Measurement and estimation

We are asked what unit results when a measurement written in metres ($m$) is multiplied by $10^{-3}$. This is about metric (SI) prefixes and powers of ten.

Multiplying by $10^{-3}$ is the same as dividing by $1000$: $10^{-3}=\frac{1}{10^3}=\frac{1}{1000}$

In SI units, the prefix kilo- (k) means a factor of $10^3$ (which is $1000$). So: $1\text{ km}=10^3\text{ m}$

To express $1$ metre in kilometres, divide both sides by $10^3$:

Check why the other choices don’t fit a $10^{-3}$ factor:

• Millimetres (milli-): $1\text{ mm}=10^{-3}\text{ m}$. Going from metres to millimetres would require multiplying by $10^3$.
• Micrometres (micro-): $1\text{ mum}=10^{-6}\text{ m}$. Going from metres to micrometres would require multiplying by $10^6$.
• Centimetres (centi-): $1\text{ cm}=10^{-2}\text{ m}$. Going from metres to centimetres would require multiplying by $10^2$.

Because $1\text{ m}=10^{-3}\text{ km}$, multiplying any metre measurement by $10^{-3}$ converts it into kilometres.

We want the unit produced when a value in metres (m) is multiplied by $10^{-3}$. Since $10^{-3}=0.001$, this means dividing by $1000$.

$\text{Value in m}\times 10^{-3}=\text{Value in new unit}$

So the new unit must be larger than a metre (you use a smaller number to describe the same distance).

Option A says millimetres. But $1$ metre equals $1000$ millimetres ($1\text{ m}=10^3\text{ mm}$).

That conversion needs multiplying by $10^3$ (or $1000$), not multiplying by $10^{-3}$. So this choice cannot be correct.

Option B is micrometres. The prefix micro- corresponds to $10^{-6}$, so there are $1,000,000$ micrometres in $1$ metre ($1\text{ m}=10^6\mu \text{m}$).

Converting metres to micrometres requires multiplying by $10^6$, which does not match $10^{-3}$.

Option D is centimetres. The prefix centi- means $10^{-2}$, and $1$ metre is $100$ centimetres ($1\text{ m}=10^2\text{ cm}$).

So metres to centimetres uses a factor of $10^2$, not $10^{-3}$, so this option is wrong.

Option C is kilometres. Since $1$ kilometre equals $1000$ metres ($1\text{ km}=10^3\text{ m}$), converting metres to kilometres means dividing by $1000$, which is the same as multiplying by $10^{-3}$:

$1\text{ m}=\frac{1}{1000}\text{ km}=10^{-3}\text{ km}$

This matches exactly.

Start with the circle area rule. If the original radius is $r$, then the starting area $A$ is:

$A=\pi r^2$

The radius is tripled, so the updated radius (call it $r_{new}$) is three times the original:

$r_{new}=3r$

Substitute $r_{new}=3r$ into the area formula to get the new area $A_{new}$:

$\begin{array}{rl}{A}_{new}& =\pi (r_{new}{)}^2\\ A_{new}& =\pi (3r{)}^2\end{array}$

Using the exponent rule $(ab{)}^n=a^n{b}^n$, square both factors:

$\begin{array}{rl}{A}_{new}& =\pi \cdot 3^2\cdot r^2\\ A_{new}& =9\pi r^2\end{array}$

Now compare the new area to the original area:

$\begin{array}{rl}{A}_{new}& =9\cdot (\pi r^2)\\ A_{new}& =9\cdot A\end{array}$

So the area is multiplied by $9$.

Check why the other answers don’t match the mathematics:

• Option B ($12$) could come from incorrectly doing something like $3\times 4$ instead of squaring the scale factor.
• Option C ($27$) would be the result of using $3^3$, which relates to volume scaling, not area.
• Option D ($3$) is only the change in radius; area depends on $r^2$, not just $r$.

The surface area increases by a factor of $9$.

The radius is a length measurement, so we treat the change as a linear scale factor $k$.

Because the radius is tripled: $k=3$

A key geometry idea is that when all lengths scale by $k$, areas scale by $k^2$.

For circles, this matches the fact that $A\propto r^2$. So if the radius is multiplied by $3$, the area is multiplied by $3^2$.

Square the scale factor:

$\text{Area factor}=k^2=3^2=9$

Use the scaling idea to judge each wrong option:

• Option D ($3$) is only $k$, not $k^2$.
• Option B ($12$) reflects a non-squaring mistake (treating the change as something like “times 4” rather than $3^2$).
• Option C ($27$) equals $k^3=3^3$, which would be for volume changes in 3D shapes, not for area.

Since $k^2=9$, the area increases by a factor of 9.

We’ll use the standard area formula for a circle (from Geometry of Shapes, Surface Area and Volume):

Here, $A$ is the area and $r$ is the radius.

To compare areas by a factor, pick a convenient value. Let the original radius be $r=1$ unit.

With radius $1$:

And using a calculator, $\pi \approx 3.14$.

Tripling the radius gives $1\times 3=3$ units.

Now compute the new area:

Using a calculator, $9\times \pi \approx 28.27$.

The scale factor for the area is the ratio:

So the area becomes 9 times as large.

Compare with the options:

• Option D ($3$) is only the radius multiplier.
• Option B ($12$) could come from not squaring the 3 correctly.
• Option C ($27$) would match $3^3$ (volume-style thinking), not area.

Therefore, the correct factor is 9.

We need to choose which option is an SI base unit that measures electric current.

The SI system has seven core base units, and every other SI unit is built from these. Knowing this is essential for units and measurement.

There are exactly seven SI base units you’re expected to know for IB math and science:

1. Mass: kilogram (kg)
2. Length: metre (m)
3. Time: second (s)
4. Electric current: ampere (A)
5. Thermodynamic temperature: kelvin (K)
6. Amount of substance: mole (mol)
7. Luminous intensity: candela (cd)

Choices A, B, and C are all electrical units, but they are derived, meaning they can be written using combinations of base units.

• watt (W): Measures power ($1\text{ W}=1\text{ J/s}$).
• ohm ($\Omega$): Measures electrical resistance.
• volt (V): Measures potential difference ($1\text{ V}=1\text{ J/C}$).

From the list of seven SI base units, ampere (A) is the one defined for measuring electric current. So Option D is correct.

We are looking for the option that is an SI base unit and that measures electric current.

Because SI has seven base units and everything else is derived from them, we can solve this by crossing out the derived units first.

The watt (W) measures power, which is how fast energy is transferred or work is done. It is not a base unit since it can be expressed using joules and seconds:

$1\text{ W}=\frac{1\text{ J}}{1\text{ s}}$

Since it’s built from other units, it must be derived.

The ohm ($\Omega$) measures resistance. Using Ohm’s Law, resistance is defined as volts divided by amperes:

$1\Omega =\frac{1\text{ V}}{1\text{ A}}$

Because it is defined using other units, it is a derived unit, not a base unit.

The volt (V) measures electric potential difference. It can be written as joules per coulomb:

$1\text{ V}=\frac{1\text{ J}}{1\text{ C}}$

So the volt is also derived and is not one of the seven SI base units.

After eliminating the derived units, the remaining choice is ampere (A). The ampere is one of the seven SI base units and is specifically used to measure the flow of charge, i.e., electric current.

The seven SI base units are: second (time), metre (length), kilogram (mass), ampere (electric current), kelvin (temperature), mole (amount of substance), and candela (luminous intensity).