A pendulum is formed by attaching a hollow sphere of radius 2 cm to...

The distance from the pivot to the center of mass of the hollow sphere is $d = L + R = 18 + 2 = 20 \text{ cm} = 0.2 \text{ m}$.

The moment of inertia of the hollow sphere about the pivot is given by the parallel axis theorem:

$I = I_{cm} + md^2 = \dfrac{2}{3}mR^2 + md^2$

The time period of the physical pendulum is:

$T = 2\pi \sqrt{\dfrac{I}{mgd}} = 2\pi \sqrt{\dfrac{\dfrac{2}{3}mR^2 + md^2}{mgd}} = 2\pi \sqrt{\dfrac{d}{g} \left(1 + \dfrac{2R^2}{3d^2}\right)}$

The time period of a simple pendulum of length $d$ is $T_0 = 2\pi \sqrt{\dfrac{d}{g}}$.

Using the binomial expansion for $R \ll d$, we get:

T \approx T_0 \left(1 +\dfrac{R^2}{3d^2}\right)

Substituting $R = 2 \text{ cm}$ and $d = 20 \text{ cm}$:

\dfrac{R^2}{3d^2} = \dfrac{2^2}{3 \times 20^2} = \dfrac{4}{1200} = \dfrac{1}{300} \approx$0.0033$

Thus, the time period $T$ is larger than $T_0$ by approximately $0.33\%$, which is about $0.3\%$.

Calculating $T_0$ using $g \approx \pi^2 \text{ m/s}^2$:

$T_0 \approx 2\pi$\sqrt{\dfrac{0.2}{\pi^2}} =$2$\sqrt{0.2} \approx$0.894 \text{ s}$

Therefore, the time period is approximately $0.89 \text{ s}$, and it is about $0.3\%$ larger than the simple pendulum value.