JEE Advanced Model Paper 7 with solutions for online practice

Torque due to applied force τ = Fr = Ipα
Where Ip is the moment of inertia about an axis passing through point of contact

I0 = IG + m(OG)2 = mr2 ⇒ IG = mr2 - m(OG)2
Ip = IG + m(GP)2 = mr2 - m(OG)2 + m(r−OG)2
= 3mr2 -

4mr2

π

where OG =

2r

π

=

2mr2

π

(π - 2)
Angular acceleration
α =

Frπ

2mr2(π−2)

=

Fπ

2mr2(π−2)

in the clockwise direction
At p , mg + F = N
Friction force f = µM = µ (mg + F)
For pure rolling , a =

µ(mg+F)

m

= α (PG)
µ =

mα(PG)

(mg+F)

= m[

Fπ

2mr(π−2)

]

r(π−2)

π(mg+F)

=

F

2(mg+F)