Young's modulus of steel is $1.9 \times 10^{11}$ N/m$^2$. When expressed in CGS units of dynes/cm$^2$, it will be equal to ($1$ N $= 10^5$ dyne, $1$ m$^2 = 10^4$ cm$^2$) (a) $1.9 \times 10^{10}$ (b) $1.9 \times 10^{11}$ (c) $1.9 \times 10^{12}$ (d) $1.9 \times 10^{13}$

> [!TLDR] > Unit conversion is just multiplying by the appropriate conversion factors, each equal to 1. Young's modulus in CGS is $1.9 \times 10^{12}$ dyne/cm², so the correct option is **(C)**. ## The core idea: why this works Unit conversion is not magic — it’s just multiplying by 1. Every conversion factor like $1\ \text{N} = 10^5\ \text{dyne}$ can be written as a fraction equal to 1: either $\frac{1\ \text{N}}{10^5\ \text{dyne}}$ or $\frac{10^5\ \text{dyne}}{1\ \text{N}}$. You pick the orientation that cancels the old unit and leaves the new one. Same for area: $1\ \text{m}^2 = 10^4\ \text{cm}^2$. The trick is to treat the units as algebraic quantities — they multiply, divide, and cancel just like numbers. ## Step-by-step 1. **Write down what you have.** Young’s modulus $Y = 1.9 \times 10^{11}\ \text{N/m}^2$. We want the same physical quantity in $\text{dyne/cm}^2$. 2. **Convert newtons to dynes.** Since $1\ \text{N} = 10^5\ \text{dyne}$, the conversion factor is $\frac{10^5\ \text{dyne}}{1\ \text{N}}$. We place it so that “N” in the numerator cancels: $$1.9 \times 10^{11}\ \frac{\text{N}}{\text{m}^2} \times \frac{10^5\ \text{dyne}}{1\ \text{N}} = 1.9 \times 10^{16}\ \frac{\text{dyne}}{\text{m}^2}$$ 3. **Convert square metres to square centimetres.** $1\ \text{m}^2 = 10^4\ \text{cm}^2$, so the factor is $\frac{1\ \text{m}^2}{10^4\ \text{cm}^2}$. We want “m²” in the denominator to cancel, so: $$1.9 \times 10^{16}\ \frac{\text{dyne}}{\text{m}^2} \times \frac{1\ \text{m}^2}{10^4\ \text{cm}^2} = 1.9 \times 10^{12}\ \frac{\text{dyne}}{\text{cm}^2}$$ > [!TIP] > A faster way: combine both conversions into one step. …

Young's modulus of steel is 1.9 × 10¹¹ N/m². When expressed in CGS units of dynes/cm², it will be equal to (1

NCERT Exemplar · Q11

1.9 \times 10^{11}

(a)

1.9 \times 10^{10}

(b)

1.9 \times 10^{11}

(c)

1.9 \times 10^{12}

(d)

1.9 \times 10^{13}

Gujarat GsebMCQ· 1mImportance★★★★★est

50% · 33/66 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 →

Unit conversion is just multiplying by the appropriate conversion factors, each equal to 1. Young's modulus in CGS is $1.9 \times 10^{12}$ dyne/cm², so the correct option is (C).

The core idea: why this works

Unit conversion is not magic — it’s just multiplying by 1. Every conversion factor like $1\ \text{N} = 10^5\ \text{dyne}$ can be written as a fraction equal to 1: either $\frac{1\ \text{N}}{10^5\ \text{dyne}}$ or $\frac{10^5\ \text{dyne}}{1\ \text{N}}$ . You pick the orientation that cancels the old unit and leaves the new one. Same for area: $1\ \text{m}^2 = 10^4\ \text{cm}^2$ .

The trick is to treat the units as algebraic quantities — they multiply, divide, and cancel just like numbers.

Step-by-step

Write down what you have.

Young’s modulus $Y = 1.9 \times 10^{11}\ \text{N/m}^2$ . We want the same physical quantity in $\text{dyne/cm}^2$ .
Convert newtons to dynes.

Since $1\ \text{N} = 10^5\ \text{dyne}$ , the conversion factor is $\frac{10^5\ \text{dyne}}{1\ \text{N}}$ . We place it so that “N” in the numerator cancels:

1.9 \times 10^{11}\ \frac{\text{N}}{\text{m}^2} \times \frac{10^5\ \text{dyne}}{1\ \text{N}} = 1.9 \times 10^{16}\ \frac{\text{dyne}}{\text{m}^2}

Convert square metres to square centimetres. $1\ \text{m}^2 = 10^4\ \text{cm}^2$ , so the factor is $\frac{1\ \text{m}^2}{10^4\ \text{cm}^2}$ . We want “m²” in the denominator to cancel, so:

1.9 \times 10^{16}\ \frac{\text{dyne}}{\text{m}^2} \times \frac{1\ \text{m}^2}{10^4\ \text{cm}^2} = 1.9 \times 10^{12}\ \frac{\text{dyne}}{\text{cm}^2}

Tip

A faster way: combine both conversions into one step. …

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

Common Mistakes in Unit Conversion (N/m² → dyne/cm²)

Mistake 1: Forgetting to Square the Length Conversion

The error: Students convert $1 \text{ m} = 100 \text{ cm}$ , then treat area conversion as $1 \text{ m}^2 = 100 \text{ cm}^2$ instead of $1 \text{ m}^2 = 10^4 \text{ cm}^2$ .

