Fill in the blanks by suitable conversion of units: (a) 1kg m²s⁻² = ldotsg cm²s⁻² (b) 1m = ldotsly (c) 3.0m s⁻

Exercises · 1.2

Q. Fill in the blanks by suitable conversion of units:

(a)

1\ \text{kg m}^2\,\text{s}^{-2} = \ldots\ \text{g cm}^2\,\text{s}^{-2}

(b)

1\ \text{m} = \ldots\ \text{ly}

(c)

3.0\ \text{m s}^{-2} = \ldots\ \text{km h}^{-2}

(d)

G = 6.67 \times 10^{-11}\ \text{N m}^2\,(\text{kg})^{-2} = \ldots\ (\text{cm})^3\,\text{s}^{-2}\,\text{g}^{-1}

Kerala DhseTextbookSubjective· 3mImportance★★★★★est

11% · 7/66 Questions

✓ Free question

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	$\frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ h}}{3600 \text{ s}}$

What About Multiple Steps?

Sometimes you need more than one conversion. Convert 2 hours to seconds:

2 \text{ h} \times \frac{60 \text{ min}}{1 \text{ h}} \times \frac{60 \text{ s}}{1 \text{ min}} = 2 \times 60 \times 60 \text{ s} = 7200 \text{ s}

Each step cancels one unit and introduces the next. This is called chain conversion — it's just multiplying by a series of 1's.

Watch out

A common mistake: forgetting to square or cube conversion factors when dealing with area or volume.

1 m² = (100 cm)² = 10,000 cm², not 100 cm².

1 m³ = (100 cm)³ = 1,000,000 cm³, not 100 cm³.

Always apply the exponent to the conversion factor itself.

The Big Picture

Unit conversion is not a trick — it's a logical tool. Every conversion factor is just a statement of equality written as a fraction. As long as you multiply by 1 (in the form of that fraction), the quantity stays the same. The only thing that changes is the label.

Final takeaway: A quantity is a number times a unit. To change the unit without changing the quantity, multiply by a conversion factor that equals 1. That's all there is to it.

"Unit conversion formula physics class 11" and "dimensional analysis and unit conversion" are frequently searched terms for this topic, which is introduced early in the Units and Measurements chapter of the NCERT/CBSE Class 11 Physics syllabus. Chain conversions in particular are a recurring numerical-question type in JEE Main and various state CETs.

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:

Suppose you have a quantity $Q$ with dimensions $[L^a M^b T^c]$ . If you change the base units (say from meters to centimeters), the numerical value must change inversely to keep the physical quantity the same.

If length unit shrinks by factor $f_L$ (1 m → 100 cm, so $f_L = 100$ ), then the numerical value of a length increases by $f_L$ .
For a quantity with dimension $L^a$ , the numerical value scales by $f_L^a$ .

Reasoning: The physical quantity is invariant — only the number changes. The exponent $a$ tells you how many times the length dimension appears, so the scaling factor is raised to that power.

5. The "Why" in One Sentence

Dimensional analysis works because physical laws are independent of the units we choose — the dimensions impose constraints that any valid equation must satisfy, reducing the number of independent variables.

Key Takeaways for Exams

Principle	Why It Holds
Dimensional homogeneity	Physical equality requires same dimensions
Buckingham Pi Theorem	Dimensions act as constraints, reducing variables
Unit conversion	Physical quantity is invariant; numerical value scales inversely with unit size

Remember: Dimensional analysis can check an equation's validity, but it cannot determine dimensionless constants (like $2\pi$ or $1/2$ ). That's where experiment or deeper theory comes in.

The key idea here is Unit Conversion, which involves expressing a physical quantity in different units by using conversion factors, then rounding the final result to the number of significant figures justified by the given data.

