JEE Main 26-Aug-2021 Shift 2 Solved Paper

Given, radius of horizontal circle, r=

L

√2

Figure illustrating the particle of mass m moving in a horizontal circle, while suspended from a ceiling is shown In equilibrium condition at point A,
T‌cos‌θ=mg ...(i)
T‌sin‌θ=

mv2

r

...(ii)
Divide Eq. (ii) by Eq. (i),
tan‌θ=v2∕rg
⇒v=√rg‌tan‌θ ...(iii)
Now, from figure, we can write
sin‌θ=

L∕√2

L

=

1

√2

⇒ θ = 45°
Substituting the value of θ in Eq. (iii), we get
v=√rg‌tan‌45°=√rg
Thus, the value of speed of particle is v=√rg.