Why it happens: The square applies to the unit, not just the number. Since $1 \text{ m} = 100 \text{ cm}$ , squaring both sides gives:

1 \text{ m}^2 = (100)^2 \text{ cm}^2 = 10^4 \text{ cm}^2

How to avoid: Always write the conversion factor explicitly with exponents:

$1 \text{ m}^2 = (10^2 \text{ cm})^2 = 10^4 \text{ cm}^2$
Never drop the square from the unit.

Mistake 2: Dividing Instead of Multiplying (or Vice Versa)

The error: Students set up the conversion backwards — e.g., multiplying by $10^5$ when they should divide, or using $10^{-4}$ instead of $10^4$ .

Why it happens: Confusion about whether the new unit is larger or smaller. A newton is $10^5$ dynes (larger unit → smaller number), but a square metre is $10^4$ cm² (larger unit → smaller number). The two factors work in opposite directions.

How to avoid: Use dimensional analysis step-by-step:

1.9 \times 10^{11} \frac{\text{N}}{\text{m}^2} \times \frac{10^5 \text{ dyne}}{1 \text{ N}} \times \frac{1 \text{ m}^2}{10^4 \text{ cm}^2}

N in numerator cancels with N in denominator
m² in denominator cancels with m² in numerator
Result: $1.9 \times 10^{11} \times 10^5 \times 10^{-4} = 1.9 \times 10^{12}$

Key check: The final unit should be dyne/cm² — if your setup doesn't yield that, flip the fraction.

Mistake 3: Arithmetic Error with Powers of 10

The error: Adding exponents incorrectly: $10^{11} \times 10^5 \times 10^{-4} = 10^{11+5-4} = 10^{12}$ , but students may get $10^{10}$ , $10^{11}$ , or $10^{13}$ .

Why it happens: Rushing the exponent arithmetic or misreading the given conversion factors.

How to avoid: Write the exponent sum explicitly:

11 + 5 - 4 = 12

Then verify: $10^{12}$ matches option (C).

Mistake 4: Ignoring the Given Conversion Factors …

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

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 17 on this concept.

GUJCET 2026Set x1 markMCQ
$Q. Out of the following physical quantities which quantity has the same unit as that of Planck's constant? (A) moment of force (B) power (C) angular momentum (D) moment of inertia$
›Reveal solutionSolution

$[h]=\text{J·s}$ , identical to the unit of angular momentum.

Planck's constant appears in $E=h\nu$ , so $[h]=\dfrac{[E]}{[\nu]}=\dfrac{\text{J}}{\text{s}^{-1}}=\text{J·s}=\text{kg·m}^2\text{s}^{-1}$ .

Angular momentum $L=mvr$ has units $\text{kg}\cdot(\text{m/s})\cdot\text{m}=\text{kg·m}^2\text{s}^{-1}=\text{J·s}$ . …
Start your 14-day free trial to see the full solution →
GUJCET 2025Set 031 markMCQ
$[L^{2}]$
›Reveal solutionSolution
[!TLDR]

Current sensitivity $=\theta/I$ ; since $\theta$ is dimensionless, its dimension is $[A^{-1}]$ .

Concept

The current sensitivity of a moving-coil galvanometer is the deflection produced per unit current, $S_I=\dfrac{\theta}{I}=\dfrac{NBA}{k}$ . An angular deflection is dimensionless.

Solution …
Start your 14-day free trial to see the full solution →
GUJCET 2024Set 131 markMCQ
$Q. Js is the unit of ________ physical quantity. (A) Angular momentum (B) Work function (C) Moment of Inertia (D) Rydberg constant$
›Reveal solutionSolution

Angular momentum $L=I\omega$ has SI unit $\text{kg m}^2\text{s}^{-1} = \text{J·s}$ . …

Start your 14-day free trial to see the full solution →
GUJCET 2023Set 091 markMCQ
$kg^{-1}s^{-2}A$
›Reveal solutionSolution
[!TLDR]

Reducing $\text{m}^2\text{V}^{-1}\text{s}^{-1}$ to fundamental units gives $\text{kg}^{-1}\text{s}^2\text{A}$ .

Concept

Mobility $\mu=\dfrac{v_d}{E}$ has units of $\dfrac{\text{m/s}}{\text{V/m}}=\text{m}^2\text{V}^{-1}\text{s}^{-1}$ . Convert the volt to base units to express $\mu$ in fundamental units.