To convert $1\ \text{kg m}^2\,\text{s}^{-2}$ to $\text{g cm}^2\,\text{s}^{-2}$ : We know that $1\ \text{kg} = 10^3\ \text{g}$ and $1\ \text{m} = 10^2\ \text{cm}$ . The unit for time (seconds) remains unchanged. $1\ \text{kg m}^2\,\text{s}^{-2} = (10^3\ \text{g}) \times (10^2\ \text{cm})^2 \times \text{s}^{-2}$ $= 10^3\ \text{g} \times 10^4\ \text{cm}^2 \times \text{s}^{-2} = 10^{3+4}\ \text{g cm}^2\,\text{s}^{-2}$ $= 10^7\ \text{g cm}^2\,\text{s}^{-2}$
✓Final answer
$1\ \text{kg m}^2\,\text{s}^{-2}$
To convert $1\ \text{m}$ to light-years ( $\text{ly}$ ): A light-year is the distance light travels in one year. We use the speed of light $c = 3 \times 10^8\ \text{m/s}$ and $1\ \text{year} = 365.25\ \text{days} = 365.25 \times 24 \times 60 \times 60\ \text{s} = 3.15576 \times 10^7\ \text{s}$ . $1\ \text{ly} = c \times (1\ \text{year}) = (3 \times 10^8\ \text{m/s}) \times (3.15576 \times 10^7\ \text{s})$ $1\ \text{ly} = 9.46728 \times 10^{15}\ \text{m}$ Therefore, $1\ \text{m} = \frac{1}{9.46728 \times 10^{15}}\ \text{ly} \approx 1.0569 \times 10^{-16}\ \text{ly}$ , which rounds to 3 significant figures: $1\ \text{m} \approx 1.06 \times 10^{-16}\ \text{ly}$
✓Final answer
$1\ \text{m}$
To convert $3.0\ \text{m s}^{-2}$ to $\text{km h}^{-2}$ : We use the conversions $1\ \text{m} = 10^{-3}\ \text{km}$ and $1\ \text{s} = \frac{1}{3600}\ \text{h}$ . $3.0\ \text{m s}^{-2} = 3.0 \times (10^{-3}\ \text{km}) \times \left(\frac{1}{3600}\ \text{h}\right)^{-2}$ $= 3.0 \times 10^{-3}\ \text{km} \times (3600)^2\ \text{h}^{-2}$ $= 3.0 \times 10^{-3} \times 12960000\ \text{km h}^{-2}$ $= 3.0 \times 12960\ \text{km h}^{-2} = 38880\ \text{km h}^{-2} = 3.888 \times 10^4\ \text{km h}^{-2}$ The given value $3.0\ \text{m s}^{-2}$ has only 2 significant figures, so the result must be rounded to 2 significant figures: $3.0\ \text{m s}^{-2} \approx 3.9 \times 10^4\ \text{km h}^{-2}$
✓Final answer
$3.0\ \text{m s}^{-2}$
To convert $G = 6.67 \times 10^{-11}\ \text{N m}^2\,(\text{kg})^{-2}$ to $\text{cm}^3\,\text{s}^{-2}\,\text{g}^{-1}$ : First, express Newton (N) in fundamental units: $1\ \text{N} = 1\ \text{kg m s}^{-2}$ . $G = 6.67 \times 10^{-11}\ (\text{kg m s}^{-2})\ \text{m}^2\,(\text{kg})^{-2}$ $G = 6.67 \times 10^{-11}\ \text{kg}^{1-2}\ \text{m}^{1+2}\ \text{s}^{-2} = 6.67 \times 10^{-11}\ \text{kg}^{-1}\ \text{m}^3\ \text{s}^{-2}$ Now, convert kg to g and m to cm: $1\ \text{kg} = 10^3\ \text{g}$ and $1\ \text{m} = 10^2\ \text{cm}$ . $G = 6.67 \times 10^{-11} \times (10^3\ \text{g})^{-1} \times (10^2\ \text{cm})^3 \times \text{s}^{-2}$ $G = 6.67 \times 10^{-11} \times 10^{-3}\ \text{g}^{-1} \times 10^6\ \text{cm}^3 \times \text{s}^{-2}$ $G = 6.67 \times 10^{-11-3+6}\ \text{cm}^3\,\text{s}^{-2}\,\text{g}^{-1}$ $G = 6.67 \times 10^{-8}\ \text{cm}^3\,\text{s}^{-2}\,\text{g}^{-1}$
✓Final answer
$G = 6.67 \times 10^{-11}\ \text{N m}^2\,(\text{kg})^{-2}$

