JEE Advanced Model Paper 4 with solutions for online practice

Given OP = 5 m and OC = 2.5 m
Therefore cosθ =

OC

OP

=

1

2

⇒ θ = 60°
Radius of circle is r = CP = OPsinθ = 5 sin 60° =

5×√3

2

m
The forces acting on the sphere, N is the normal reaction.
Net force towards centre C of the circle = N sin θ
⇒ mrω2 = Nsinθ (i)
Also mg = Ncosθ (ii)
From (ii) , N =

mg

cosθ

=

0.5×10

cos60°

= 10N
From (i) mrω2 Rsinθ = Nsinθ
⇒ ω = √

N

mR

= √

10

0.5×5

= 10 rads−1
∴ Time period T =

2π

ω

= π second = 3.14 s