ARG! I had a whole post written out, and then I lost it all...
I don't think #20 is a very good question, personally. Those two compounds look like enantiomers, but are really the same compound. One is just rotated 180 degrees from the other.
In fact, that's my little mnemonic for remembering meso compounds: enantiomers that are really the same molecule.
in #26 I and III have the same exact structure and arrangment as the ones in #20 so I don't understand why can't they form a recemic mixture too ie be optically inactive just like the molecules in #20?
Because, like the molecules in #20, I and III are really the same compound. I and III are the same meso compound. II and IV are enantiomers, which mixed together would give a racemic mixture. I and III, as a meso compound, WILL be optically inactive, but because they are not enantiomers (they're identical) they cannot give a racemic mixture.
I think you may be getting confused as to WHY both racemic mixtures and meso compounds are optically inactive. It is true that they are, but for different reasons.
In a racemic mixture, there are two compounds (enantiomers) that are present in a 1:1 ratio. It's not always easy, but it is possible to separate those two compounds into two batches of enantiomerically pure compounds. If you take a sample of one of the enantiomerically pure compounds and measure its optical rotation, the plane polarized light would interact with one molecule and rotate. It would interact with the next molecule and rotate a bit more, but in the same direction thereby reinforcing the rotation from the previous molecule. The plane polarized light would interact with the millions of molecules in the enantiomerically pure sample and the light would be rotated a bit more each time - in the same direction - to give a non-zero value for the optical rotation at the end of the experiment.
You can do the same with a sample of the other enantiomer. The plane polarized light would interact with one molecule of the other enantiomer and rotate. But because this sample is the opposite enantiomer, the plane polarized light would rotate in the OPPOSITE direction. It would interact with the next molecule and rotate again. Same situation for the millions of molecules in this sample of the OTHER enantiomer. At the end, if the concentration of the two samples was the same, you would have another non-zero value for the optical rotation. This value would have the same magnitude as the first sample, but the opposite sign (i.e. if the first enantiomerically pure sample gave an optical rotation of -75 degrees, the second sample (if it is the same concentration) will give an optical rotation value of +75 degrees. Same magnitude, opposite sign).
Now, if you mix the two enantiomers together again in a 1:1 ratio, you will regenerate the racemic mixture. Now there is an equal number of molecules of one enantiomer as there are of the other enantiomer. When you introduce plane polarized light, the plane polarized light will interact with one molecule of one enantiomer and rotate. Then it has an equal probability of interacting with a molecule of the OTHER enantiomer. If it interacts with a molecule of the OTHER enantiomer, the plane polarized light will rotate back the OTHER way, canceling out the first rotation. After the plane polarized light has interacted with all the millions of molecules of one enantiomer and all the millions of molecules of the other enantiomer, the net effect on the optical rotation will be nil. That is, for all the rotations in one direction, the plane polarized light will have as many rotations in the other direction and the end value for the optical rotation will be zero. Thus racemic mixtures are optically inactive.
NOW, for meso compounds, the story is the same but different. Take a look at question 20 again. The two compounds look like enantiomers. If they really were enantiomers and you took the optical rotation of a mixture of the compounds, the net effect would be nil just like the racemic mixture above. BUT, if they really were enantiomers, you might be able to separate them and find the non-zero values for each enantiomer separately. The difference here is that while the two compounds LOOK like enantiomers, they are really the SAME compound. Because they are the same compound they cannot be separated.
When you attempt to take an optical rotation of a meso compound, the plane polarized light interacts with one molecule that might happen to be rotated to look like the compound on the left in question 20. This might cause the plane polarized light to rotate. But then the plane polarized light will interact with a molecule that might be rotated to look like the compound on the right. This might rotate the plane polarized light back in the OTHER direction. In bulk, the net effect of a meso compound is an optical rotation value of zero. When the plane polarized light gets to the end of the experiment, it will have rotated a net value of zero degrees.
Thus, meso compounds are optically inactive and can never be made to be optically active by separating compounds like a racemic mixture can. A racemic mixture, being a mixture of two different enantiomers, can be separated to give 2 batches of enantiomerically pure compound. The separate enantiomerically pure compounds will give a non-zero value for optical rotation. Mixed together, the racemic mixture of 2 separate enantiomers will have the effect of canceling out the rotation of the separate enantiomers and a racemic mixture will give an optical rotation value of zero. A meso compound cannot be separated and will always give an optical rotation value of zero. Both meso compounds and racemic mixtures are optically inactive.