Q.A new system of units is proposed in which unit of mass is $\alpha$ kg, unit of length $\beta$ m and unit of time $\gamma$ s. How much will 5 J measure in this new system?

The key idea is Unit Conversion, which involves expressing a physical quantity's value from one system of units to another by relating their fundamental dimensions.

1. The dimensional formula for energy (Joule) is $[E] = ML^2T^{-2}$. In SI units, $1 \text{ J} = 1 \text{ kg} \cdot (1 \text{ m})^2 \cdot (1 \text{ s})^{-2}$.
2. From the problem statement, the new units are $M' = \alpha \text{ kg}$, $L' = \beta \text{ m}$, and $T' = \gamma \text{ s}$. This implies $1 \text{ kg} = \frac{1}{\alpha} M'$, $1 \text{ m} = \frac{1}{\beta} L'$, and $1 \text{ s} = \frac{1}{\gamma} T'$.
3. Substitute these relations into the expression for 1 Joule: …

When we change the system of units, the fundamental physical quantity itself does not change; only its numerical measure and the unit used to express it change. For instance, 1 meter is the same length as 100 centimeters. The length is constant, but the number changed from 1 to 100 because the unit changed from meter to centimeter.

This principle is captured by the relation $n_1 U_1 = n_2 U_2$, where $n_1$ is the numerical value in the first system with unit $U_1$, and $n_2$ is the numerical value in the second system with unit $U_2$. Our goal is to find $n_2$.

Let's break down the problem:

1. Identify the physical quantity and its dimensional formula.

   The quantity in question is Energy, given as 5 J.

   The dimensional formula for energy is $[M L^2 T^{-2}]$. This means that the unit of energy is derived from the unit of mass, the square of the unit of length, and the inverse square of the unit of time.

2. Define the base units in the old (SI) system.

   In the standard SI system:

   • Unit of mass ($M_{old}$) = 1 kg
   • Unit of length ($L_{old}$) = 1 m
   • Unit of time ($T_{old}$) = 1 s The unit of energy in this system is the Joule (J), where $1 \text{ J} = 1 \text{ kg} \cdot (1 \text{ m})^2 \cdot (1 \text{ s})^{-2}$. The given value is $n_{old} = 5$ and $U_{old} = \text{J}$.

3. Define the base units in the new system.

   As per the proposal:

   • New unit of mass ($M_{new}$) = $\alpha$ kg
   • New unit of length ($L_{new}$) = $\beta$ m
   • New unit of time ($T_{new}$) = $\gamma$ s

4. Express the new unit of energy ($U_{new}$) in terms of the new base units.

   Since the dimensions of energy are $[M L^2 T^{-2}]$, the new unit of energy will be:

   $U_{new} = (M_{new}) (L_{new})^2 (T_{new})^{-2}$

5. Substitute the definitions of the new base units to express $U_{new}$ in terms of the old (SI) base units.

   Substitute the expressions from Step 3 into the formula from Step 4:

   $U_{new} = (\alpha \text{ kg}) (\beta \text{ m})^2 (\gamma \text{ s})^{-2}$

   $U_{new} = \alpha \text{ kg} \cdot \beta^2 \text{ m}^2 \cdot \gamma^{-2} \text{ s}^{-2}$

   Group the numerical factors and the old units:

   $U_{new} = (\alpha \beta^2 \gamma^{-2}) (\text{kg m}^2 \text{ s}^{-2})$

   From Step 2, we know that $1 \text{ kg m}^2 \text{ s}^{-2} = 1 \text{ J}$.

   So, $U_{new} = (\alpha \beta^2 \gamma^{-2}) \text{ J}$. …

Concept: Conversion of Units (Dimensional Analysis)

When we change from one system of units to another, the numerical value of a physical quantity changes inversely with the size of the unit. The key formula is:

where $a, b, c$ are the dimensions of the quantity in mass, length, and time respectively.

Method: Dimensional Conversion Method

Step 1: Write the dimensional formula of the quantity

Energy (Joule) has dimensions:

So, $a = 1$, $b = 2$, $c = -2$.

Step 2: Identify the units in both systems

• Old system (SI): $M_1 = 1\ \text{kg}$, $L_1 = 1\ \text{m}$, $T_1 = 1\ \text{s}$
• New system: $M_2 = \alpha\ \text{kg}$, $L_2 = \beta\ \text{m}$, $T_2 = \gamma\ \text{s}$ …

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

1. Forgetting the Dimensional Formula of Energy

The Mistake:

Students often try to convert joules directly using only the given units (mass, length, time) without first recalling that energy has a specific dimensional formula.

Why it’s wrong:

Joule is a derived unit. You cannot convert it using just mass, length, and time individually — you must use the correct combination:

How to avoid:

Always write the dimensional formula of the physical quantity before starting the conversion. For energy, it’s always $M^1 L^2 T^{-2}$.

2. Mixing Up the Conversion Factor Direction

The Mistake:

Students incorrectly set up the ratio — for example, writing

instead of raising each factor to the correct power.

Why it’s wrong:

The conversion formula is:

where $a, b, c$ are the exponents from the dimensional formula. For energy, $a=1$, $b=2$, $c=-2$.

How to avoid:

Memorise the standard conversion formula:

New numerical value = Old numerical value × (Old unit / New unit) raised to respective powers.

Then plug in the exponents carefully.

3. Incorrectly Substituting the Given Units

The Mistake:

Students write $M_1 = 1\ \text{kg}$, $M_2 = \alpha\ \text{kg}$, but then treat $\alpha$ as a number without units — or worse, invert the ratio.

Why it’s wrong:

The new unit of mass is $\alpha$ kg, meaning 1 new mass unit = $\alpha$ kg. So:

How to avoid:

Write clearly:

• Old system: 1 kg, 1 m, 1 s
• New system: $\alpha$ kg, $\beta$ m, $\gamma$ s

Then the ratios become:

4. Forgetting the Negative Exponent for Time

The Mistake:

Students treat the time exponent as positive, writing $(\gamma)^2$ instead of $(\gamma)^{-2}$.

Why it’s wrong: …