JEE PYQ: Units & Measurements - Question ID d367a95f34ce (JEE Main 2018)

In the chapter Units & Measurements, we study how errors propagate through calculations involving measured quantities. When a physical quantity depends on another measured quantity raised to some power, the relative error in the derived quantity is related to the relative error in the measured quantity by the exponent.

For a sphere:

• The surface area $S$ is given by: $S = 4\pi r^2$ where $r$ is the radius.
• The volume $V$ is given by: $V = \frac{4}{3}\pi r^3$

The key concept here is error propagation in power functions. If a quantity $Q$ depends on a measured quantity $x$ as $Q = kx^n$, where $k$ is a constant, then the relative error in $Q$, denoted $\frac{\Delta Q}{Q}$, is related to the relative error in $x$, $\frac{\Delta x}{x}$, by: $\frac{\Delta Q}{Q} = |n| \cdot \frac{\Delta x}{x}$ This means the relative error scales linearly with the exponent.

In this problem, we are given the relative error in the surface area, $\alpha = \frac{\Delta S}{S}$, and we are to find the relative error in the volume, $\frac{\Delta V}{V}$.

Let’s denote:

• $r$: radius of the sphere
• $\Delta r$: absolute error in radius
• $S = 4\pi r^2$: surface area
• $V = \frac{4}{3}\pi r^3$: volume

Step 1: Express relative error in surface area

We are given that $\frac{\Delta S}{S} = \alpha$. So, $\alpha = \frac{2 \Delta r}{r} \quad \Rightarrow \quad \frac{\Delta r}{r} = \frac{\alpha}{2}$ This gives us the relative error in the radius.

Step 2: Express relative error in volume

We already found $\frac{\Delta r}{r} = \frac{\alpha}{2}$. Substituting: $\frac{\Delta V}{V} = 3 \cdot \frac{\alpha}{2} = \frac{3}{2} \alpha$

Step 3: Match with given options

The relative error in volume is $\frac{3}{2} \alpha$, which corresponds to option A.

Students often make the following mistakes:

• Incorrect exponent application: Some confuse the exponent in surface area (2) with that in volume (3) and apply the wrong scaling factor. For example, they might think $\frac{\Delta V}{V} = \frac{2}{3} \alpha$, which is option B — a common distractor.
• Forgetting to relate errors through radius: Instead of expressing both $\frac{\Delta S}{S}$ and $\frac{\Delta V}{V}$ in terms of $\frac{\Delta r}{r}$, some try to relate $\Delta S$ and $\Delta V$ directly, leading to confusion.
• Sign errors or absolute values: While relative errors are positive, students sometimes carry negative signs or forget that errors are magnitudes.

Exam Tip: Always express the relative error of the derived quantity in terms of the relative error of the measured quantity using the power rule. In this case, since $S \propto r^2$ and $V \propto r^3$, the relative errors scale as 2 and 3 times $\frac{\Delta r}{r}$, respectively. Use this to connect $\alpha$ to $\frac{\Delta V}{V}$.

Final Answer: A: $\frac{3}{2} \alpha$