In an experiment to determine the period of a simple pendulum of length 1 m, it is attached to different spherical bobs of radii $$r_1$$ and $$r_2$$. The two spherical bobs have uniform mass distribution. If the relative difference in the periods, is found to be $$5 \times 10^{-4}$$ s, the difference in radii, $$|r_1 - r_2|$$ is best-given by

We know that for small-amplitude oscillations the time-period of a simple pendulum is given by the formula

$$T = 2\pi \sqrt{\dfrac{L}{g}}\;,$$

where $$L$$ is the distance between the point of suspension and the centre of mass of the bob, and $$g$$ is the acceleration due to gravity.

In the laboratory the length of the string is adjusted to be exactly $$1\ \text{m}$$. If the string is tied to the surface of a spherical bob of radius $$r$$, the centre of the sphere lies a distance $$r$$ below the tie-point. Hence the effective pendulum lengths are

$$L_1 = 1 + r_1,\qquad L_2 = 1 + r_2.$$

Because the radii are very small compared with $$1\ \text{m}$$, the two periods $$T_1$$ and $$T_2$$ differ only slightly. To connect a small change in period with a small change in length, we differentiate the period formula. Writing $$T = 2\pi \sqrt{L/g}$$, we get

$$\dfrac{dT}{dL} = 2\pi \left(\dfrac{1}{2}\right)\dfrac{1}{\sqrt{gL}} = \dfrac{\pi}{\sqrt{gL}}.$$

For a small finite difference we can therefore write

$$\Delta T \;=\; \dfrac{\pi}{\sqrt{gL}}\;\Delta L,$$

with $$\Delta T = |T_1 - T_2|$$ and $$\Delta L = |L_1 - L_2| = |r_1 - r_2|.$$

The problem states that the observed difference in periods is

$$\Delta T = 5 \times 10^{-4}\ \text{s}.$$

Because the radii are very small, we may put $$L \approx 1\ \text{m}$$ in the square root. Using $$g \approx 9.8\ \text{m s}^{-2}$$ we have

$$\sqrt{gL} = \sqrt{9.8 \times 1} \approx 3.13\ \text{m}^{1/2}\text{s}^{-1}.$$

Substituting into the linearised relation gives

$$|r_1 - r_2| \;=\; \Delta L = \dfrac{\sqrt{gL}}{\pi}\;\Delta T = \dfrac{3.13}{3.1416}\;\bigl(5 \times 10^{-4}\bigr)\ \text{m}.$$

Carrying out the numerical calculation,

$$|r_1 - r_2| \approx 0.000998 \times 5 \times 10^{-4}\ \text{m} \approx 4.98 \times 10^{-4}\ \text{m}.$$

Converting to centimetres (remembering that $$1\ \text{m} = 100\ \text{cm}$$),

$$|r_1 - r_2| \approx 4.98 \times 10^{-4}\ \text{m} \times 100 = 4.98 \times 10^{-2}\ \text{cm} \approx 0.05\ \text{cm}.$$

Hence, the correct answer is Option B.