(a) The earth-moon distance is about 60 earth radius. What will be the diameter of the earth (approximately in

NCERT Exemplar · Q25

Q. (a) The earth-moon distance is about 60 earth radius. What will be the diameter of the earth (approximately in degrees) as seen from the moon?

(b) Moon is seen to be of

(\frac{1}{2})^\circ

diameter from the earth. What must be the relative size compared to the earth?

(c) From parallax measurement, the sun is found to be at a distance of about 400 times the earth-moon distance. Estimate the ratio of sun-earth diameters.

CBSEShort· 5mImportance★★★★★est

71% · 47/66 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 — Atomic Volume Calculation

Atomic Volume: Meaning and Calculation

Atoms are mostly empty space — a tiny dense nucleus wrapped in a fuzzy electron cloud. So "atomic volume" does not mean the volume of a solid ball; it means the average space one atom occupies when many atoms are packed together in a solid or liquid.

Think of a crowded hall: to find the space per person you divide the hall's volume by the number of people. Atomic volume does exactly that for atoms.

Definition

Atomic volume is the volume occupied by one mole of atoms of an element in its solid or liquid state. It is found from the element's molar mass and density:

V_{\text{atomic}} = \frac{M}{\rho}

$V_{\text{atomic}}$ = atomic volume (cm³/mol)
$M$ = molar mass (g/mol)
$\rho$ = density (g/cm³)

This gives the volume per mole. Dividing by Avogadro's number gives the space per single atom:

V_{\text{one atom}} = \frac{V_{\text{atomic}}}{N_A}, \qquad N_A = 6.022\times10^{23}\ \text{mol}^{-1}

Worked example — aluminium

Molar mass $M = 26.98$ g/mol, density $\rho = 2.70$ g/cm³:

V_{\text{atomic}} = \frac{26.98}{2.70} = 9.99\ \text{cm}^3/\text{mol}

So one mole of Al atoms occupies about 10 cm³. Per atom:

V_{\text{one Al atom}} = \frac{9.99}{6.022\times10^{23}} = 1.66\times10^{-23}\ \text{cm}^3

Periodic trends

Down a group: atomic volume increases — more electron shells make atoms larger.
Across a period: it generally decreases — rising nuclear charge pulls the electrons in tighter.
Allotropes differ: diamond is denser than graphite, so diamond's atomic volume is smaller, though both are carbon.

These trends help explain why alkali metals (large atomic volume) are soft and reactive, while transition metals (smaller atomic volume) are hard and dense. …

Why this formula?

Atomic Volume Calculation: Understanding the "Why" Behind the Formula

What Is Atomic Volume?

Atomic volume is not the volume of a single atom — it's the volume occupied by one mole of atoms of an element in its solid state. This is a macroscopic quantity that helps us understand how tightly atoms pack together.

The key formula is:

\text{Atomic Volume} = \frac{\text{Atomic Mass}}{\text{Density}}

Let's break down why this works.

The Core Reasoning: From Mass to Volume

Step 1: What does density tell us?

Density ( $\rho$ ) is defined as:

\rho = \frac{\text{Mass}}{\text{Volume}}

For a pure solid element, if we take one mole of atoms:

Mass of one mole = Atomic mass (in g/mol)
Volume of one mole = Atomic volume (in cm³/mol)

So:

\rho = \frac{\text{Atomic mass}}{\text{Atomic volume}}

Step 2: Rearranging to find atomic volume

\text{Volume} = \frac{\text{Mass}}{\text{Density}}

Therefore:

\boxed{\text{Atomic Volume} = \frac{\text{Atomic Mass}}{\text{Density}}}

Why This Makes Physical Sense

Atomic mass tells you how heavy one mole of atoms is; density tells you how much mass fits in a given space. Dividing mass by density gives the space that mass occupies.

Example intuition: If iron has atomic mass ≈ 56 g/mol and density 7.87 g/cm³, then:

\text{Atomic volume} = \frac{56}{7.87} \approx 7.1 \text{ cm}^3/\text{mol}

This means one mole of iron atoms (about $6.022 \times 10^{23}$ atoms) occupies roughly 7.1 cm³ of space.

Important Exam Points

Concept	Why It Matters
Units	Atomic mass in g/mol, density in g/cm³ → atomic volume in cm³/mol
Solid state only	The formula assumes atoms are closely packed; gases/liquids have different packing

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

Using the small-angle relation $\theta \approx D/d$ :

the Earth's angular diameter from the Moon is about $2^\circ$ .
the Moon's diameter is about $\tfrac{1}{4}$ the Earth's.
the Sun's diameter is about $100$ times the Earth's.

