A particle attached to an ideal string is project from position B (lowest position). At position A, tension in string becomes zero. Find speed in string at B.

Correct Option: The correct answer is √(7gl/2).

1. Image Analysis & System Setup:
Based on the provided diagram, the particle moves in a vertical circle of radius $l$ (the length of the string) under acceleration due to gravity $g$ (pointing downwards).
The diagram shows that the angle between the string at the lowest position $B$ (pointing straight down) and the position $A$ is $120°$.
Therefore, the angle that the string makes with the upward vertical at position $A$ is:
$180°-120°=60°$

2. Dynamics at Position A:
At position $A$, the forces acting on the particle along the radial direction (towards the center) are the tension $T$ and the radial component of the gravitational force.
The radial component of gravity pointing towards the center is:
$mgcos60°$
The equation of motion in the radial direction at $A$ is:
$T+mgcos60°=\frac{m{v}_A^2}{l}$
Since the tension in the string becomes zero at $A$ ($T=0$), we get:
$mgcos60°=\frac{m{v}_A^2}{l}$
Simplifying for the square of the velocity at $A$ ($v_A^2$):
$v_A^2=glcos60°=\frac{gl}{2}$

3. Conservation of Mechanical Energy:
Applying the principle of conservation of mechanical energy between the lowest position $B$ and position $A$:
Let the potential energy at the lowest point $B$ be zero. The height $h$ of position $A$ above $B$ is:
$h=l+lcos60°=l(1+\frac{1}{2})=\frac{3l}{2}$
Equating the total energy at $B$ and $A$:
$\frac{1}{2}m{v}_B^2=\frac{1}{2}m{v}_A^2+mgh$
Multiply the entire equation by $\frac{2}{m}$:
$v_B^2=v_A^2+2gh$
Substitute the values of $v_A^2$ and $h$ into the equation:
$v_B^2=\frac{gl}{2}+2g\left(\frac{3l}{2}\right)$
$v_B^2=\frac{gl}{2}+3gl$
$v_B^2=\frac{7gl}{2}$
Taking the square root on both sides:
$v_B=\sqrt{\frac{7gl}{2}}$