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(1)  u = 13 m/s
a = -2 m/s²
formula of distance travelled in nth second by an object is 
Sn = u + a(2n-1)/2
distance travelled in 7th second is
S = 13 - 2(2*7-1)/2
S = 0 m
hence option (1) is correct.
(2)  let distance between them is r
 formula of force is 
F = kQ1Q2/r²
where k = 1/4π∈
force between 3Q and Q is
F = 3Q²×k/r²             (repulsive)  -----------------------(1) 
let charge distribution on the sphere is as after connecting them by a thin wire and let there is no loss of charge 
charge on the both sphere is same after connecting them
 3Q-q = Q+q
q = Q
so charge on each sphere is 2Q
so force 
F1 = 2Q×2Qk/r²
F1 = 4Q²×k/r²               (repulsive)             ---------------------(2)
(2) ÷ (1) 
F1/F  = 4/3
F1 = 4F/3 
hence repulsive force between them is 4F/3 
hence option (1) is correct. 
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In first question north and south are mentioned ...are they not important for solving this question
Your second ans is correct but 1st on is wrong corrects ans is option 4
please see once again plz

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U = 13m/s j
a = -2 m/s² j
velocity will be zero at t = 13/2 = 6.5s
We need to find distance travelled in 7th second.
It is equal to distance travelled in 7 seconds minus distance travelled in 6 seconds.
Since it is changing its direction at t=6.5s, displacement will be 0 if we find the displacement in the 7th second. To find the distance travelled, find the distance travelled from 6 to 6.5 second and 6.5s to 7 second. Both will be same as time is same. So find the displacement from 6 to 6.5s and multiply it with 2.

S = (ut+0.5at²)₆.₅ - (ut+0.5at²)₆
⇒ S = (13×6.5 + 0.5×(-2)×6.5²) - (13×6 + 0.5×(-2)×6²) 
⇒ S = 13(6.5-6) - 0.5×2×(6.5²-6²)
⇒ S = 13×0.5 - 1×(42.25-36)
⇒S = 6.5 - 6.25
⇒ S = 0.25m
So distance travelled in 7th second = 2S = 2×0.25 = 0.5m

Initial charges = Q and 3Q
F=k \frac{3Q \times Q}{r^2} =3 \frac{kQ^2}{r^2}

When they are connected by a wire, charge distribution will happen according to the potential of the spheres. Since both are identical spheres, charge will be equally divided between the two.
The charges on both spheres will be (Q+3Q)/2 = 2Q
F' = k \frac{2Q \times 2Q}{r^2} = 4\frac{kQ^2}{r^2}

F'=4 \frac{kQ^2}{r^2}=  \frac{4}{3} \times (3 \frac{kQ^2}{r^2})= \frac{4}{3} F
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