The tool: small-angle angular diameter

\theta \text{ (radians)} \approx \frac{D}{d}

where $D$ is the object's actual diameter and $d$ is its distance from the observer — valid whenever $D \ll d$ , true for every body in this problem.

(a) Earth's angular diameter seen from the Moon

Earth's diameter is $D_E = 2R_E$ ; the Earth–Moon distance is $d_{EM} = 60R_E$ .

\theta_E = \frac{D_E}{d_{EM}} = \frac{2R_E}{60R_E} = \frac{1}{30}\ \text{rad}

Converting to degrees ( $1\ \text{rad} = 180/\pi$ degrees):

\theta_E = \frac{1}{30} \times \frac{180^\circ}{\pi} = \frac{6^\circ}{\pi} \approx 1.9^\circ \approx 2^\circ

(b) Relative size of the Moon

The Moon's angular diameter as seen from Earth, over the same distance $d_{EM}$ , is given as $\theta_M = \left(\tfrac{1}{2}\right)^\circ$ . Since both $\theta_E$ and $\theta_M$ use the same distance $d_{EM}$ , their ratio equals the ratio of the actual diameters:

\frac{D_M}{D_E} = \frac{\theta_M}{\theta_E} = \frac{0.5^\circ}{2^\circ} = \frac{1}{4}

(c) Ratio of Sun's diameter to Earth's diameter

A total solar eclipse tells us the Sun and Moon have (almost) the same angular diameter as seen from Earth: $\theta_S \approx \theta_M$ . Since $\theta = D/d$ for both:

$\frac{D_S}{d_{SE}} \approx \frac{D_M}{d_{EM}} \implies \frac{D_S}{D_M} \approx \frac{d_{SE}}{d_{EM}} = 400$ …

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

Concept: Angular Size and Small-Angle Approximation

When an object's physical size is much smaller than its distance from the observer, the angular diameter (in radians) is given by:

\theta \ (\text{radians}) \approx \frac{\text{actual diameter}}{\text{distance}}

To convert radians to degrees: $1 \text{ radian} \approx 57.3^\circ$

(a) Method: Small-Angle Formula

Step 1: Identify given data

Distance from Moon to Earth = $60 \, R_E$ (where $R_E$ = Earth's radius)
Earth's actual diameter = $2 R_E$

Step 2: Apply the small-angle formula

\theta \ (\text{rad}) = \frac{\text{diameter}}{\text{distance}} = \frac{2 R_E}{60 R_E} = \frac{1}{30} \ \text{radians}

Step 3: Convert to degrees

\theta = \frac{1}{30} \times 57.3^\circ \approx 1.91^\circ

Answer: The Earth's diameter as seen from the Moon is approximately $2^\circ$ .

(b) Method: Ratio Using Angular Size Formula

Step 1: For Moon as seen from Earth

\theta_M = \frac{1}{2}^\circ = \frac{0.5 \times \pi}{180} \ \text{radians} \approx 0.00873 \ \text{rad}

Step 2: Write the angular size relation

\theta_M = \frac{D_M}{d_{EM}} \quad \text{and} \quad \theta_E = \frac{D_E}{d_{EM}}

where $D_M$ , $D_E$ are diameters, $d_{EM}$ is Earth-Moon distance.

Step 3: Take the ratio

\frac{D_M}{D_E} = \frac{\theta_M}{\theta_E} = \frac{0.5^\circ}{2^\circ} = \frac{1}{4}

Answer: The Moon's diameter is $\frac{1}{4}$ of Earth's diameter.

(c) Method: Combined Angular Size and Distance Ratios

Step 1: Given

Sun's distance $d_S = 400 \times d_{EM}$
From (a), Earth's angular size from Moon = $2^\circ$
From (b), Moon's angular size from Earth = $0.5^\circ$ …

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

Here are the common mistakes students make on this classic angular size and astronomical scaling problem, and how to avoid each.

Mistake 1: Confusing Diameter with Radius in the Angular Size Formula

The Mistake:

For part (a), students often plug the Earth’s radius ( $R_E$ ) into the formula for angular diameter instead of the Earth’s diameter ( $D_E = 2R_E$ ). They might write:

\theta \approx \frac{R_E}{60 R_E} = \frac{1}{60} \text{ rad}

This gives half the correct answer.