Solution
$\text{V}=\frac{\text{J}}{\text{C}}=\frac{\text{kg}\,\text{m}^2\text{s}^{-2}}{\text{A}\,\text{s}}=\text{kg}\,\text{m}^2\text{A}^{-1}\text{s}^{-3}.$
Then …
Start your 14-day free trial to see the full solution →
GUJCET 2023Set 091 markMCQ
$M^1L^1T^{-2}A^{-2}$
›Reveal solutionSolution
[!TLDR] $[L]=M^1L^2T^{-2}A^{-2}$ → (B).

Concept

Self-inductance is defined by $\varepsilon=-L\dfrac{dI}{dt}$ , so $L=\dfrac{\varepsilon\,dt}{dI}$ . (NCERT Electromagnetic Induction.)

Solution …
Start your 14-day free trial to see the full solution →
GUJCET 2023Set 091 markMCQ
$Q. The earth takes 24 h to rotate once about its axis. How much time does the Sun takes to shift by 1 minute viewed from the earth. (A) 4 minutes (B) 40 s (C) 4 s (D) 40 minutes$
›Reveal solutionSolution
[!TLDR]

The Sun moves 1° in 4 minutes, so 1 arcminute of apparent shift takes 4 seconds.

Concept

Earth's rotation makes the Sun appear to sweep $360^\circ$ in 24 hours; a small angular shift corresponds to a proportional time.

Solution

Angular rate of the Sun's apparent motion:

$\frac{360^\circ}{24\ \text{h}} = 15^\circ/\text{h} = \frac{1^\circ}{4\ \text{min}}$ …
Start your 14-day free trial to see the full solution →
GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQ
$Q. The dimensional formula of electric flux is ___.$
(a) M^1 L^-3 T^-3 A^-1
(b) M^1 L^3 T^-3 A^-1
(c) M^-1 L^3 T^-3 A^-1
(d) M^1 L^3 T^3 A^-1
›Reveal solutionSolution

Flux = E x area, i.e. (V/m) x m^2 = volt-metre; expressing volt in base units gives M L^3 T^-3 A^-1.

Electric flux Phi = E x A, with units (V/m)(m^2) = V.m.

…

Start your 14-day free trial to see the full solution →
GUJCET 2021Set 151 markMCQ
$[M^{-1} L^{2} T^{-3} A^{1}]$
›Reveal solutionSolution

Electric potential is energy per unit charge.

Concept: …
Start your 14-day free trial to see the full solution →
GUJCET 2020Set 071 markMCQ
$1^\circ$
›Reveal solutionSolution

$24$ hours per $360^\circ$ gives $4$ minutes per degree.

Concept: Earth turns $360^\circ$ in $24$ h, so the angular rate is $360^\circ/24\,\text{h}=15^\circ/\text{h}$ . …
Start your 14-day free trial to see the full solution →
GUJCET 2020Set 071 markMCQ
$V\cdot s\cdot A^{-1}$
›Reveal solutionSolution

$1\,\text{H}=\text{Wb/A}=\text{V·s/A}$ ; $\text{Wb·s·A}^{-1}$ has an extra second and is not inductance. …

Start your 14-day free trial to see the full solution →
GUJCET 2019Set 131 markMCQ
$M^0L^0T^0$
›Reveal solutionSolution

Torsional constant has the dimensions of torque, $M^1L^2T^{-2}$ .

Concept: For a torsion spring, restoring torque $\tau = C\theta$ , so $C = \tau/\theta$ . Angle $\theta$ is dimensionless, hence $C$ has the dimensions of torque (energy).

Steps: …
Start your 14-day free trial to see the full solution →
GUJCET 2019Set 131 markMCQ
$M^1L^2T^1Q^{-2}$
›Reveal solutionSolution
The dimensional formula asked for is that of resistance (equivalently inductive reactance $\omega L$ ): $M^1L^2T^{-1}Q^{-2}$ .

Concept: Resistance $R=\dfrac{V}{I}$ . Expressing potential and current in terms of charge $Q$ : $V=\dfrac{\text{energy}}{Q}=\dfrac{ML^2T^{-2}}{Q}$ and $I=\dfrac{Q}{T}$ .

Steps:

$[R]=\dfrac{[V]}{[I]}=\dfrac{ML^2T^{-2}Q^{-1}}{Q\,T^{-1}}$ . …
Start your 14-day free trial to see the full solution →

←Previous questionQ10 — The mean length of an object is 5 cm. Which of the following measureme…→Next questionQ12 — If momentum (P), area (A) and time (T) are taken to be fundamental qua…

Q.Young's modulus of steel is 1.9×10111.9 \times 10^{11}1.9×1011 N/m2^22. When expressed in CGS units of dynes/cm2^22, it will be equal to (111 N =105= 10^5=105 dyne, 111 m2=104^2 = 10^42=104 cm2^22)

The core idea: why this works

Step-by-step

🎓Unlock everything free for 14 days

Q.Young's modulus of steel is $1.9 \times 10^{11}$ N/m $^2$ . When expressed in CGS units of dynes/cm $^2$ , it will be equal to ( $1$ N $= 10^5$ dyne, $1$ m $^2 = 10^4$ cm $^2$ )

Unlock everything free for 14 days