1. Given data

• Mass of ball: $m = 100\,\text{g} = 0.1\,\text{kg}$
• Dropping height: $h = 10\,\text{cm} = 0.1\,\text{m}$
• Additional depression of spring: $x = \dfrac{h}{2} = 5\,\text{cm} = 0.05\,\text{m}$
• Acceleration due to gravity: $g = 10\,\text{m s}^{-2}$

2. Physical idea

The ball is dropped from height $h$ above the platform. After touching the platform, it continues to move downward and compresses the spring by $x = h/2$.

At the instant of maximum compression, the ball momentarily comes to rest, so kinetic energy is zero.

Hence, using conservation of mechanical energy:

$\text{Loss in gravitational potential energy} = \text{Gain in spring potential energy}$

3. Total downward displacement of the ball

The ball first falls through height $h$, then further moves down by $x$ while compressing the spring.

So total loss in gravitational potential energy is

$mg(h+x)$

4. Spring potential energy at maximum compression

$U_s = \frac{1}{2}kx^2$

Thus,

$mg(h+x)=\frac{1}{2}kx^2$

5. Substitute values

Here,

$m=0.1, \quad g=10, \quad h=0.1, \quad x=0.05$

So,

$0.1\cdot 10\cdot (0.1+0.05)=\frac{1}{2}k(0.05)^2$

$1\cdot 0.15 = \frac{1}{2}k(0.0025)$

$0.15 = 0.00125k$

Therefore,

$k=\frac{0.15}{0.00125}=120\,\text{N m}^{-1}$

6. Final answer

$\boxed{120}$

7. Comparison with stored correct answer

Stored correct answer = $120$

My derived answer = $120$

So the answer matches.