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Step-by-Step Solution
Step 1: Recall the formula for the period of a simple pendulum
The period $T$ of a simple pendulum of (effective) length $l$ is given by
$$
T = 2\pi \sqrt{\frac{l}{g}}.
$$
Step 2: Recognize the relationship for a small change in length
For small changes in $l$, the resulting small change in $T$ obeys the proportional relationship:
$$
\frac{\Delta T}{T} \;=\; \frac{1}{2}\,\frac{\Delta l}{l}.
$$
This comes from differentiating $T = 2\pi\sqrt{l/g}$ with respect to $l$.
Step 3: Identify the given data
The length of the pendulum (from pivot to the center of the spherical bob) is approximately $l = 1\,\text{m}$.
The measured difference in the period is given as $\Delta T = 5 \times 10^{-4}\,\text{s}$.
The actual period for $l \approx 1\,\text{m}$ is about $T \approx 2\,\text{s}$ (since $2\pi \sqrt{1/9.8} \approx 2\,\text{s}$).
Step 4: Plug values into the relationship
Using
$$
\frac{\Delta T}{T} \;=\; \frac{1}{2}\,\frac{\Delta l}{l},
$$
substitute $\Delta T = 5 \times 10^{-4}\,\text{s}$ and $T \approx 2\,\text{s}$, $l = 1\,\text{m}$.
$$
\frac{5 \times 10^{-4}}{2} \;=\; \frac{1}{2} \times \frac{\Delta l}{1}.
$$
Step 5: Solve for the difference in length $\Delta l$
Simplify the left-hand side:
$$
2.5 \times 10^{-4} \;=\; \frac{1}{2}\,\Delta l.
$$
Hence,
$$
\Delta l \;=\; 2 \times \bigl(2.5 \times 10^{-4}\bigr)
\;=\; 5 \times 10^{-4}\;\text{m}.
$$
Step 6: Convert the length difference into centimeters
Since $1\,\text{m} = 100\,\text{cm}$, we get
$$
\Delta l
\;=\; 5 \times 10^{-4}\,\text{m}
\;\times\; 100\,\frac{\text{cm}}{\text{m}}
\;=\; 5 \times 10^{-2}\,\text{cm}
\;=\; 0.05\,\text{cm}.
$$
Therefore, the difference in the radii, $\bigl|r_1 - r_2\bigr|$, is $0.05\,\text{cm}.$
Final Answer
The best estimate for the difference in the radii of the bobs is 0.05 cm.