Unit conversion involves multiplying by conversion factors (ratios equal to 1) to change units without altering the physical quantity's magnitude, then rounding the final result to match the precision of the given data. The results are:

$1\ \text{kg m}^2\,\text{s}^{-2} = \mathbf{10^7\ \text{g cm}^2\,\text{s}^{-2}}$
$1\ \text{m} = \mathbf{1.06 \times 10^{-16}\ \text{ly}}$
$3.0\ \text{m s}^{-2} = \mathbf{3.9 \times 10^4\ \text{km h}^{-2}}$
$G = 6.67 \times 10^{-11}\ \text{N m}^2\,(\text{kg})^{-2} = \mathbf{6.67 \times 10^{-8}\ (\text{cm})^3\,\text{s}^{-2}\,\text{g}^{-1}}$

Unit conversion is a fundamental skill in physics and chemistry, allowing us to express a physical quantity in different units while preserving its actual value. The core idea is to multiply the given quantity by one or more "conversion factors". A conversion factor is a ratio of two equivalent quantities expressed in different units, making the ratio itself equal to 1. For example, since $1 \text{ kg}$ is the same as $1000 \text{ g}$ , the ratio $\frac{1000\ \text{g}}{1\ \text{kg}}$ is equal to 1. Multiplying any quantity by such a factor changes its units without changing its magnitude.

The process involves:

Identifying the initial units and the target units.
Finding the appropriate conversion factors that relate these units.
Multiplying the original quantity by these factors, ensuring that the unwanted units cancel out and the desired units remain. Pay close attention to powers of units (e.g., $\text{m}^2$ , $\text{s}^{-2}$ ).
Rounding the final result to the number of significant figures justified by the given data.

Let's apply this to each part of the problem.

(a) $1\ \text{kg m}^2\,\text{s}^{-2} = \ldots\ \text{g cm}^2\,\text{s}^{-2}$

Here, we need to convert kilograms (kg) to grams (g) and meters (m) to centimeters (cm). The unit of time (seconds, s) remains the same.

Convert kg to g:

We know that $1\ \text{kg} = 1000\ \text{g}$ .

The conversion factor is $\frac{1000\ \text{g}}{1\ \text{kg}}$ .
Convert m $^2$ to cm $^2$ :

We know that $1\ \text{m} = 100\ \text{cm}$ .

Therefore, $1\ \text{m}^2 = (100\ \text{cm})^2 = 100^2\ \text{cm}^2 = 10000\ \text{cm}^2$ .

The conversion factor is $\frac{10000\ \text{cm}^2}{1\ \text{m}^2}$ .
Perform the conversion:

Multiply the given quantity by the conversion factors:

1\ \text{kg m}^2\,\text{s}^{-2} \times \left(\frac{1000\ \text{g}}{1\ \text{kg}}\right) \times \left(\frac{10000\ \text{cm}^2}{1\ \text{m}^2}\right)

Notice how 'kg' and 'm$^2$' units cancel out:

1 \times 1000 \times 10000\ \text{g cm}^2\,\text{s}^{-2}

1000 \times 10000 = 10^3 \times 10^4 = 10^7

So, $1\ \text{kg m}^2\,\text{s}^{-2} = 10^7\ \text{g cm}^2\,\text{s}^{-2}$.

(b) $1\ \text{m} = \ldots\ \text{ly}$

We need to convert meters (m) to light-years (ly). A light-year is the distance light travels in one Julian year (365.25 days) in a vacuum.