Why it happens:

The formula $\theta \approx \frac{\text{arc length}}{\text{radius}}$ is usually taught for small angles, where the "arc length" is the linear size of the object. Students forget that the linear size of a sphere seen from a distance is its diameter, not its radius.

How to Avoid:

Always define the variables clearly before plugging in.

Linear size of the object = Diameter ( $D$ ).
Distance to the object = $d$ .
For small angles (in radians): $\theta \approx \frac{D}{d}$ .

For part (a):

$D_{\text{Earth}} = 2 R_E$
$d = 60 R_E$
$\theta \approx \frac{2 R_E}{60 R_E} = \frac{1}{30} \text{ rad}$

Then convert to degrees:

\theta = \frac{1}{30} \times \frac{180^\circ}{\pi} \approx \frac{180^\circ}{30 \times 3.14} \approx 1.9^\circ

Key takeaway: The angular diameter of a sphere is determined by its full width, not its half-width.

Mistake 2: Forgetting to Convert Radians to Degrees

The Mistake:

Students stop at $\theta = \frac{1}{30}$ rad and write the answer as $\frac{1}{30}^\circ$ or just leave it in radians without converting.

Why it happens:

The problem explicitly asks for the answer "in degrees," but students often treat the radian measure as if it were already in degrees.

How to Avoid:

Always check the unit requested in the question. Use the conversion factor:

1 \text{ radian} = \frac{180^\circ}{\pi} \approx 57.3^\circ

So $\frac{1}{30}$ rad $\approx 1.9^\circ$ .

Pro tip: Memorize that $1$ rad $\approx 57^\circ$ . For quick checks, $\frac{1}{60}$ rad $\approx 1^\circ$ .

Mistake 3: Mixing Up Which Object is the "Observer" and Which is the "Observed"

The Mistake:

In part (b), students might calculate the Moon's diameter as seen from Earth, but the question asks for the relative size of the Moon compared to Earth. They might invert the ratio or use the wrong angular diameter.

Why it happens:

The problem has three parts with shifting perspectives:

(a) Earth seen from Moon.
(b) Moon seen from Earth.
(c) Sun seen from Earth.

Students lose track of which angular diameter belongs to which object.

How to Avoid:

Draw a simple diagram or write a clear statement for each part.

For part (b):

Given: Angular diameter of Moon as seen from Earth = $\theta_M = 0.5^\circ$ .
From part (a): Angular diameter of Earth as seen from Moon = $\theta_E \approx 2^\circ$ .
Distance between Earth and Moon is the same in both directions ( $d$ ).

Use the angular size formula for both:

\theta_E \approx \frac{D_E}{d}, \quad \theta_M \approx \frac{D_M}{d}

Divide the two equations:

\frac{D_M}{D_E} = \frac{\theta_M}{\theta_E} = \frac{0.5^\circ}{2^\circ} = \frac{1}{4}

Key takeaway: When distances are equal, the ratio of diameters equals the ratio of angular diameters.

Mistake 4: Using the Wrong Distance in Part (c)

The Mistake:

For part (c), students use the Earth-Moon distance ( $d_{EM}$ ) as the distance to the Sun, instead of the given "400 times the earth-moon distance."

Why it happens:

The problem states: "the sun is found to be at a distance of about 400 times the earth-moon distance." Students might misread this as "the Sun is 400 times farther than the Moon" but then forget to multiply the distance correctly in the angular size formula.

How to Avoid:

Write the given data explicitly:

$d_{\text{Sun}} = 400 \times d_{\text{Earth-Moon}}$
Angular diameter of Sun as seen from Earth = $\theta_{\text{Sun}} \approx 0.5^\circ$ (same as Moon, a famous coincidence).

Now, for the Sun:

\theta_{\text{Sun}} \approx \frac{D_{\text{Sun}}}{d_{\text{Sun}}}

For the Moon (from part b):

$\theta_{\text{Moon}} \approx \frac{D_{\text{Moon}}}{d_{\text{EM}}}$ …

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.

KCET 2024Set D-21 markMCQ
$^{27}$
›Reveal solutionSolution
The ratio of volume to surface area for any spherical nucleus is $R/3$ , where $R = R_0 A^{1/3}$ . For Al $^{27}$ , $A=27$ , so $R = 1.2 \times 10^{-15} \times 3 = 3.6 \times 10^{-15}$ m, giving a ratio of $1.2 \times 10^{-15}$ m. The correct option is (D).

