Elimination of OH from spirocyclic compounds
This question has been unanswered for quite long, and meanwhile, I have found an answer, so I wanted to share it.

On treating the compound with sulfuric acid, the $\ce{OH}$ groups will be protonated to $\ce{OH_2^+}$.



The $\ce{OH_2^+}$ groups are good leaving groups, so they will leave as $\ce{H_2O}$, forming carbocation.(one by one)


As cyclopropane is under significant ring strain, it will undergo ring expansion to form cyclobutane ring, which is somewhat less strained. At the same time, $\ce{H^+}$ ions will be expelled, forming double bonds. (It should also take place step by step.)


In this case I don't think that the double bond will not cause any significant barrier to the formation of the molecules, as the double bond is here the bridging bond, not at the bridgehead[formation of double bonds at bridgehead is not possible for smaller rings(&lt;8 carbon rings), as it causes very high strain: Bredt's rule]

I have considered only elimination in this reaction scheme. Other compounds may also be formed. Feel free to post answers if you think other compounds will be formed or if you think the above mechanism is wrong in any way.