Q.Fill in the blanks:

Concept: Significant Figures Calculation — each conversion must preserve the number of significant figures from the given data.

(a) Side $1\ \text{cm} = 0.01\ \text{m}$. Volume $= (0.01\ \text{m})^3 = 1 \times 10^{-6}\ \text{m}^3$.

(b) Radius $2.0\ \text{cm} = 20\ \text{mm}$, height $10.0\ \text{cm} = 100\ \text{mm}$.

Surface area $= 2\pi r h + 2\pi r^2 = 2\pi(20)(100) + 2\pi(20)^2 = 4000\pi + 800\pi = 4800\pi\ \text{mm}^2$.

With $\pi \approx 3.14$, $4800 \times 3.14 = 15072\ \text{mm}^2$. Given two significant figures in radius, result is $1.5 \times 10^4\ \text{mm}^2$.

(c) $18\ \text{km h}^{-1} = 18 \times \frac{1000}{3600} = 5\ \text{m s}^{-1}$. In 1 s, distance $= 5\ \text{m}$.

(d) Relative density 11.3 means density $= 11.3\ \text{g cm}^{-3}$.

In $\text{kg m}^{-3}$: $11.3 \times 1000 = 11300\ \text{kg m}^{-3}$.

The Concept: Unit Conversion with Significant Figures

Every measurement has a number and a unit. To convert between units, you multiply by a conversion factor — a fraction equal to 1 (e.g., $1\ \text{m} / 100\ \text{cm} = 1$). The trick is to arrange the factor so the old unit cancels and the new unit remains.

But there’s a second layer: significant figures. The given numbers (1 cm, 2.0 cm, 10.0 cm, 18 km/h, 11.3) tell you how many digits are reliable. Your answer should not pretend to be more precise than the data. For example, “1 cm” has 1 significant figure, so $1\ \text{cm}^3$ is exactly $1\ \text{cm}^3$ — but when we convert, we keep the result as $10^{-6}\ \text{m}^3$ (which is exact, since 1 cm = 0.01 m exactly). For parts (b) and (d), the given numbers have 2 or 3 significant figures, so the answer must match.

Let’s go through each part.

(a) Volume of a cube of side 1 cm

Step 1: Volume in cm³

A cube of side $1\ \text{cm}$ has volume $V = (1\ \text{cm})^3 = 1\ \text{cm}^3$.

Step 2: Convert cm to m

We know $1\ \text{cm} = 10^{-2}\ \text{m}$. So $1\ \text{cm}^3 = (10^{-2}\ \text{m})^3 = 10^{-6}\ \text{m}^3$.

That’s it. The conversion is exact because the definition of “centi” is exact. No significant figure issue here — the answer is simply $10^{-6}\ \text{m}^3$.

(b) Surface area of a solid cylinder: radius $2.0\ \text{cm}$, height $10.0\ \text{cm}$

Step 1: Formula for total surface area

A solid cylinder has two circular ends and a curved side.

Total surface area $S = 2\pi r^2 + 2\pi r h = 2\pi r (r + h)$.

Step 2: Plug in values (in cm)

$r = 2.0\ \text{cm}$, $h = 10.0\ \text{cm}$.

$S = 2\pi (2.0)(2.0 + 10.0) = 2\pi (2.0)(12.0) = 2\pi \times 24.0 = 48.0\pi\ \text{cm}^2$.

Step 3: Significant figures

Both $2.0$ and $10.0$ have 2 significant figures. So $48.0$ has 3 digits, but the product $48.0\pi$ should be reported with 2 significant figures because the least precise input has 2.

$48.0\pi \approx 150.796...$ — rounding to 2 significant figures gives $1.5 \times 10^2\ \text{cm}^2$.

Step 4: Convert cm² to mm²

$1\ \text{cm} = 10\ \text{mm}$, so $1\ \text{cm}^2 = (10\ \text{mm})^2 = 100\ \text{mm}^2$.

Thus $S = 1.5 \times 10^2\ \text{cm}^2 \times 100\ \frac{\text{mm}^2}{\text{cm}^2} = 1.5 \times 10^4\ \text{mm}^2$.

(c) Speed $18\ \text{km/h}$ — distance covered in 1 second

Step 1: Convert km/h to m/s

The standard conversion: $1\ \text{km/h} = \frac{1000\ \text{m}}{3600\ \text{s}} = \frac{5}{18}\ \text{m/s}$.

So $18\ \text{km/h} = 18 \times \frac{5}{18}\ \text{m/s} = 5\ \text{m/s}$.

Step 2: Distance in 1 second

Distance = speed × time = $5\ \text{m/s} \times 1\ \text{s} = 5\ \text{m}$.