Determine the value of 1 light-year in meters:

We use the formula: Distance = Speed $\times$ Time.
- Speed of light ( $c$ ) is approximately $2.99792458 \times 10^8\ \text{m s}^{-1}$ .
- Time in one Julian year: $1\ \text{year} = 365.25\ \text{days} \times \frac{24\ \text{h}}{1\ \text{day}} \times \frac{60\ \text{min}}{1\ \text{h}} \times \frac{60\ \text{s}}{1\ \text{min}}$ $1\ \text{year} = 365.25 \times 24 \times 60 \times 60\ \text{s} = 31557600\ \text{s}$
Now, calculate $1\ \text{ly}$ in meters:

$1\ \text{ly} = c \times 1\ \text{year}$

$1\ \text{ly} = (2.99792458 \times 10^8\ \text{m s}^{-1}) \times (31557600\ \text{s})$

$1\ \text{ly} \approx 9.4607 \times 10^{15}\ \text{m}$
Perform the conversion from m to ly:

We want to find how many light-years are in $1\ \text{m}$ . We use the conversion factor $\frac{1\ \text{ly}}{9.4607 \times 10^{15}\ \text{m}}$ .

1\ \text{m} \times \left(\frac{1\ \text{ly}}{9.4607 \times 10^{15}\ \text{m}}\right)

1\ \text{m} = \frac{1}{9.4607 \times 10^{15}}\ \text{ly}

1\ \text{m} \approx 1.057 \times 10^{-16}\ \text{ly}

Round to the correct significant figures. The quantities used ( $c$ and the year length) are known to at least 3 significant figures, so the conventional result for this conversion is reported to 3 significant figures:

1\ \text{m} \approx 1.06 \times 10^{-16}\ \text{ly}

(c) $3.0\ \text{m s}^{-2} = \ldots\ \text{km h}^{-2}$

Here, we need to convert meters (m) to kilometers (km) and seconds (s) to hours (h).

Convert m to km:

We know that $1\ \text{km} = 1000\ \text{m}$ .

The conversion factor is $\frac{1\ \text{km}}{1000\ \text{m}}$ .
Convert s $^{-2}$ to h $^{-2}$ :

We know that $1\ \text{h} = 3600\ \text{s}$ .

This means $1\ \text{s} = \frac{1}{3600}\ \text{h}$ .

So, $1\ \text{s}^{-2} = (1\ \text{s})^{-2} = \left(\frac{1}{3600}\ \text{h}\right)^{-2} = (3600)^2\ \text{h}^{-2}$ .

The conversion factor is $\frac{(3600)^2\ \text{h}^{-2}}{1\ \text{s}^{-2}}$ .

Watch out

When converting units with negative exponents (like $\text{s}^{-2}$ ), remember that the conversion factor is also raised to that power. A common mistake is to simply divide by the conversion factor for the base unit. For example, $1\ \text{s}^{-2}$ is NOT $\frac{1}{3600}\ \text{h}^{-2}$ . Instead, $1\ \text{s}^{-2} = (3600)^2\ \text{h}^{-2}$ .
Perform the conversion:

Multiply the given quantity by the conversion factors:

3.0\ \text{m s}^{-2} \times \left(\frac{1\ \text{km}}{1000\ \text{m}}\right) \times \left(\frac{(3600)^2\ \text{h}^{-2}}{1\ \text{s}^{-2}}\right)

Notice how 'm' and 's$^{-2}$' units cancel out:

3.0 \times \frac{1}{1000} \times (3600)^2\ \text{km h}^{-2}

3.0 \times \frac{1}{1000} \times 12960000\ \text{km h}^{-2}

3.0 \times 12960\ \text{km h}^{-2}

38880\ \text{km h}^{-2}

Expressing in scientific notation: $3.888 \times 10^4\ \text{km h}^{-2}$.

