In the diagram shown in figure, match the following columns (Take, g = 10 ms-2)Column I(A) Normal reaction(8) Force of friction(C) Acceleration of blockColumn II(p) 12 SI unit(q) 20 SI unit(r ) zero(s) 2 SI unit
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a
A B Cp q r
b
A B Cq p r
c
A B Cq p s
d
A B Cp s r
answer is C.
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Detailed Solution
N=mg-202sin45°=20 N,f=μkN=12 NSince, 202cos45°>μN block will move and its acceleration will bea=202cos45°-μkNm=20-124=2 ms-2Hence, A→q,B→p,C→s.