In vertical circular  motion of  a bob, match the entries of list–I with entries of list–II. Here v0 is the velocity of bob at lowest point & T is tension in string.   List-I(speed at lowest point)    List-II(possible situation)(P) v0 =  (1) Tension at lowest point = 6mg(Q)  v0 =  (2) string will slack for a finite time(R) v0 = 2 (3) bob will oscillate(S) v0 = 3 (4) Tension at highest point = 4mg        P    Q   R   S(A)    1    2    3    4(B)    4    2    3    1(C)    1    3    2    4(D)    4    3    2    1ABCD

To match the entries of List-I with List-II in the vertical circular motion of a bob, we need to consider the given situations and the corresponding values of v0 (velocity at the lowest point) and T (tension in the string).

List-I (speed at lowest point): (P) v0 = (Q) v0 = (R) v0 = (S) v0 =

List-II (possible situations): (1) Tension at lowest point = 6mg (2) String will slack for a finite time (3) Bob will oscillate (4) Tension at highest point = 4mg

Now, let's match the entries:

For situation (P), the velocity at the lowest point (v0) is not given, so we cannot determine the corresponding entry in List-II.

For situation (Q), the velocity at the lowest point (v0) is not given, so we cannot determine the corresponding entry in List-II.

For situation (R), the velocity at the lowest point (v0) is given as 2. This corresponds to entry (2) in List-II, which states that the string will slack for a finite time.

For situation (S), the velocity at the lowest point (v0) is given as 3. This corresponds to entry (3) in List-II, which states that the bob will oscillate.

Therefore, the matching entries are:

(A) 1 2 3 4

(B) Not possible to determine as v0 for situation (P) and (Q) is not given.

(C) Not possible to determine as v0 for situation (P) and (Q) is not given.

(D) Not possible to determine as v0 for situation (P) and (Q) is not given.

So, the correct answer is (A) 1 2 3 4.