4. Round to the correct significant figures.

The given acceleration, $3.0\ \text{m s}^{-2}$ , has only 2 significant figures, so the final answer must be rounded to 2 significant figures:

3.0\ \text{m s}^{-2} \approx 3.9 \times 10^4\ \text{km h}^{-2}

(d) $G = 6.67 \times 10^{-11}\ \text{N m}^2\,(\text{kg})^{-2} = \ldots\ (\text{cm})^3\,\text{s}^{-2}\,\text{g}^{-1}$

This conversion requires an extra step: breaking down the Newton (N) unit into its base SI units.

Express Newton (N) in base SI units:

From Newton's second law, Force = mass $\times$ acceleration ( $F=ma$ ).

So, $1\ \text{N} = 1\ \text{kg} \times 1\ \text{m s}^{-2} = 1\ \text{kg m s}^{-2}$ .
Substitute N into the expression for G:

G = 6.67 \times 10^{-11}\ (\text{kg m s}^{-2})\ \text{m}^2\,(\text{kg})^{-2}

Combine the powers of identical units:

G = 6.67 \times 10^{-11}\ \text{kg}^{(1-2)}\ \text{m}^{(1+2)}\ \text{s}^{-2}

G = 6.67 \times 10^{-11}\ \text{kg}^{-1}\ \text{m}^3\ \text{s}^{-2}

Now, the units are in terms of kg, m, and s, which are easier to convert to g, cm, and s.

3. Convert kg $^{-1}$ to g $^{-1}$ :

We know $1\ \text{kg} = 1000\ \text{g}$ .

So, $1\ \text{kg}^{-1} = (1000\ \text{g})^{-1} = \frac{1}{1000}\ \text{g}^{-1}$ .

Convert m $^3$ to cm $^3$ :

We know $1\ \text{m} = 100\ \text{cm}$ .

So, $1\ \text{m}^3 = (100\ \text{cm})^3 = 100^3\ \text{cm}^3 = 1000000\ \text{cm}^3 = 10^6\ \text{cm}^3$ .
Perform the conversion:

G = 6.67 \times 10^{-11} \times \frac{1}{1000} \times 10^6\ \text{g}^{-1}\ \text{cm}^3\,\text{s}^{-2}

G = 6.67 \times 10^{-11} \times 10^{-3} \times 10^6\ \text{cm}^3\,\text{s}^{-2}\,\text{g}^{-1}

G = 6.67 \times 10^{(-11 - 3 + 6)}\ \text{cm}^3\,\text{s}^{-2}\,\text{g}^{-1}

G = 6.67 \times 10^{-8}\ \text{cm}^3\,\text{s}^{-2}\,\text{g}^{-1}

✓Final answer

1\ \text{kg m}^2\,\text{s}^{-2} = \mathbf{10^7\ \text{g cm}^2\,\text{s}^{-2}}

Method: the conversion-factor chain — for each unit that needs changing, write a fraction equal to 1 (new unit over its equivalent old unit, raised to whatever power the original unit carried), then multiply the given quantity by one such factor per unit. The old units cancel algebraically, leaving only the target units; round the final numeric result to the number of significant figures justified by the given data. The same chain technique handles all four parts below.

