The correct option is C 32
Tr=124Cr−1(√3)125−r(4√5)r−1
When both the terms are rational , Tr will be rational.
Hence, 125−r2 and r−14 both must be integers.
Therefore, r must be of the form 4k+1, where k is an integer.
There are 125 terms in the expansion. Hence, k can assume values from 0 to 31. Hence, there are 32 values of k and 32 rational terms in the expansion.
Hence, option B is correct.