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}

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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: …

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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. …

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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 …

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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 39 on this concept.

CBSE 2026Set ANNUAL1 mark
$Q. 1 Joule is equal to how many ergs?$
›Reveal solutionSolution

1 J = 10⁷ erg.

The joule (SI) and erg (CGS) are both units of energy/work, defined as force × distance. 1 J = 1 kg·m²·s⁻², while 1 erg = 1 g·cm²·s⁻². Converting: 1 kg = 10³ g and 1 m = 10² cm, so 1 J = 1 kg …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. In SI system:$
(a) All derived units are obtained by multiplying (or) dividing the fundamental units.
(b) All derived units are obtained by adding the fundamental units.
(c) All derived units are obtained by subtracting the fundamental units.
(d) Depends on the physical quantity.
›Reveal solutionSolution

Derived SI units always come from multiplying/dividing base units, never from adding them.

Every physical quantity's dimensional formula is built by raising the base dimensions (mass $M$ , length $L$ , time $T$ , ...) to powers and combining them by multiplication/division. For example: velocity $= L/T$ , force $= MLT^{-2}$ , energy $= ML^2T^{-2}$ . Addition or subtraction is only defined between quantities of the same dimension (you cannot add a length to a time), so it can never be the rule used to build a new unit from the base units. The general construction rule for the whole SI system of derived units is therefore multiplication/division of the fundament …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The submultiple 10^-2 has the prefix :$
(a) Centi
(b) Hecto
(c) Tera
(d) Zepto
›Reveal solutionSolution

The prefix for the submultiple 10^-2 is centi.

SI prefixes are used to express very large or very small quantities as multiples or submultiples of a base unit, each prefix corresponding to a specific power of ten. Centi (symbol c) corresponds to 10^-2, hecto (h) corresponds to 10^2, tera (T) corresponds to 10^12, and zepto (z) corresponds to 10^-21. A familiar exa …

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CBSE 2026Set ANNUAL1 mark
$Q. Write the answer in one word/one sentence: Write the value of 108 km/hour in m/s.$
›Reveal solutionSolution

108 km/h converts to 30 m/s using the standard factor 5/18.

1 km/h = 1000 m / 3600 s = (5/18) m/s

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Unit of time is:$
(a) second
(b) ampere
(c) kelvin
(d) meter
›Reveal solutionSolution

The SI base unit of time is the second (s).

The International System of Units (SI) defines seven base quantities, each with its own base unit. Ampere is the unit of electric current, kelvin is the unit of thermodynamic temperature, and metre is the unit of length. None of these measure time.

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Which of the following is a derived unit ?$
(a) ampere
(b) mole
(c) kelvin
(d) joule
›Reveal solutionSolution
Ampere, mole and kelvin are base units; joule is derived. Answer (D).

The seven SI base units are: metre, kilogram, second, ampere, kelvin, mole and candela.

ampere, mole, kelvin -> base units. …
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CBSE 2026Set ANNUAL1 markMCQ
$Q. The prefix used for the multiple 10^-6 is$
(a) a) macro
(b) b) micro
(c) c) nano
(d) d) milli
›Reveal solutionSolution
[!TLDR]

b) micro

Why

The SI prefix for the multiplying factor 10^-6 is 'micr …
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CBSE 2025Set ANNUAL1 markMCQ
$Q. SI unit of energy joule is equivalent to (A) 10^6 erg (B) 10^-7 erg (C) 10^7 erg (D) 10^5 erg$
›Reveal solutionSolution

1 joule equals 10^7 erg, found by converting mass and length between SI (kg, m) and CGS (g, cm) units.

Energy has dimensional formula $[ML^2T^{-2}]$ . In SI, mass is measured in kg and length in m; in CGS, mass is in g and length in cm.

Conversion factors:

$1\text{ kg} = 10^3\text{ g}$

$1\text{ m} = 10^2\text{ cm} \Rightarrow 1\text{ m}^2 = 10^4\text{ cm}^2$

So: …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. How many scientific fundamental quantities are given in SI units?$
(a) 5
(b) 7
(c) 3
(d) 9
›Reveal solutionSolution
The SI system recognises 7 base physical quantities; every other unit (like N, J, Pa, C) is derived from these.

The International System of Units (SI) is built on a small set of base quantities that are chosen to be mutually independent — no one can be expressed in terms of the others. NCERT Class 11 Chemistry (Unit 1) lists these seven base quantities and their SI units:

Length — metre (m)

Mass — kilogram (kg)

Time — second (s)

Electric current — ampere (A)

Thermodynamic temperature — kelvin (K)

Amount of substance — mole (mol)

Luminous intensity — candela (cd) …
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CBSE 2025Set hz1 markMCQ
$Q. The correct relation between Light year and metre is:$
(a) 1 Light year = 7.469 x 10^15 m
(b) 1 Light year = 4.2 m
(c) 1 Light year = 9.467 x 10^15 m
(d) None of them
›Reveal solutionSolution

1 light year = speed of light x time in 1 year approx 9.467 x 10^15 m.

A light year is defined as the distance travelled by light in vacuum in one year. To compute it:

Speed of light, c = 3 x 10^8 m/s

1 year = 365.25 days x 24 hours x 3600 seconds approx 3.156 x 10^7 s

Distance = c x t = (3 x 10^8 m/s) x (3.156 x 10^7 s) approx 9.467 x 10^15 m

…

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$Q. Fill in the blank: The volume of a cube of side 2 cm is equal to ......... m^3.$
›Reveal solutionSolution

The volume of a 2 cm side cube is 8 cm^3, which equals 8 x 10^-6 m^3.

Side of cube, a = 2 cm.

Volume, V = a^3 = (2 cm)^3 = 8 cm^3. …

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CBSE 2025Set ANNUAL1 mark
$Q. Fill in the blank: G = 6.67 x 10^-11 Nm^2 kg^-2 = ......... cm^3 s^-2 g^-1.$
›Reveal solutionSolution

Converting G = 6.67 x 10^-11 Nm^2 kg^-2 into cgs units gives 6.67 x 10^-8 cm^3 s^-2 g^-1.

First express G in base SI units: since N = kg m s^-2, Nm^2 kg^-2 = (kg m s^-2)(m^2) kg^-2 = m^3 kg^-1 s^-2.

So G = 6.67 x 10^-11 m^3 kg^-1 s^-2.

Now convert m to cm: 1 m^3 = (10^2 cm)^3 = 10^6 cm^3.

Convert kg^-1 to g^-1: 1 kg^-1 = (10^-3)^-1 g^-1... more directly, since 1 kg = 10^3 g, 1 kg^-1 = 10^-3 g^-1. …

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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

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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 $^2$ )

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