Given data

• Length of pendulum: $L = 250\text{ cm} = 2.5\text{ m}$
• Mass of bob: $m = 200\text{ g} = 0.2\text{ kg}$
• Angular displacement: $\theta = 60^\circ$
• Acceleration due to gravity: $g = 10\text{ m/s}^2$

Concept used The bob is released from rest at $60^\circ$ from the vertical. Its maximum speed occurs at the lowest point.

We use conservation of mechanical energy: $mgh = \frac{1}{2}mv^2$

Find the vertical height fallen by the bob For a pendulum displaced by angle $\theta$ from vertical, the rise in height above the lowest point is: $h = L(1-\cos\theta)$

Substituting values: $h = 2.5(1-\cos60^\circ)$ $\cos60^\circ = \frac{1}{2}$ $h = 2.5\left(1-\frac{1}{2}\right) = 2.5 \times 0.5 = 1.25\text{ m}$

Apply energy conservation $mg(1.25) = \frac{1}{2}mv^2$

Mass cancels out: $g(1.25) = \frac{v^2}{2}$

$10 \times 1.25 = \frac{v^2}{2}$ $12.5 = \frac{v^2}{2}$ $v^2 = 25$ $v = 5\text{ m/s}$

Final answer The maximum velocity attained by the bob is: $\boxed{5}$

Comparison with stored answer Stored correct answer = $5$

This matches the derived answer.