The key idea here is that a nucleus is modeled as a uniform sphere. For any sphere, the ratio of volume to surface area is simply $R/3$ — a clean geometric fact. Once you know the nuclear radius formula, the rest is arithmetic.

The nuclear radius is given by $R = R_0 A^{1/3}$ , where $R_0 = 1.2 \times 10^{-15}$ m and $A$ is the mass number. For Al $^{27}$ , $A = 27$ , so $A^{1/3} = 3$ . That gives $R = 1.2 \times 10^{-15} \times 3 = 3.6 \times 10^{-15}$ m.

Now, for a sphere:

Volume $V = \frac{4}{3}\pi R^3$

Surface area $S = 4\pi R^2$

The ratio $\frac{V}{S} = \frac{\frac{4}{3}\pi R^3}{4\pi R^2} = \frac{R}{3}$ .

So the ratio is simply $\frac{3.6 \times 10^{-15}}{3} = 1.2 \times 10^{-15}$ m. …
Start your 14-day free trial to see the full solution →
TG EAPCET 2023Set ap-2023-05-10-AN1 markMCQ
$\left(P + \dfrac{a}{V^2}\right)(V-b) = RT$
›Reveal solutionSolution
The key is to find the dimensions of $a$ and $b$ separately from the Van der Waals equation, then combine them with $RT$ to get the dimensions of $\frac{ab}{RT}$ . The result is $\left[M^0L^6T^0\right]$ , which is option (D).

The Van der Waals equation corrects the ideal gas law for real gas behaviour. The term $\frac{a}{V^2}$ accounts for intermolecular attraction, and $b$ accounts for the finite volume of molecules. Because the equation is dimensionally consistent, we can extract the dimensions of $a$ and $b$ by looking at how they appear.

Find the dimensions of $b$ .

In the term $(V - b)$ , we subtract $b$ from $V$ . Only quantities with the same dimensions can be added or subtracted. So $b$ must have the same dimensions as volume $V$ .

Volume has dimensions $[L^3]$ .

Hence, $[b] = [L^3]$ .

Find the dimensions of $a$ .

Look at the term $\left(P + \frac{a}{V^2}\right)$ . Again, $P$ and $\frac{a}{V^2}$ must have the same dimensions because they are added.

Pressure $P$ has dimensions $[ML^{-1}T^{-2}]$ (force per unit area).

So $\left[\frac{a}{V^2}\right] = [ML^{-1}T^{-2}]$ .

Since $[V^2] = [L^6]$ , we get $[a] = [ML^{-1}T^{-2}] \times [L^6] = [ML^5T^{-2}]$ .

Find the dimensions of $RT$ .

From the ideal gas law $PV = nRT$ , for one mole ( $n=1$ ) we have $PV = RT$ .

So $[RT] = [P][V] = [ML^{-1}T^{-2}] \times [L^3] = [ML^2T^{-2}]$ .

This is the same as energy (work), which makes sense — $RT$ is energy per mole.

Combine to get $\frac{ab}{RT}$ .

Now put the dimensions together:

$[ab] = [a][b] = [ML^5T^{-2}] \times [L^3] = [ML^8T^{-2}]$ . …
Start your 14-day free trial to see the full solution →
TG EAPCET 2023Set ap-2023-05-11-AN1 markMCQ
$\log_e 10 = 2.303$
›Reveal solutionSolution
Radioactive decay follows an exponential law governed by the half-life. Using the relation between activity and time, we find that reducing to 1% of initial activity requires approximately 66.5 days.

The activity of a radioactive substance measures how many nuclei decay per unit time. Because decay is a random process governed by probability, the activity decreases exponentially with a characteristic time scale set by the half-life.

The key insight is that after each half-life period, exactly half of the remaining active nuclei have decayed. So if we want the activity to drop to some small fraction like 1%, we need to count how many half-lives fit into that reduction.

The mathematical relationship is:
$A(t) = A_0 \left(\frac{1}{2}\right)^{t/T_{1/2}}$
where $A(t)$ is the activity at time $t$ , $A_0$ is the initial activity, and $T_{1/2}$ is the half-life.

