(a) How many astronomical units (A.U.) make 1 parsec? (b) Consider a sunlike star at a distance of 2 parsecs.

NCERT Exemplar · Q43

Q. (a) How many astronomical units (A.U.) make 1 parsec?

(b) Consider a sunlike star at a distance of 2 parsecs. When it is seen through a telescope with 100 magnification, what should be the angular size of the star? Sun appears to be

(1/2)^\circ

from the earth. Due to atmospheric fluctuations, eye can't resolve objects smaller than 1 arc minute.

(c) Mars has approximately half of the earth's diameter. When it is closest to the earth it is at about 1/2 A.U. from the earth. Calculate what size it will appear when seen through the same telescope. (Comment: This is to illustrate why a telescope can magnify planets but not stars.)

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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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(a) $1\ \text{parsec} \approx 206265\ \text{A.U.}$ (b) A sun-like star at 2 parsecs, even magnified 100×, still appears as an unresolved point ( $0.436''$ , far below the eye's $1$ -arcmin resolution). (c) Mars, magnified 100×, appears $\approx 29.3$ arc minutes across — large enough to be resolved as a disk.

(a) Astronomical units in one parsec

A parsec is defined as the distance at which 1 A.U. subtends an angle of exactly $1''$ (one arcsecond). Using the small-angle relation $\theta = D/d$ :

1\ \text{pc} = \frac{1\ \text{A.U.}}{1''\ \text{(in radians)}}

Converting $1''$ to radians: $1'' = \dfrac{1}{3600}^\circ = \dfrac{\pi}{648000}\ \text{rad}$ .

1\ \text{pc} = \frac{648000}{\pi}\ \text{A.U.} \approx 206265\ \text{A.U.}

(b) Angular size of a sun-like star at 2 parsecs

The Sun's angular diameter from Earth is $\theta_{\odot} = (1/2)^\circ$ at a distance of 1 A.U. — so a sun-like star has the same actual diameter as the Sun. At $2\ \text{pc} = 2 \times 206265\ \text{A.U.} = 412530\ \text{A.U.}$ , its angular size shrinks with distance:

\theta_{\text{star}} = \theta_{\odot} \times \frac{1\ \text{A.U.}}{412530\ \text{A.U.}} = 0.5^\circ \times \frac{1}{412530} \approx 1.21\times10^{-6\circ} \approx 0.00436''

Magnified $100\times$ by the telescope: $100 \times 0.00436'' = 0.436''$ .

Since the eye cannot resolve anything smaller than $1' = 60''$ , and $0.436'' \ll 60''$ , the star remains an unresolved point of light even at 100× magnification.

(c) Angular size of Mars at closest approach

Mars's diameter is about half the Earth's: taking $D_{\text{Earth}} \approx 1.274\times10^4\ \text{km}$ , $D_{\text{Mars}} \approx 6.37\times10^3\ \text{km}$ . Mars is at $d = \tfrac12\ \text{A.U.} = 7.48\times10^7\ \text{km}$ at closest approach. …

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Concept: Angular Size, Parallax, and Astronomical Distance Scales

The core idea is that angular size depends on both the actual physical size of an object and its distance from the observer. For stars, their immense distance makes them point-like even through telescopes, while planets (being closer and larger) show a measurable disk.

(a) How many A.U. make 1 parsec?

Method: Definition of Parsec (Parallax of 1 arcsecond)

Steps:

Recall the definition: 1 parsec is the distance at which 1 Astronomical Unit (A.U.) subtends an angle of 1 arcsecond ( $1''$ ).
Use the small-angle formula: For a small angle $\theta$ in radians:

\theta (\text{radians}) = \frac{\text{baseline}}{\text{distance}}

Here, baseline = 1 A.U., distance = 1 parsec, and $\theta = 1''$ .

Convert 1 arcsecond to radians:

1'' = \frac{1}{3600} \text{ degrees} = \frac{1}{3600} \times \frac{\pi}{180} \text{ radians}

1'' = \frac{\pi}{648000} \text{ radians} \approx 4.848 \times 10^{-6} \text{ rad}

Apply the formula:

\frac{\pi}{648000} = \frac{1 \text{ A.U.}}{1 \text{ parsec}}

\Rightarrow 1 \text{ parsec} = \frac{648000}{\pi} \text{ A.U.}

Calculate:

1 \text{ parsec} \approx \frac{648000}{3.14159} \approx 206265 \text{ A.U.}

Final Answer:

1 parsec = $206265$ A.U. (or exactly $\frac{648000}{\pi}$ A.U.)

(b) Angular size of a sunlike star at 2 parsecs through 100× telescope

Method: Small-angle formula + Magnification effect

Steps:

Find the star's actual angular size (without telescope): Sun's distance = 1 A.U. = $\frac{1}{206265}$ parsec. Sun's angular diameter from Earth = $0.5^\circ = 30$ arcminutes = $1800$ arcseconds. Angular size $\propto \frac{1}{\text{distance}}$ , so at 2 parsecs:

\theta_{\text{star}} = 1800'' \times \frac{1/206265}{2} = 1800'' \times \frac{1}{412530}

\theta_{\text{star}} \approx 0.00436'' \text{ (arcseconds)}

Apply telescope magnification: Magnification increases angular size:

\theta_{\text{through telescope}} = 100 \times 0.00436'' = 0.436''

Compare with atmospheric limit: Eye cannot resolve objects smaller than 1 arcminute = 60 arcseconds. Since $0.436'' \ll 60''$ , the star remains a point — no disk is visible.

Final Answer:

Angular size = $0.436$ arcseconds — still far below the 1 arcminute resolution limit.

