Hello arjun1110,
Beware of analogies, specially if you dont use the correct one. The electrical capacitance is better modeled as tank's diameter than its height.
In your derivation, you state that Q1=Q2, when you must say V1=V2, being the capacitors in parallel.
If you want convince yourself, you can carge a 1uF capacitor and discharge it over a 1000uF.
If you prefer a more mathematical way, proceed calculating the initial stored energy in the small capacitor and the sum of energies on both after discharging.
With the given values, E1= (1 e-6 F * 10V**2)/2 = 50uJoule
After discharging, and assuming as you say that both capacitors have 5 V across them, the energy stored in the big capacitor will be:
E2 = (1 e-3 F * 5V**2)2 = 12.5 mJoule
Even desregarding the energy still stored in the small capacitor, the final energy in the system is orders of magnitude greater than the initial energy.
Regards