1. Effective length of the pendulum

For a simple pendulum, the time period is

$T = 2\pi \sqrt{\frac{L}{g}}$

Here, the bob is spherical, so the pendulum length is measured from the point of suspension to the center of the bob.

If the string length is $1\,\text{m}$ and bob radius is $r$, then effective length is

$L = 1 + r$

So for the two bobs,

$T_1 = 2\pi \sqrt{\frac{1+r_1}{g}}, \qquad T_2 = 2\pi \sqrt{\frac{1+r_2}{g}}$

2. Use small change approximation

Since $r_1$ and $r_2$ are small compared to $1\,\text{m}$, we use

$\Delta T \approx \frac{dT}{dL}\,\Delta L$

From

$T = 2\pi \sqrt{\frac{L}{g}}$

we get

$\frac{dT}{dL} = 2\pi \cdot \frac{1}{2\sqrt{gL}} = \frac{\pi}{\sqrt{gL}}$

At $L \approx 1\,\text{m}$,

$\Delta T \approx \frac{\pi}{\sqrt{g}}\,|r_1-r_2|$

Given the difference in periods is

$|T_1-T_2| = 5\times 10^{-4}\,\text{s}$

therefore,

$5\times 10^{-4} = \frac{\pi}{\sqrt{g}}\,|r_1-r_2|$

So,

$|r_1-r_2| = \frac{5\times 10^{-4}\sqrt{g}}{\pi}$

Taking $g \approx 9.8\,\text{m/s}^2$,

$\sqrt{g} \approx 3.13$

Thus,

$|r_1-r_2| \approx \frac{5\times 10^{-4}\times 3.13}{3.14} \approx 5\times 10^{-4}\,\text{m}$

Converting to cm:

$5\times 10^{-4}\,\text{m} = 0.05\,\text{cm}$

3. Check options

• A: $1\,\text{cm}$
• B: $0.05\,\text{cm}$
• C: $0.5\,\text{cm}$
• D: $0.01\,\text{cm}$

Hence, the best answer is

$\boxed{0.05\,\text{cm}}$

So, Option B is correct.