is the string aabb in the language of this Turing machine ?? as upon reading aa it goes to the final state but upon b there is no transition defined so won't it result in a dead configuration?
When a Turing machine enters an accepting state, it immediately stops running and accepts whatever the original input string was (regardless of the state of the tape). So given the input
aabb, the Turing machine will read the two
as and enter the accepting state $q_2$, at which point it will halt and accept
In this case it doesn't matter that there's no transition for reading a
b from $q_2$ because we've already accepted (though you are correct in general that whenever we're running a Turing machine and we arrive at a state with no transition defined for the character we're reading, we implicitly go to a dead state and reject).