(a) $\text{kg m}^2\text{s}^{-2} \to \text{g cm}^2\text{s}^{-2}$ : chain two factors, $\dfrac{10^3\,\text{g}}{1\,\text{kg}}$ and $\left(\dfrac{10^2\,\text{cm}}{1\,\text{m}}\right)^2$ (squared because the original unit is $\text{m}^2$ ). Multiplying through: $1 \times 10^3 \times 10^4 = 10^7\ \text{g cm}^2\text{s}^{-2}$ .
(b) $\text{m} \to \text{ly}$ : first pin down the chain's anchor value, $1\ \text{ly} = c \times 1\ \text{yr} \approx 9.467\times10^{15}\ \text{m}$ , then invert it into a conversion factor $\dfrac{1\,\text{ly}}{9.467\times10^{15}\,\text{m}}$ and multiply: $1\ \text{m} \times$ that factor $\approx 1.057\times10^{-16}\ \text{ly}$ , which rounds to $1.06\times10^{-16}\ \text{ly}$ (3 significant figures).
(c) $\text{m s}^{-2} \to \text{km h}^{-2}$ : chain $\dfrac{1\,\text{km}}{10^3\,\text{m}}$ with the squared time factor $\left(\dfrac{3600\,\text{s}}{1\,\text{h}}\right)^{2}$ (squared because the original unit is $\text{s}^{-2}$ — a negative-power unit needs its conversion factor raised to that same power). Multiplying through: $3.0 \times 10^{-3} \times 3600^2 = 38880 = 3.888\times10^4\ \text{km h}^{-2}$ , which rounds to $3.9\times10^4\ \text{km h}^{-2}$ (2 significant figures, matching the given $3.0\ \text{m s}^{-2}$ ).
(d) $\text{N m}^2\text{kg}^{-2} \to \text{cm}^3\text{s}^{-2}\text{g}^{-1}$ : first break $\text{N}$ into base units, $1\ \text{N} = 1\ \text{kg m s}^{-2}$ , reducing $G$ to $\text{kg}^{-1}\text{m}^3\text{s}^{-2}$ . Then chain $\left(10^3\right)$ for $\text{kg}^{-1}\to\text{g}^{-1}$ and $10^6$ for $\text{m}^3\to\text{cm}^3$ : $6.67\times10^{-11} \times 10^{-3} \times 10^{6} = 6.67\times10^{-8}\ \text{cm}^3\text{s}^{-2}\text{g}^{-1}$ .

✓Final answer

\boxed{10^7\ \text{g cm}^2\text{s}^{-2}}

Common Mistakes in Unit Conversion Problems

Students often lose marks in unit conversion despite knowing the basic method. Here are the most frequent errors and how to avoid them.

Mistake 1: Forgetting to Square or Cube the Conversion Factor

Example from (a): $1\ \text{kg m}^2\,\text{s}^{-2} = \ldots\ \text{g cm}^2\,\text{s}^{-2}$

Wrong approach: Convert $1\ \text{kg} = 1000\ \text{g}$ and $1\ \text{m} = 100\ \text{cm}$ , then write $1000 \times 100 = 10^5$
Why it's wrong: The unit is $\text{m}^2$ , so the conversion factor for length must be squared

Correct method:

$1\ \text{kg} = 10^3\ \text{g}$
$1\ \text{m}^2 = (100\ \text{cm})^2 = 10^4\ \text{cm}^2$
Multiply: $10^3 \times 10^4 = 10^7$

Answer: $\boxed{10^7\ \text{g cm}^2\,\text{s}^{-2}}$

How to avoid: Write the conversion factor with its exponent explicitly:

For $\text{m}^2$ , use $(100\ \text{cm})^2$ , not $100\ \text{cm}$
For $\text{m}^3$ , use $(100\ \text{cm})^3$ , etc.

Mistake 2: Mixing Up Time Unit Conversions (Especially in (c))

Example from (c): $3.0\ \text{m s}^{-2} = \ldots\ \text{km h}^{-2}$

Wrong approach: Convert $1\ \text{s} = \frac{1}{3600}\ \text{h}$ , then $1\ \text{s}^{-2} = \frac{1}{(3600)^2}\ \text{h}^{-2}$ — but students often invert this

Correct method:

$1\ \text{m} = 10^{-3}\ \text{km}$
$1\ \text{s}^{-2} = \left(\frac{1}{3600}\ \text{h}\right)^{-2} = (3600)^2\ \text{h}^{-2} = 1.296 \times 10^7\ \text{h}^{-2}$
Multiply: $3.0 \times 10^{-3} \times 1.296 \times 10^7 = 3.888 \times 10^4$

Answer: $\boxed{3.9 \times 10^4\ \text{km h}^{-2}}$ (rounded to 2 significant figures)

How to avoid: Remember: seconds in denominator → multiply by $(3600)^2$ when converting to hours. A quick check: acceleration in $\text{km h}^{-2}$ should be a large number compared to $\text{m s}^{-2}$ .