The number 18 has 2 significant figures, but the conversion factor $\frac{5}{18}$ is exact (since 1 km = 1000 m and 1 h = 3600 s are definitions). So the answer 5 m is exact in this context.

(d) Relative density of lead = 11.3

Step 1: What is relative density?

Relative density (specific gravity) is the ratio of the density of a substance to the density of water at 4°C.

Density of water = $1\ \text{g/cm}^3$ exactly (by definition of the gram).

So density of lead = $11.3 \times 1\ \text{g/cm}^3 = 11.3\ \text{g/cm}^3$.

Step 2: Convert to kg/m³

$1\ \text{g/cm}^3 = 1000\ \text{kg/m}^3$ because:

$1\ \text{g} = 10^{-3}\ \text{kg}$, $1\ \text{cm}^3 = 10^{-6}\ \text{m}^3$, so

$\frac{1\ \text{g}}{1\ \text{cm}^3} = \frac{10^{-3}\ \text{kg}}{10^{-6}\ \text{m}^3} = 10^3\ \text{kg/m}^3$.

Thus $11.3\ \text{g/cm}^3 = 11.3 \times 1000\ \text{kg/m}^3 = 11300\ \text{kg/m}^3$.

Step 3: Significant figures

11.3 has 3 significant figures. So $11300$ should be written as $1.13 \times 10^4\ \text{kg/m}^3$ to show 3 significant figures (11300 could be ambiguous — it might look like 3, 4, or 5 sig figs). Scientific notation removes the ambiguity.

Method: Dimensional Analysis & Unit Conversion with Significant Figures

This method uses conversion factors (ratios equal to 1) to change units, while respecting the significant figures of the given data.

General Steps:

1. Identify given value and its units.
2. Choose conversion factors (e.g., $1\ \text{cm} = 10^{-2}\ \text{m}$).
3. Multiply so unwanted units cancel.
4. Round final answer to the least number of significant figures in the original data.

(a) Volume of cube: $1\ \text{cm}$ side → $\text{m}^3$

• Step 1: Volume $= (1\ \text{cm})^3 = 1\ \text{cm}^3$
• Step 2: $1\ \text{cm} = 10^{-2}\ \text{m}$ → $(1\ \text{cm})^3 = (10^{-2}\ \text{m})^3$
• Step 3: $= 10^{-6}\ \text{m}^3$

Answer: $\boxed{10^{-6}\ \text{m}^3}$

(1 significant figure, matching the side length)

(b) Surface area of cylinder: $r = 2.0\ \text{cm}$, $h = 10.0\ \text{cm}$ → $(\text{mm})^2$

• Step 1: Surface area $= 2\pi r h + 2\pi r^2$ $= 2\pi (2.0)(10.0) + 2\pi (2.0)^2$ $= 40\pi + 8\pi = 48\pi\ \text{cm}^2$
• Step 2: $1\ \text{cm} = 10\ \text{mm}$ → $1\ \text{cm}^2 = 100\ \text{mm}^2$
• Step 3: $48\pi\ \text{cm}^2 = 48\pi \times 100\ \text{mm}^2 = 4800\pi\ \text{mm}^2$
• Step 4: $\pi \approx 3.14$ → $4800 \times 3.14 = 15072\ \text{mm}^2$ Given data has 2 significant figures (2.0, 10.0) → round to $1.5 \times 10^4\ \text{mm}^2$

Answer: $\boxed{1.5 \times 10^4\ \text{mm}^2}$

(c) Speed $18\ \text{km h}^{-1}$ → m in 1 s

• Step 1: $18\ \text{km h}^{-1}$ means $18\ \text{km}$ in $1\ \text{h}$
• Step 2: Convert km → m: $18\ \text{km} = 18 \times 1000\ \text{m} = 18000\ \text{m}$ Convert h → s: $1\ \text{h} = 3600\ \text{s}$
• Step 3: Distance in 1 s $= \frac{18000\ \text{m}}{3600\ \text{s}} = 5\ \text{m}$

Answer: $\boxed{5\ \text{m}}$

(1 significant figure, matching 18)

(d) Relative density of lead = 11.3 → density in $\text{g cm}^{-3}$ and $\text{kg m}^{-3}$

• Step 1: Relative density = density of substance / density of water Density of water = $1\ \text{g cm}^{-3} = 1000\ \text{kg m}^{-3}$
• Step 2: In $\text{g cm}^{-3}$: $11.3 \times 1 = 11.3\ \text{g cm}^{-3}$
• Step 3: In $\text{kg m}^{-3}$: $11.3 \times 1000 = 11300\ \text{kg m}^{-3}$ (11.3 has 3 significant figures → keep 3)

Answer: $\boxed{11.3\ \text{g cm}^{-3}}$ and $\boxed{1.13 \times 10^4\ \text{kg m}^{-3}}$

(a) Volume of a cube of side 1 cm → m³

Common mistakes:

• Forgetting to cube the conversion factor

  Students often write:

  $1\ \text{cm} = 0.01\ \text{m}$, so volume = $0.01\ \text{m}^3$ ✗

  Correct: volume = $(0.01\ \text{m})^3 = 1 \times 10^{-6}\ \text{m}^3$ ✓

• Mixing up cm³ and m³

  They might recall $1\ \text{m}^3 = 10^6\ \text{cm}^3$ but then invert it.

