A particle is moving in a vertical circle. The tensions in the string when passing through two positions at angles 30° and 60° from vertical (lowest position) are T1 and T2 respectively, Then

The correct option is T1>T2.

Let us analyze the motion of a particle of mass $m$ moving in a vertical circle of radius $R$ under gravity. Let $\theta$ be the angle made by the string with the vertical (lowest position) at any instant.

When the particle is at an angle $\theta$ from the lowest position, two forces act on it in the radial direction:
1. Tension in the string ($T$) pointing radially inwards towards the center.
2. A component of gravity ($mgcos\theta$) pointing radially outwards from the center.

The net centripetal force is provided by the difference of these radial forces:

$T-mgcos\theta =\frac{m{v}^2}{R}$

Rearranging the terms, we get the expression for tension:

$T=\frac{m{v}^2}{R}+mgcos\theta$

By the law of conservation of mechanical energy, the velocity $v$ of the particle at angle $\theta$ is related to its velocity $v_0$ at the lowest point ($\theta =0$) by the relation:

$\frac{1}{2}m{v}^2+mgh=\frac{1}{2}m{v}_0^2$

Here, the height raised $h$ is given by $h=R(1-cos\theta )$. Substituting this value, we obtain:

$v^2=v_0^2-2gR(1-cos\theta )$

Substitute this expression for $v^2$ back into the tension equation:

$T=\frac{m}{R}[v_0^2-2gR(1-cos\theta \left)\right]+mgcos\theta$

$T=\frac{m{v}_0^2}{R}-2mg+3mgcos\theta$

Since $m$, $g$, $R$, and $v_0$ are constants, we can see that the tension $T$ depends directly on the value of $cos\theta$. As the angle $\theta$ increases from $0°$ to $90°$, the value of $cos\theta$ decreases.

Given the two angles:
For position 1: ${\theta }_1=30°$
For position 2: ${\theta }_2=60°$

Since $cos(30°)>cos(60°)$, it directly follows that:

$T_1>T_2$