Alternatively, using the decay constant $\lambda = \frac{\ln 2}{T_{1/2}}$ :
$A(t) = A_0 e^{-\lambda t}$
Let me work through this step by step:

Set up the equation for 1% activity We want $A(t) = 0.01 \, A_0$ , so:

$0.01 \, A_0 = A_0 e^{-\lambda t}$
Dividing both sides by $A_0$ :
$0.01 = e^{-\lambda t}$

Take the natural logarithm

$\ln(0.01) = -\lambda t$
Since $0.01 = \frac{1}{100} = 10^{-2}$ :
$\ln(10^{-2}) = -\lambda t$ $-2 \ln 10 = -\lambda t$ $2 \ln 10 = \lambda t$

Express the decay constant in terms of half-life …
Start your 14-day free trial to see the full solution →
TG EAPCET 2022Set eng-2022-07-20-FN1 markMCQ
$C$
›Reveal solutionSolution
We treat density as a product of powers of $C$ , $G$ , and $h$ , solve the system of dimensional equations, and find that density has dimensions $C^5 G^{-2} h^{-1}$ , which corresponds to option (B).

The key idea here is dimensional analysis — a powerful tool that lets us express any physical quantity in terms of chosen fundamental units. When we pick $C$ (velocity), $G$ (gravitational constant), and $h$ (Planck’s constant) as base units, we need to find how density $\rho$ (mass per volume) relates to them. The trick is to write density as $[ \rho ] = C^a G^b h^c$ and solve for $a$ , $b$ , $c$ using the known dimensions of each quantity.

Let’s recall the dimensions in the standard MLT (mass, length, time) system:

Velocity $C$ : $[C] = L T^{-1}$

Gravitational constant $G$ : from Newton’s law $F = G \frac{m_1 m_2}{r^2}$ , we get $[G] = M^{-1} L^3 T^{-2}$

Planck’s constant $h$ : from $E = h \nu$ , we have $[h] = M L^2 T^{-1}$

Density $\rho$ : $[\rho] = M L^{-3}$

Now we set up the equation:

Write the dimensional equation We assume $[\rho] = [C]^a [G]^b [h]^c$ . Substituting dimensions:

$M^1 L^{-3} T^0 = (L T^{-1})^a \cdot (M^{-1} L^3 T^{-2})^b \cdot (M L^2 T^{-1})^c$

Expand and collect powers Right side becomes:

$M^{-b + c} \cdot L^{a + 3b + 2c} \cdot T^{-a - 2b - c}$

Equate exponents for M, L, T

For mass: $1 = -b + c$

For length: $-3 = a + 3b + 2c$

For time: $0 = -a - 2b - c$

Solve the system

From the mass equation: $c = 1 + b$

From the time equation: $a = -2b - c = -2b - (1 + b) = -3b - 1$

Substitute into the length equation:

$-3 = (-3b - 1) + 3b + 2(1 + b)$
Simplify: $-3 = -3b - 1 + 3b + 2 + 2b$ …
Start your 14-day free trial to see the full solution →
KCET 2019Set A-11 markMCQ
$\alpha$
›Reveal solutionSolution
In a head-on collision, the $\alpha$ -particle is aimed directly at the nucleus, so the perpendicular distance between the initial velocity line and the nucleus — the impact parameter — is exactly zero.

The key idea here is the definition of impact parameter in Rutherford's scattering experiment. The impact parameter $b$ is the perpendicular distance between the initial velocity vector of the $\alpha$ -particle and the centre of the target nucleus. It tells you how "off-centre" the collision is.

For a head-on collision, the $\alpha$ -particle is aimed straight at the nucleus. That means the line of the initial velocity passes directly through the centre of the nucleus. The perpendicular distance from that line to the centre is therefore zero.

Recall the definition: Impact parameter $b$ = distance of closest approach if the nucleus were not there — it's the miss distance. Mathematically, if the initial velocity is along a line, $b$ is the perpendicular distance from the nucleus centre to that line.

Head-on means zero miss distance: When the $\alpha$ -particle is fired straight at the nucleus, there is no "miss" — it's coming right at it. So $b = 0$ .

Consequences: For $b = 0$ , the $\alpha$ -particle experiences the maximum repulsive force (Coulomb force) and comes to rest momentarily at the distance of closest approach before being repelled back. This is the case that gives the largest scattering angle ( $180^\circ$ ). …
Start your 14-day free trial to see the full solution →

←Previous questionQ24 — Why length, mass and time are chosen as base quantities in mechanics?→Next questionQ26 — Which of the following time measuring devices is most precise? (a) A w…

Q.(a) The earth-moon distance is about 60 earth radius. What will be the diameter of the earth (approximately in degrees) as seen from the moon?

The tool: small-angle angular diameter

(a) Earth's angular diameter seen from the Moon

(b) Relative size of the Moon

(c) Ratio of Sun's diameter to Earth's diameter

🎓Unlock everything free for 14 days

Unlock everything free for 14 days