Chemistry - Heats of combustion and stability of rings
Solution 1:
Compounds A-D all have the same molecular formula, $\ce{C6H12}$. We can burn each compound and measure the heat given off (heat of combustion). Since they are isomers, they will each burn according to the same equation
$$\ce{C6H12 + 9O2 -> 6CO2 + 6H2O + heat}$$
Any differences in the heat given off can be used to say that a compound is more stable (it had a lower energy to begin with, so less heat is given off) or less stable (it had a higher energy to begin with, so more heat is given off).
This link provides the heats of combustion for some useful model compounds. Look at the last column ("Total Strain") in Table I, it shows that cyclopropane is slightly more strained than cyclobutane, while cyclopentane and cyclohexane are both much less strained. The strain energy (SE) in cyclopropane will not change appreciably when we add a propyl group to the ring, nor will the SE in cyclobutane change appreciably when we add an ethyl group to the ring.
Therefore, since cyclopropane has the most ring strain and since propyl, ethyl and methyl groups don't contain any SE, the correct answer to the question is "A".
Solution 2:
The products of combustion are the same so the compound with the highest potential energy (least stable) will have the highest enthalpy of combustion. All have the same number of carbons. The cyclopropane has the most ring strain and therefore the highest potential energy.
If you use metacognition you can see what's being asked. They can only write questions asking "Which has the most ring strain?" so many times. This question is supposed to involve more thinking but can be distilled down to "Which has the most ring strain?"
A good strategy in answering multiple choice questions is to determine what is the same and what is different about the answer choices. In this case the number of carbons is the same so the products are the same. What is different is the number of carbons in the rings and therefore the angle strain is different.