(c) Angular size of Mars through the same telescope

Method: Angular diameter formula + Magnification

Steps:

Given data:
- Earth's diameter $\approx 12742$ km
- Mars diameter = half of Earth's = $6371$ km
- Distance at closest approach = $0.5$ A.U.
- 1 A.U. $\approx 1.496 \times 10^8$ km
Calculate Mars' angular size (without telescope):

$\theta_{\text{Mars}} = \frac{\text{diameter}}{\text{distance}} \text{ (in radians)}$ …

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Here are the common mistakes students make on this problem, broken down by part, with the concept-first reasoning and how to avoid each.

Part (a): The Parsec Definition

Common Mistake 1: Confusing the formula for 1 parsec.

Students often write $1 \text{ parsec} = \frac{1 \text{ A.U.}}{\theta}$ without checking the units of $\theta$ , or they use $\theta = 1^\circ$ instead of $\theta = 1 \text{ arcsecond}$ .

Why this happens: The definition of a parsec is based on parallax. One parsec is the distance at which 1 A.U. subtends an angle of 1 arcsecond. Students forget that the angle must be in radians for the small-angle formula $D = \frac{d}{\theta}$ to work directly.

How to avoid:

Always convert the angle to radians before using $D = \frac{d}{\theta}$ .
$1 \text{ arcsecond} = \frac{1}{3600} \text{ degrees} = \frac{1}{3600} \times \frac{\pi}{180} \text{ radians}$ .
Then:

1 \text{ parsec} = \frac{1 \text{ A.U.}}{1 \text{ arcsec in radians}} = \frac{1}{\frac{1}{3600} \times \frac{\pi}{180}} \text{ A.U.} = \frac{3600 \times 180}{\pi} \text{ A.U.} \approx 206265 \text{ A.U.}

Key result: $1 \text{ parsec} \approx 2.06 \times 10^5 \text{ A.U.}$

Part (b): Angular Size of a Star Through a Telescope

Common Mistake 2: Using the magnification formula incorrectly.

Students often think magnification multiplies the actual angular size of the star. They calculate the star's angular size at 2 parsecs, then multiply by 100.

Why this happens: They confuse angular magnification (which applies to extended objects like planets) with the fact that stars are point sources. A star's angular size is smaller than the telescope's diffraction limit or the atmosphere's blurring limit. Magnification does not make a point source appear larger — it just makes the blur spot bigger.

How to avoid:

First, calculate the star's actual angular size. The Sun's angular diameter from Earth is $(1/2)^\circ$ . At 2 parsecs, the star is much farther.
Distance to Sun: $1 \text{ A.U.}$ (by definition). Distance to star: $2 \text{ parsecs} = 2 \times 2.06 \times 10^5 \text{ A.U.} = 4.12 \times 10^5 \text{ A.U.}$
Using the small-angle formula (angular size $\propto 1/\text{distance}$ ):

\text{Angular size of star} = \frac{1/2^\circ}{4.12 \times 10^5} \approx 1.21 \times 10^{-6} \text{ degrees}

Convert to arcminutes: $1^\circ = 60 \text{ arcminutes}$ , so:

\text{Angular size} \approx 1.21 \times 10^{-6} \times 60 \approx 7.3 \times 10^{-5} \text{ arcminutes}

Atmospheric limit: Eye cannot resolve objects smaller than 1 arcminute. The star's actual angular size is far smaller than this limit.
Conclusion: Even with 100× magnification, the star's image is still a point (blurred by atmosphere). Magnification does not help resolve a point source.

Part (c): Apparent Size of Mars Through the Telescope

Common Mistake 3: Forgetting to convert units or using wrong distance.

Students sometimes use 1/2 A.U. as the distance but forget to convert the Earth-Mars distance into the same units as the planet's diameter.

Why this happens: They mix A.U. and kilometers without converting, or they use the Earth-Sun distance instead of the Earth-Mars distance.

How to avoid:

Step 1: Get the actual angular size of Mars without magnification.
- Mars diameter $\approx$ half of Earth's diameter $\approx 0.5 \times 12742 \text{ km} \approx 6371 \text{ km}$ .
- Distance from Earth when closest: $0.5 \text{ A.U.} = 0.5 \times 1.496 \times 10^8 \text{ km} \approx 7.48 \times 10^7 \text{ km}$ .
- Angular size (in radians) = $\frac{\text{diameter}}{\text{distance}} = \frac{6371}{7.48 \times 10^7} \approx 8.52 \times 10^{-5} \text{ radians}$ .
- Convert to arcminutes: $1 \text{ radian} = 3437.75 \text{ arcminutes}$ , so:

\text{Angular size} \approx 8.52 \times 10^{-5} \times 3437.75 \approx 0.293 \text{ arcminutes}

Step 2: Apply magnification. For an extended object, angular magnification multiplies the angular size: …

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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 37 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 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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CBSE 2025Set ANNUAL1 mark
$Q. Fill in the blank: The volume of a cube of side 2 cm is equal to ......... m^3.$
›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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CBSE 2025Set ANNUAL1 mark
$Q. Answer in one word or one sentence: How many metre is equivalent to 1 Fermi?$
›Reveal solutionSolution

1 Fermi equals 10^-15 metre, the typical scale of atomic nuclei.

The fermi (symbol fm), also called the femtometre, is a unit of length used to express nuclear and sub-nuclear distances, since nuclear radii are of …

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Q.(a) How many astronomical units (A.U.) make 1 parsec?

(a) Astronomical units in one parsec

(b) Angular size of a sun-like star at 2 parsecs

(c) Angular size of Mars at closest approach

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