Mistake 3: Incorrect Handling of Light Year Conversion (Part b)

Example from (b): $1\ \text{m} = \ldots\ \text{ly}$

Wrong approach: Using $1\ \text{ly} = 9.46 \times 10^{15}\ \text{m}$ , students often write $1\ \text{m} = 9.46 \times 10^{15}\ \text{ly}$ (inverting the factor)

Correct method:

$1\ \text{ly} = 9.46 \times 10^{15}\ \text{m}$
Therefore, $1\ \text{m} = \frac{1}{9.46 \times 10^{15}} = 1.057 \times 10^{-16}\ \text{ly}$

Answer: $\boxed{1.06 \times 10^{-16}\ \text{ly}}$

How to avoid: Always write the known equality first, then divide both sides to get the desired conversion. If $1\ \text{ly} = 9.46 \times 10^{15}\ \text{m}$ , then $1\ \text{m} = \frac{1}{9.46 \times 10^{15}}\ \text{ly}$ .

Mistake 4: Forgetting to Convert All Units in Compound Units (Part d)

Example from (d): $G = 6.67 \times 10^{-11}\ \text{N m}^2\,(\text{kg})^{-2} = \ldots\ (\text{cm})^3\,\text{s}^{-2}\,\text{g}^{-1}$

Wrong approach: Converting only mass and length but forgetting that $\text{N}$ itself contains $\text{kg m s}^{-2}$

Correct method — break down $\text{N}$ first:

$1\ \text{N} = 1\ \text{kg m s}^{-2}$
So $\text{N m}^2\,(\text{kg})^{-2} = (\text{kg m s}^{-2})(\text{m}^2)(\text{kg}^{-2}) = \text{m}^3\,\text{s}^{-2}\,\text{kg}^{-1}$

Now convert each unit:

$1\ \text{m}^3 = (100\ \text{cm})^3 = 10^6\ \text{cm}^3$
$1\ \text{kg}^{-1} = (1000\ \text{g})^{-1} = 10^{-3}\ \text{g}^{-1}$
Multiply: $10^6 \times 10^{-3} = 10^3$

So $G = 6.67 \times 10^{-11} \times 10^3 = 6.67 \times 10^{-8}$

Answer: $\boxed{6.67 \times 10^{-8}\ (\text{cm})^3\,\text{s}^{-2}\,\text{g}^{-1}}$

How to avoid: Always expand compound units like $\text{N}$ into base units ( $\text{kg m s}^{-2}$ ) before converting. This prevents missing hidden conversions.

Mistake 5: Arithmetic Errors with Powers of 10

Students often make errors when adding/subtracting exponents during multiplication of conversion factors.

How to avoid: Write all conversion factors in scientific notation and handle exponents separately:

$10^a \times 10^b = 10^{a+b}$
$10^a \div 10^b = 10^{a-b}$

Double-check your exponent arithmetic — this is where most "silly mistakes" happen.

Quick Checklist Before Submitting

Check	What to verify
✓	Did I square/cube length conversions when needed?
✓	Did I handle time conversions correctly (especially $\text{s}^{-2}$ )?
✓	Did I expand compound units like $\text{N}$ ?
✓	Did I invert the conversion factor correctly?
✓	Did I add/subtract exponents correctly?

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

KEAM 2026Set eng-2026-04214 marksMCQ
$2.0\ ms^{-2} = X\ km\ h^{-2}$
›Reveal solutionSolution

Converting, $2.0\ ms^{-2}=2.592\times10^4\ km\,h^{-2}$ ; keeping two significant figures gives $2.6\times10^4$ .

Conversion factor: $1\ ms^{-2}=\dfrac{10^{-3}\ km}{(1/3600\ h)^2}=\dfrac{3600^2}{1000}\ km\,h^{-2}=\dfrac{12960000}{1000}=12960\ km\,h^{-2}$ .

