JEE Advanced Model Paper 8 with solutions for online practice

Force due to pressure difference is F = P × A
=

1

2

ρ(V22−V12) × πR2
F =

1

2

ρ (14 - 7) × πR2 =

7πρR2

2

For translational motion F = f = ma .....(1)
For rotational motion
f × R = Iα =

Ia

R

f =

Ia

R2

=

2

5

mR2 ×

a

R2

=

2

5

ma
F = ma + f = ma +

2

5

ma =

7

5

ma
ma =

5

7

F
f =

2

5

×

5

7

F =

2F

7

=

2

7

×

7πρR2

2

f = πρ R2
µMg = πρ R2
µ

πρR2

mg

=

1

4

⇒ µ =

1

4

=

1

β

⇒ β = 4