Animated Solution for Physics - Physics and Measurement: Comprehension Passage

In the realm of experimental physics, no measurement is absolutely perfect. Every instrument has a least count, and every observation carries a tiny shadow of uncertainty. But what happens when we use these imperfect measurements to calculate something else? How does the error "propagate" through our mathematical formulas?

This is where the true elegance of calculus shines. By treating small errors as differentials, we can use the machinery of derivatives to see exactly how an error in one variable amplifies or diminishes when it passes through a function. The passage provided introduces us to this concept using series expansions, but as we will see, logarithmic differentiation is often the most powerful weapon in our arsenal.

A Dimensionless Ratio

Imagine you are in a lab, and you have measured a dimensionless quantity $a$ with a small uncertainty $\Delta a$. You need to calculate a ratio $r$ defined by the equation:

Our mission is to find the resulting error $\Delta r$. While we could use the quotient rule directly, there is a much more elegant path. Whenever you see complex fractions, products, or powers, taking the natural logarithm is like a magic wand that simplifies the landscape.

Let's take the natural logarithm on both sides:

Notice how the division has beautifully transformed into a simple subtraction. Now, we differentiate both sides. Remember the chain rule: the derivative of $\ln (1-a)$ is $\frac{1}{1-a}$ multiplied by the derivative of $(1-a)$, which is $-1$.

In the language of error analysis, we replace the exact differentials $dr$ and $da$ with our small, discrete errors $\Delta r$ and $\Delta a$.

Now, we just need to perform some basic algebra. Taking the lowest common multiple for the terms inside the bracket:

The $+a$ and $-a$ in the numerator cancel out perfectly, leaving a $2$. The denominator becomes a difference of squares, $1-a^2$.

To find the absolute error $\Delta r$, we multiply this relative error by the original expression for $r$:

The $(1-a)$ term in the numerator cancels with the one in the denominator, yielding our final, pristine result:

This perfectly matches option (b). The negative sign simply indicates that an increase in $a$ leads to a decrease in $r$, which makes perfect sense if you look at the original function!

Radioactive Decay

Now, let's shift gears from pure algebra to the fascinating world of modern physics. We are dealing with a sample of radioactive nuclei.

We are given that the initial number of nuclei is exactly $N_0=3000$. In the first $1.0\text{ s}$, $1000\pm 40$ nuclei decay.

Here is the crucial catch where many students make a silly mistake: The radioactive decay law, $N=N_0{e}^{-\lambda t}$, uses $N$ to represent the number of nuclei remaining, not the number that have decayed!

So, we must first calculate the remaining active nuclei:

Since the initial count of 3000 is an exact number, the uncertainty in the remaining nuclei $N$ is entirely due to the uncertainty in the decayed amount. Therefore, $\Delta N=40$.

We need to find the error in the decay constant, $\Delta \lambda$. Let's write down our master equation:

Once again, the natural logarithm comes to our rescue to bring the exponent down:

Now, we differentiate this equation. Since $N_0$ and $t$ are constants in this specific context (we are looking at the error in $\lambda$ caused by the error in $N$ at a fixed time $t=1$), their derivatives are zero.

Rearranging to isolate the differential of the decay constant:

We take the absolute value because, in error analysis, we are interested in the maximum possible uncertainty. Replacing the differentials with our discrete errors:

This is our final working formula. It tells a beautiful physical story: the fractional error in the decay constant is directly proportional to the fractional error in the remaining nuclei.

Let's substitute the values we carefully extracted from the problem:

The uncertainty in the decay constant is $0.02{\text{ s}}^{-1}$, which matches option (c).

By mastering logarithmic differentiation, you transform terrifying error propagation problems into simple, elegant algebraic steps. Keep practicing, and you will start seeing these patterns everywhere in physics!