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Question

In column-I, some situations are shown and in column-II, information about their resulting motion is given. Select the correct answer using the codes given below the columns

Column I Column II
p. A uniform solid sphere of mass 1 kg and radius 1 m, μs=0.05 1. Friction will be in the direction of acceleration of centre of mass of body.
q. A uniform body of mass 1 kg, r=12 m, R=1 m, I (about axis passing through centre and perpendicular to the plane of paper) = 2 kgm2,μs=0.3 2. Friction will be opposite to the direction of acceleration of centre of mass of the body.
r. A uniform solid cylinder released on a fixed incline plane
m=2 kg,R=1 m,μs=25
3. Body rotates clockwise
s. A uniform body of mass 1 kg,r=12 m,R=1 m, I(about axis passing through centre and perpendicular to the plain of the paper) =2 kgm2,μs=0.5
(String tightly wound on inner radius is pulled).
4. Body rotates anticlockwise

A
p-1; q- 2; r-4; s-3
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B
p-2; q- 1; r-3; s-4
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C
p-4; q- 1; r-2; s-3
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D
p-4; q- 2; r-3; s-1
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Solution

The correct option is A p-1; q- 2; r-4; s-3
p.
Let friction be static
F+f =maFRfR=25mR2αFf = 25mRα =25ma(a=Rα)F+fFf=52f=3F7=67NBut fL= μmg = 0.5 N
Therefore, friction is kinetic

q.
4f=ma = a4×12+f×1=2af=2 NfL=μmg=3 N

r. a=gsinθ1+ImR2=10×121+12=103m/s2mgsinθ f=maf=mg2ma=10203=103NfL=μsN=μsmg sinθ=25×2×10×32=43 N
Therefore, friction is static

s.
fL=5Nf=103N Static frictionFf=ma 4f=1af×14×12=2×aR f2=2a a=23m/s2

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