JEE PYQ: Units & Measurements - Question ID 0ce94ca62841 (JEE Main 2019)

The question tests the fundamental concept of dimensional homogeneity and unit conversion. Density is defined as mass per unit volume: $\rho = \frac{m}{V}$ where $\rho$ is density, $m$ is mass, and $V$ is volume.

When units change, the numerical value of a physical quantity changes, but the physical quantity itself remains invariant. To find the new numerical value in a different unit system, we use the principle: $\text{Numerical value} \times \text{Unit} = \text{Constant}$ Thus, if the unit changes, the numerical value must adjust to keep the product constant.

The key formula for converting the numerical value of density from one unit system to another is: $$\rho_{\text{new}} = \rho_{\text{old}} \times \left(\frac{\text{Unit of mass}_{\text{old}}}{\text{Unit of mass}_{\text{new}}}\right) \times \left(\frac{\text{Unit of length}_{\text{new}}}{\text{Unit of length}_{\text{old}}}\right)^3$$ This accounts for the fact that volume depends on the cube of length.

Step 1: Identify given data
- Density in SI units: $\rho_{\text{SI}} = 128\ \text{kg m}^{-3}$
- New unit of length: $L_{\text{new}} = 25\ \text{cm}$
- New unit of mass: $M_{\text{new}} = 50\ \text{g}$
- We need to find the numerical value of density in the new unit system.

Step 2: Convert all units to SI for consistency
- $1\ \text{cm} = 10^{-2}\ \text{m} \Rightarrow L_{\text{new}} = 25\ \text{cm} = 25 \times 10^{-2}\ \text{m} = 0.25\ \text{m}$
- $1\ \text{g} = 10^{-3}\ \text{kg} \Rightarrow M_{\text{new}} = 50\ \text{g} = 50 \times 10^{-3}\ \text{kg} = 0.05\ \text{kg}$

Step 3: Express density in terms of new units
Density in new units is: $\rho_{\text{new}} = \frac{\text{Mass in new units}}{\text{Volume in new units}} = \frac{M_{\text{new}}}{(L_{\text{new}})^3}$ But we want the numerical value of $\rho$ when expressed in these new units. Let this numerical value be $n$. Then: $\rho = n \times \left(\frac{M_{\text{new}}}{(L_{\text{new}})^3}\right)$ But $\rho$ is also equal to $128\ \text{kg m}^{-3}$. So: $128\ \text{kg m}^{-3} = n \times \left(\frac{0.05\ \text{kg}}{(0.25\ \text{m})^3}\right)$

Step 4: Solve for $n$
First, compute $(0.25\ \text{m})^3$: $(0.25)^3 = 0.015625\ \text{m}^3$ Now, compute the denominator: $\frac{0.05}{0.015625} = 3.2\ \text{kg m}^{-3}$ So: $128 = n \times 3.2 \Rightarrow n = \frac{128}{3.2} = 40$

Step 5: Conclusion
The numerical value of the density in the new unit system is $40$, which corresponds to option A.

Trap 1: Forgetting to cube the length conversion factor.
Since volume depends on length cubed, students often mistakenly use the length conversion factor linearly instead of cubing it. This leads to incorrect results like $640$ (option B), which is a common distractor.

Trap 2: Incorrect unit conversion.
Mixing up grams and kilograms or centimeters and meters without proper conversion factors can lead to wrong values. Always convert all units to a consistent system (preferably SI) before applying formulas.

Exam Tip:
When converting units, write down each step clearly. Use dimensional analysis to verify that the units cancel out correctly. For density, ensure that mass and volume units are consistent and that the length unit is raised to the power of 3.