Hard cone question

Can someone talk me through this question step by step please

I got a weird answer and haven't done differential modelling in a while but here was my method

1 - Find the relationship between h and r (Triangles mean that r = htan(30), therefore r = (1/sqrt(3))h
2 - Use this relationship to find one between V and h by subbing r = (1/sqrt(3))h into V = (1/3)(pi)(r^2)(h) to get V = (1/9)(pi)(h^3)
3 - Now differentiate this with respect to h to get dV/dh = (1/3)(pi)(h^2)
4 - dV/dt = dV/dh x dh/dt, therefore we can now sub in dV/dh = (1/3)(pi)(h^2) to get dV/dt = (1/3)(pi)(h^2) x dh/dt
5 - Sub this into the equation the question gives us ( dV/dt = -2h ) to get (1/3)(pi)(h^2) x dh/dt = -2h
6 - divide both sides by h to get all of the h's on one side : (1/3)(pi)(h) x dh/dt = -2
7 - "Multiply" both sides by dt to get (1/3)(pi)(h) dh = -2 dt and add integration signs
8 - Integrate both sides with their respective variables to get (1/6)(pi)(h^2) = -2t + C
9 - Sub in the given initial value of h = 50 when t = 0 to get (1250/3)pi = C, now you have the complete equation (1/6)(pi)(h^2) = -2t + (1250/3)pi
10 - We want to find the time taken for the tank to become empty, so sub in h = 0 to get 0 = -2t + (1250/3)pi
11 - Add 2t to both sides to get 2t = (1250/3)pi
12 - Divide by 2 to solve for T and get T = (625/3)pi

I may have made a mistake somewhere and my explanation isn't the best but i'm pretty sure thats the answer