So $2.0\ ms^{-2}=2.0\times12960=25920\ km\,h^{-2}=2.592\times10^4$ .

The datum $2.0$ has two significant figures, so the answer must be reported to two: $2.6\times10^4$ .

✓Final answer
$The correct option is (A) .$
KEAM 2025Set eng-2025-04264 marksMCQ
$Q. One torr is (A) 1 mm of Hg (B) 1 cm of Hg (C) 76 mm of Hg (D) 100 mm of Hg (E) 76 cm of Hg$
›Reveal solutionSolution

By definition, $1\ \text{torr}=1\ \text{mm Hg}$ .

The torr is a unit of pressure defined as $\dfrac{1}{760}$ of a standard atmosphere, which equals the pressure exerted by a $1\ \text{mm}$ column of mercury. Thus $1$ torr $=1$ mm of Hg.

✓Final answer
$The correct option is (A) .$
KEAM 2025Set pha-2025-0424A4 marksMCQ
$10^{-28}m^2$
›Reveal solutionSolution

The barn is a unit of area for nuclear cross-sections: $1\,\text{barn} = 10^{-28}\,m^2$ (equal to $100\,fm^2$ ).

It is roughly the cross-sectional area of a uranium nucleus. Converting: $1\,\text{barn} = 10^{-24}\,cm^2 = 10^{-28}\,m^2$ . Hence option (A).

✓Final answer
$The correct option is (A) .$
KEAM 2024Set pha-2024-06104 marksMCQ
$Q.In the measurement of length 6 $\mu$m is equal to $x$ pm. Then the value of $x$ is (A) $1.5 \times 10^{-5}$ (B) $1.2 \times 10^{6}$ (C) $3 \times 10^{-6}$ (D) $6 \times 10^{6}$ (E) $2 \times 10^{-12}$$
›Reveal solutionSolution

Convert micrometres to picometres.

$1\,\mu m = 10^{-6}$ m and $1$ pm $= 10^{-12}$ m.
$6\,\mu m = 6\times10^{-6}\text{ m} = \frac{6\times10^{-6}}{10^{-12}}\text{ pm} = 6\times10^{6}\text{ pm}.$
✓Final answer
$x = 6\times10^{6}$

←Previous question1.1 — Fill in the blanks: (a) The volume of a cube of side 1 cm is equal to …→Next question1.3 — A calorie is a unit of heat (energy in transit) and it equals about 4.…

Q.Fill in the blanks by suitable conversion of units:

(a) 1 kg m2 s−2=… g cm2 s−21\ \text{kg m}^2\,\text{s}^{-2} = \ldots\ \text{g cm}^2\,\text{s}^{-2}1 kg m2s−2=… g cm2s−2

(b) 1 m=… ly1\ \text{m} = \ldots\ \text{ly}1 m=… ly

(c) 3.0 m s−2=… km h−23.0\ \text{m s}^{-2} = \ldots\ \text{km h}^{-2}3.0 m s−2=… km h−2

(d) G=6.67×10−11 N m2 (kg)−2=… (cm)3 s−2 g−1G = 6.67 \times 10^{-11}\ \text{N m}^2\,(\text{kg})^{-2} = \ldots\ (\text{cm})^3\,\text{s}^{-2}\,\text{g}^{-1}G=6.67×10−11 N m2(kg)−2=… (cm)3s−2g−1

🎓Unlock everything free for 14 days

(a) $1\ \text{kg m}^2\,\text{s}^{-2} = \ldots\ \text{g cm}^2\,\text{s}^{-2}$

(b) $1\ \text{m} = \ldots\ \text{ly}$

(c) $3.0\ \text{m s}^{-2} = \ldots\ \text{km h}^{-2}$

(d) $G = 6.67 \times 10^{-11}\ \text{N m}^2\,(\text{kg})^{-2} = \ldots\ (\text{cm})^3\,\text{s}^{-2}\,\text{g}^{-1}$

Unlock everything free for 14 days