JEE Main 16 March 2021 Shift 1 Solved Paper

The normal force by the side walls of the groove will be equal to the centripetal force acting on it.
i.e. N=‌

mv2

r

.....(i)
where,
r‌‌=20cm=0.2m
m‌‌=200g=200×10−3kg and v=rω=‌

2πr

T

=‌

2π×0.2

40

m∕s
Substituting the given values in Eq. (i), we get
N‌‌=‌

(200×10−3)×( 2π×0.2 40 )2

0.2

‌‌≈9.859×10−4N