How to avoid:

Always write the conversion with the exponent outside the bracket:

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

$\Rightarrow$ Volume $= (1\ \text{cm})^3 = (10^{-2}\ \text{m})^3 = 10^{-6}\ \text{m}^3$

Final answer:

$\boxed{1 \times 10^{-6}\ \text{m}^3}$

(b) Surface area of a solid cylinder: radius $2.0\ \text{cm}$, height $10.0\ \text{cm}$ → (mm)²

Common mistakes:

• Using wrong formula

  Some use only curved surface area ($2\pi rh$) and forget the two circular ends ($2\pi r^2$).

  Total surface area of a solid cylinder = $2\pi r h + 2\pi r^2 = 2\pi r (r + h)$.

• Significant figures error

  Radius is given as $2.0\ \text{cm}$ (2 significant figures), height as $10.0\ \text{cm}$ (3 significant figures).

  Students often report too many digits in the final answer.

• Unit conversion mistake

  They convert cm to mm correctly ($1\ \text{cm} = 10\ \text{mm}$) but forget to square the conversion factor:

  $1\ \text{cm}^2 = (10\ \text{mm})^2 = 100\ \text{mm}^2$.

How to avoid:

1. Write the formula first:

   $A = 2\pi r (r + h)$

2. Plug in values in cm:

   $r = 2.0\ \text{cm}$, $h = 10.0\ \text{cm}$

   $A = 2\pi (2.0)(2.0 + 10.0) = 2\pi (2.0)(12.0) = 48\pi\ \text{cm}^2$

3. Convert to mm²:

   $1\ \text{cm}^2 = 100\ \text{mm}^2$

   $A = 48\pi \times 100 = 4800\pi\ \text{mm}^2$

4. Apply significant figures:

   Least precise measurement is $2.0$ (2 s.f.), so answer should have 2 significant figures.

   $4800\pi \approx 15079.6\ \text{mm}^2$ → round to $1.5 \times 10^4\ \text{mm}^2$

Final answer:

$\boxed{1.5 \times 10^4\ \text{mm}^2}$

(c) Speed $18\ \text{km h}^{-1}$ → metres in 1 second

Common mistakes:

• Using wrong conversion factor

  $1\ \text{km} = 1000\ \text{m}$, $1\ \text{h} = 3600\ \text{s}$

  Some students multiply by 1000 and divide by 60 (thinking 1 h = 60 min only).

• Forgetting to convert both units

  They convert km to m but forget to convert hour to seconds.

How to avoid:

Use the standard conversion:

$18\ \text{km h}^{-1} = 18 \times \frac{1000\ \text{m}}{3600\ \text{s}} = 18 \times \frac{5}{18} = 5\ \text{m s}^{-1}$

So in 1 second, distance = 5 m.

Final answer:

$\boxed{5\ \text{m}}$

(d) Relative density of lead = 11.3 → density in $\text{g cm}^{-3}$ and $\text{kg m}^{-3}$

Common mistakes:

• Confusing relative density with density

  Relative density has no units — it’s the ratio of density of substance to density of water.

  Density of water = $1\ \text{g cm}^{-3} = 1000\ \text{kg m}^{-3}$.

• Wrong conversion between g cm⁻³ and kg m⁻³

  Common error: $1\ \text{g cm}^{-3} = 100\ \text{kg m}^{-3}$ ✗

  Correct: $1\ \text{g cm}^{-3} = 1000\ \text{kg m}^{-3}$ ✓

How to avoid:

• In $\text{g cm}^{-3}$:

  Density = relative density × density of water

  $= 11.3 \times 1\ \text{g cm}^{-3} = 11.3\ \text{g cm}^{-3}$

• In $\text{kg m}^{-3}$:

  $11.3\ \text{g cm}^{-3} = 11.3 \times 1000\ \text{kg m}^{-3} = 11300\ \text{kg m}^{-3}$

Significant figures note:

Relative density is given as 11.3 (3 s.f.), so both answers should have 3 s.f.

Final answers:

$\boxed{11.3\ \text{g cm}^{-3}}$ and $\boxed{1.13 \times 10^4\ \text{kg m}^{-3}}$

Summary of key pitfalls to avoid

Would you like practice problems on this exact set of conversions?