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Question

Two particles A & B are projected as shown in fig in x-y plane under the effect of force which provide a constant acceleration a=11 m/s2 in negative y- direction. Then match situation in Column-I with the corresponding result in column-II (All positions are given in metre) (VAB=velocity of A with respect to B; rAB position of A with respect to B)

Column-IColumn-II(A) Seperation between the two particles is minimum at ...sec(p) 0(B) Minimum seperation between two particles is ...m(q) 0.5(C) Time when velocities of both particles are perpendicular to each other is ....sec(r) 0.9(D) At the time of minimum seperationVAB.rAB(s) 2(t) 25

A
a-t, b-p, c-p,s, d-q
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B
a-p, b-t, c-s, d-q
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C
a-p, b-t, c-p,s, d-q
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D
a-r, b-t, c-p,s, d-p
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Solution

The correct option is D a-r, b-t, c-p,s, d-p

tanθ=520=14
tanϕ=1020=12
A,B:
dmin=425sinα=425sin(ϕθ)
dmin=517[sinϕcosθcosϕsinθ]dmin=517[15×41725×117]dmin=55[2]=25

t=425cosα500=517105[25×417+15×117]t=110[8+1]=910 sec


C:
VA=(8^i+6^j)(11t)^j
VB=(12^i+16^j)(11t)^j
VA=8^i+(611t)^j
VB=12^i+(1611t)^j
VA.VB=96+9666t176t+121t2
0=242t+121t2=0
t=0 and t=242121=2

D:
At minimum separation relative velocity is perpendicular to displacement, hence dot product is zero.

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