1. For a simple pendulum, the time period is

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

where $L$ is the length and $g$ is acceleration due to gravity.

2. Initially, the time period is given as $T_0$:

$T_0 = 2\pi \sqrt{\frac{L}{g}}$

3. The new length is reduced to $\frac{1}{16}$ times the original length:

$L' = \frac{L}{16}$

4. The new time period becomes

$T' = 2\pi \sqrt{\frac{L'}{g}} = 2\pi \sqrt{\frac{L/16}{g}}$

$T' = 2\pi \sqrt{\frac{L}{g}} \cdot \sqrt{\frac{1}{16}}$

$T' = T_0 \cdot \frac{1}{4}$

5. Therefore,

$T' = \frac{T_0}{4}$

6. Checking options:

• A: $4T_0$ ❌
• B: $\frac{1}{4}T_0$ ✅
• C: $T_0$ ❌
• D: $8\pi T_0$ ❌

Hence, the correct answer is Option B.