A body of mass m is tied to one end of a string of length l and revolves vertically in a circular path. At the lowest point of circle, what must be the K.E. of the body so as to complete the circle

The correct option is 2.5 mgl.

To find the minimum kinetic energy (K.E.) required at the lowest point of a vertical circle so that the body can complete the circular path, we apply the principles of circular motion and conservation of mechanical energy.

Let the mass of the body be $m$ and the length of the string be $l$, which acts as the radius of the circular path.

At the highest point of the vertical circle, the minimum velocity $v_h$ required to just maintain the circular path without the string going slack is given by:

$v_h=\sqrt{gl}$

The kinetic energy at this highest point is:

$K.E{.}_h=\frac{1}{2}m{v}_h^2$

Substituting the minimum velocity $v_h$ into the kinetic energy formula:

$K.E{.}_h=\frac{1}{2}m\left(gl\right)=0.5mgl$

By choosing the lowest point of the circle as the reference level (height = 0) for gravitational potential energy, we have:

$P.E{.}_l=0$

The highest point is at a height of $2l$ (diameter of the circle) above the lowest point. Therefore, the potential energy at the highest point is:

$P.E{.}_h=mg\left(2l\right)=2mgl$

According to the law of conservation of mechanical energy, the total mechanical energy at the lowest point must equal the total mechanical energy at the highest point:

$K.E{.}_l+P.E{.}_l=K.E{.}_h+P.E{.}_h$

Substituting the calculated values into the conservation equation:

$K.E{.}_l+0=0.5mgl+2mgl$

$K.E{.}_l=2.5mgl$

Therefore, the kinetic energy of the body at the lowest point must be 2.5 mgl to complete the circle.