A block of mass 250 g is placed on a vertical spring with a spring...

Let $h$ be the maximum height the block rises from its initial compressed position.

According to the law of conservation of mechanical energy, the initial elastic potential energy of the spring is converted into the gravitational potential energy of the block at the maximum height.

$\dfrac{1}{2}kx^2 = mgh$

Given:
$m = 250 \text{ g} = 0.25 \text{ kg}$
$k = 100 \text{ N/m}$
$x = 10 \text{ cm} = 0.1 \text{ m}$
$g = 10 \text{ m/s}^2$

Substituting the values into the energy equation:

$\dfrac{1}{2} \times 100 \times (0.1)^2 = 0.25 \times 10 \times h$

$50 \times 0.01 = 2.5h$

$0.5 = 2.5h$

$h = \dfrac{0.5}{2.5} \text{ m}$

$h = 0.2 \text{ m} = 20 \text{ cm}$