Let's consider the ABO blood group system and the Rh factor separately.
ABO blood groups:
Mother (A+): Possible genotypes are $I^AI^A$ or $I^Ai$.
Father (B+): Possible genotypes are $I^BI^B$ or $I^Bi$.
Child (A+): Possible genotypes are $I^AI^A$ or $I^Ai$.
Since the child is A+, they must have received at least one $I^A$ allele. Since the father is B+, if the child received an $I^A$ allele from the mother, then they must have received an $i$ allele from the father. This means both the mother and father must carry the $i$ allele, so both must be heterozygous. Therefore, for the ABO blood group system alone, the genotypes must be $I^Ai$, $I^Bi$, and $I^Ai$ for mother, father, and child, respectively.
Rh factor:
Mother (+): Possible genotypes are ++ or +-.
Father (+): Possible genotypes are ++ or +-.
Child (+): Possible genotypes are ++ or +-.
Since both parents are Rh-positive, they can either be homozygous (+++) or heterozygous (+-). The child can inherit either a + or – allele from each parent, resulting in either ++ or +-.
Therefore, since the child is A+, and both parents are positive, the only possible genotype combination from the provided options that yields an A+ child is: $I^Ai$ | $I^Bi$ | $I^Ai$, which are options B and E.
A bob of heavy mass \(m\) is suspended by a light string of length \(l\). The bob is given a horizontal velocity \(v_0\) as shown in figure. If the string gets slack at some point P making an angle \( \theta \) from the horizontal, the ratio of the speed \(v\) of the bob at point P to its initial speed \(v_0\) is :