1. Charge left on the sphere after detachment

Initially, the solid sphere has total charge $Q+q$. A tiny piece carrying charge $q$ gets detached, so the remaining spherical portion has charge $Q.$

Since the detached piece is very small, we can treat it as a point charge $q$.

2. Electric force on the detached piece

After detachment, the remaining body is assumed spherical with radius $R$ and charge $Q$. For a point outside a uniformly charged sphere, the electric field is the same as if all charge were concentrated at the center.

If the piece has fallen vertically downward by distance $y$, its distance from the center is $r=R+y.$

So the electric force on charge $q$ is $F_e = \frac{1}{4\pi\varepsilon_0}\frac{qQ}{(R+y)^2}.$

Since the piece is below the sphere and both charges are positive, this force acts downward, same as gravity.

Gravity also acts downward with force $F_g=mg.$

3. Use work-energy theorem

Initially, the piece starts from rest at the bottom of the sphere, so initial kinetic energy is zero. After falling through height $y$, its speed is $v$.

Hence, $\frac12 mv^2 = W_g + W_e.$

Work done by gravity

$W_g = mgy.$

Work done by electric force

Electric force varies with position, so $W_e = \int_{R}^{R+y} \frac{1}{4\pi\varepsilon_0}\frac{qQ}{r^2}\,dr.$

Now, $W_e = \frac{qQ}{4\pi\varepsilon_0}\int_R^{R+y} \frac{dr}{r^2}$ $= \frac{qQ}{4\pi\varepsilon_0}\left[-\frac{1}{r}\right]_R^{R+y}$ $= \frac{qQ}{4\pi\varepsilon_0}\left(\frac{1}{R}-\frac{1}{R+y}\right).$

Simplify: $\frac{1}{R}-\frac{1}{R+y} = \frac{y}{R(R+y)}.$

So, $W_e = \frac{qQ}{4\pi\varepsilon_0}\cdot \frac{y}{R(R+y)}.$

4. Substitute into energy equation

$\frac12 mv^2 = mgy + \frac{qQ}{4\pi\varepsilon_0}\cdot \frac{y}{R(R+y)}.$

Multiply by $2/m$: $v^2 = 2gy + \frac{2}{m}\cdot \frac{qQ}{4\pi\varepsilon_0}\cdot \frac{y}{R(R+y)}.$

Factor out $2y$: $v^2 = 2y\left[g + \frac{qQ}{4\pi\varepsilon_0 R(R+y)m}\right].$

5. Match with options

This exactly matches: $\boxed{v^2 = 2y\left[ \frac{qQ}{4\pi\varepsilon_0 R(R+y)m} + g \right]}$

So the correct option is $\boxed{\text{C}}.$

6. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

They agree.