Calculate the length of the arc of a circle of radius 31.0 cm which subtends an angle of (π)/(6) at the centre

NCERT Exemplar · Q32

\dfrac{\pi}{6}

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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

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The arc length is found by multiplying the radius by the central angle (in radians). For $r = 31.0$ cm and $\theta = \pi/6$ , the result is $31.0 \times \pi/6 \approx 16.2$ cm.

The key idea here is that arc length is a direct proportion of the radius and the angle. When the angle is given in radians, the formula is beautifully simple: $s = r\theta$ . This works because one radian is defined as the angle that subtends an arc equal in length to the radius. So if you have $\theta$ radians, you're essentially stacking $\theta$ such radius-length arcs together.

Let's walk through it step by step.

Identify what's given.

The radius $r = 31.0$ cm. The central angle $\theta = \dfrac{\pi}{6}$ radians. Notice the angle is already in radians — that's important. If it were in degrees, we'd have to convert first.
Recall the arc length formula.

For a circle, the length $s$ of an arc subtended by an angle $\theta$ (in radians) at the centre is:

s = r\theta

This is not a coincidence — it's the definition of the radian measure.

Plug in the values.

s = 31.0 \times \frac{\pi}{6}

Simplify the expression.

s = \frac{31.0\pi}{6}

Get a numerical value (if needed). Using $\pi \approx 3.1416$ :

s \approx \frac{31.0 \times 3.1416}{6} = \frac{97.3896}{6} \approx 16.2316

Rounding to three significant figures (since 31.0 has three), we get $s \approx 16.2$ cm. …

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Here’s a breakdown of the common mistakes students make when solving this problem, along with how to avoid each one.

1. Using the wrong formula for arc length

Mistake:

Students often confuse the formula for arc length with the formula for the area of a sector.

Wrong: $A = \frac{1}{2} r^2 \theta$ (this is area, not length).
Wrong: $l = \frac{\theta}{360^\circ} \times 2\pi r$ without checking if $\theta$ is in degrees or radians.

How to avoid:

Always recall the definition:

Arc length $l = r \theta$ , where $\theta$ is in radians.

If the angle is given in degrees, convert first:

$\theta_{\text{rad}} = \frac{\pi}{180^\circ} \times \theta_{\text{deg}}$ .

Here, $\theta = \frac{\pi}{6}$ is already in radians, so directly use:

l = r \theta = 31.0 \times \frac{\pi}{6}

2. Forgetting to keep units consistent

Mistake:

Using radius in cm but angle in degrees without conversion, or mixing cm and m.

How to avoid:

The formula $l = r\theta$ requires $\theta$ in radians (a pure number).
The unit of $l$ will be the same as the unit of $r$ .
Here $r = 31.0\ \text{cm}$ , so $l$ will be in cm. No conversion needed.

3. Incorrect simplification of $\pi$ or rounding too early

Mistake:

Using $\pi = 3.14$ too early leads to rounding errors.
Or writing $l = 31.0 \times \frac{22}{7 \times 6}$ and getting a messy fraction.

How to avoid:

Keep $\pi$ symbolic until the final step. Only substitute a numerical value (like $3.14$ or $\frac{22}{7}$ ) at the end, and round as per the problem’s requirement.

Correct approach:

l = \frac{31.0\pi}{6}

If a decimal is needed:

l \approx \frac{31.0 \times 3.1416}{6} \approx 16.23\ \text{cm}

4. Misreading the angle — thinking it’s in degrees

Mistake:

Seeing $\frac{\pi}{6}$ and treating it as $30^\circ$ but then using the degree-based formula incorrectly. …

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

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:

$[ab] = [a][b] = [ML^5T^{-2}] \times [L^3] = [ML^8T^{-2}]$ . …
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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 …
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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: $-3 = -3b - 1 + 3b + 2 + 2b$ …
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Q.Calculate the length of the arc of a circle of radius 31.0 cm which subtends an angle of $\dfrac{\pi}{6}$ at the centre.

Concept understanding — Atomic Volume Calculation

Atomic Volume: Meaning and Calculation

Definition

Worked example — aluminium

Periodic trends

Why this formula?

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

What Is Atomic Volume?

The Core Reasoning: From Mass to Volume

Step 1: What does density tell us?

Step 2: Rearranging to find atomic volume

Why This Makes Physical Sense

Important Exam Points

Concept: Arc Length of a Circle

Method: Direct Formula for Arc Length (Angle in Radians)

Steps

1. Using the wrong formula for arc length

2. Forgetting to keep units consistent

3. Incorrect simplification of $\pi$ or rounding too early

4. Misreading the angle — thinking it’s in degrees

Q.Calculate the length of the arc of a circle of radius 31.0 cm which subtends an angle of π6\dfrac{\pi}{6}6π​ at the centre.

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Q.Calculate the length of the arc of a circle of radius 31.0 cm which subtends an angle of $\dfrac{\pi}{6}$ at the centre.

Unlock everything free for 14 days