}]$
  
  Energy ( $E$ ) = Force $\times$ distance = $[MLT^{-2}] \times [L] = [ML^2T^{-2}]$
- Angular Momentum ( $l$ ):
  
  Angular momentum ( $l$ ) is given by the product of position vector and linear momentum ( $l = r \times p = r \times mv$ ).
  
  Position ( $r$ ) = $[L]$
  
  Mass ( $m$ ) = $[M]$
  
  Velocity ( $v$ ) = $[LT^{-1}]$
  
  Dimensions of $l = [L] \times [M] \times [LT^{-1}] = [ML^2T^{-1}]$
- Gravitational Constant ( $G$ ):
  
  From Newton's Law of Universal Gravitation, the force between two masses $m_1$ and $m_2$ separated by a distance $r$ is $F = \frac{Gm_1m_2}{r^2}$ .
  
  We can rearrange this to find $G$ : $G = \frac{Fr^2}{m_1m_2}$ .
  
  Dimensions of $F = [MLT^{-2}]$
  
  Dimensions of $r^2 = [L^2]$
  
  Dimensions of $m_1m_2 = [M^2]$
  
  Dimensions of $G = \frac{[MLT^{-2}] \times [L^2]}{[M^2]} = [M^{1-2}L^{1+2}T^{-2}] = [M^{-1}L^3T^{-2}]$
Tip

When determining dimensions, always start from a known formula involving the quantity and break it down into fundamental dimensions. For constants like $G$ , rearrange the formula to isolate the constant. …

Start your 14-day free trial to unlock the full solution →

🔍 Step 1: Recall the dimensions of each quantity

First, write the dimensions correctly:

Energy $E$ : $[M L^2 T^{-2}]$
Mass $m$ : $[M]$
Angular momentum $l$ : $[M L^2 T^{-1}]$
Gravitational constant $G$ : $[M^{-1} L^3 T^{-2}]$

Common Mistake #1: Wrong dimensions for angular momentum or $G$ .

✓ How to avoid: Memorise the standard formulas:

Angular momentum $l = m v r$ → $[M] \cdot [L T^{-1}] \cdot [L] = [M L^2 T^{-1}]$

$G$ from $F = \frac{G m_1 m_2}{r^2}$ → $[G] = \frac{[F][r^2]}{[m^2]} = \frac{[M L T^{-2}][L^2]}{[M^2]} = [M^{-1} L^3 T^{-2}]$

🔢 Step 2: Substitute into the expression

We have:

P = E \, l^2 \, m^{-5} \, G^{-2}

So the dimension of $P$ is:

[P] = [E] \cdot [l]^2 \cdot [m]^{-5} \cdot [G]^{-2}

Substitute:

[P] = [M L^2 T^{-2}] \cdot [M L^2 T^{-1}]^2 \cdot [M]^{-5} \cdot [M^{-1} L^3 T^{-2}]^{-2}

⚠️ Common Mistake #2: Errors in exponent arithmetic

Students often mess up the powers — especially when squaring $[l]$ or handling the negative exponent on $G$ .

✓ How to avoid: Do each dimension separately — mass, length, time — step by step.

For mass $M$ :

From $E$ : $+1$
From $l^2$ : $+2$
From $m^{-5}$ : $-5$
From $G^{-2}$ : $(-1) \times (-2) = +2$

Total: $1 + 2 - 5 + 2 = 0$

For length $L$ :

From $E$ : $+2$
From $l^2$ : $+4$
From $G^{-2}$ : $3 \times (-2) = -6$

Total: $2 + 4 - 6 = 0$

For time $T$ :

From $E$ : $-2$
From $l^2$ : $-2$
From $G^{-2}$ : $(-2) \times (-2) = +4$

Total: $-2 - 2 + 4 = 0$

✓ Final result …

Start your 14-day free trial to unlock the full solution →

←Previous questionQ38 — A physical quantity X is related to four measurable quantities a, b, c…→Next questionQ40 — If velocity of light c, Planck's constant h and gravitational constant…

Q.In the expression $P = E l^2 m^{-5} G^{-2}$ , $E$ , $m$ , $l$ and $G$ denote energy, mass, angular momentum and gravitational constant, respectively. Show that $P$ is a dimensionless quantity.

Concept understanding — Unit Conversion

Unit Conversion: Why 1 Metre and 100 Centimetres Are the Same Thing

The Intuition: Same Quantity, Different Labels

The Precise Statement

How to Convert: The Only Rule You Need

The Reverse: Metres to Centimetres

Common Conversion Factors You'll Use

Why this formula?

Dimensional Analysis: Why the Key Principles Hold

1. The Core Idea: Physical Quantities Have Dimensions

2. The Principle of Dimensional Homogeneity

Why must this hold?

3. The Buckingham Pi Theorem: Why We Can Derive Relationships

Why does this work?

4. Why We Can Convert Units Using Dimensional Analysis

Why this works: …

Method: Dimensional Analysis

Step 1: Write the dimensions of each given quantity

Step 2: Substitute into the expression for $P$

Step 3: Simplify powers of each dimension

🔍 Step 1: Recall the dimensions of each quantity

🔢 Step 2: Substitute into the expression

⚠️ Common Mistake #2: Errors in exponent arithmetic

For mass $M$ :

For length $L$ :

For time $T$ :

✓ Final result …

Q.In the expression P=El2m−5G−2P = E l^2 m^{-5} G^{-2}P=El2m−5G−2, EEE, mmm, lll and GGG denote energy, mass, angular momentum and gravitational constant, respectively. Show that PPP is a dimensionless quantity.

🎓Unlock everything free for 14 days

Q.In the expression $P = E l^2 m^{-5} G^{-2}$ , $E$ , $m$ , $l$ and $G$ denote energy, mass, angular momentum and gravitational constant, respectively. Show that $P$ is a dimensionless quantity.

Unlock everything free for 14 days