Question

Which out of $SN_1$ and $SN_2$ reaction mechanisms result in the formation of racemic mixture, why?

Answer 1

Hint: A mixture which contains two enantiomers in equal quantity with zero optical rotation. This type of a mixture is known as racemic mixture. E.g.- Racemic acid (racemic form of tartaric acid) is an equal mixture of two mirror-image of the isomers (enantiomers), optically active in opposite directions.

Complete answer:
$SN_1$ leads to a racemic mixture but the $SN_2$ will form the inverted product. Then the 2nd step is the attack of the nucleophile wherein the nucleophile eagerness to attack from any side of the leaving group in the substrate molecule i.eThe product is racemic where as in $SN_1$, it's a single step reaction will be from the transition state. In $SN_1$ racemization the reaction goes from 100% (R) or (S) to 50-50 (R) and (S). So I would assume the answer is that the rate of racemization is twice the rate of reactant incorporation. $SN_1$ reaction is a two stepped process or reaction.
step 1 formation of the carbocation
step 2 attack of the nucleophile on carbocation
carbocation formed is an intermediate and is planer so nucleophile and can attack from either side or direction. But the direction from which leaving groups usually depart the substrate is not that much supporting nucleophile to attack on it that means there is racemization but inversion product will be present in higher quantity than retention product.
$SN_1$ reactions provide racemization at the alpha carbon atom. If that is the only chiral center, one will get a racemic mixture. If there are other chiral centers, then one gets a pair of diastereomers. Racemization occurs in the $SN_1$ reaction because when the $SN_1$ reaction occurs, a group (base/nucleophile) attacks from (in front and back side) both sides.

Note: The carbocation and its substituents are all in the same plane, which means that the nucleophile can attack from either side that is right or left. As a result, both enantiomers are formed in the $SN_1$ reaction, which leads to a racemic mixture of both enantiomers.