A mass of 2kg is tied to the end of a string of length 1m. It is, then, whirled in a vertical circle with a constant speed of 5m/s . Given that g= 10 ms⁻² . At which of the following locations of tension in the string will be 70 N

The correct option is At the bottom.

To find the location where the tension in the string is 70 N, we analyze the forces acting on the mass at different points of its vertical circular motion.

Let us write down the given values:
Mass of the body,
$m=2\text{kg}$
Length of the string (radius of the circle),
$r=1\text{m}$
Constant speed,
$v=5\text{m/s}$
Acceleration due to gravity,
$g=10{\text{ms}}^{-2}$

The centripetal force required to keep the body in a circular path of radius $r$ with a constant speed $v$ is given by the formula:
$F_c=\frac{m{v}^2}{r}$

Substituting the given values to calculate this constant centripetal force:
$F_c=\frac{2\times 5^2}{1}=2\times 25=50\text{N}$

Now, let us analyze the tension in the string at the bottom of the circular path:

At the lowest point (the bottom of the circle), the tension $T$ acts vertically upwards towards the center of the circle, while the weight of the mass $mg$ acts vertically downwards. The net force pointing towards the center provides the necessary centripetal force:
$T-mg=\frac{m{v}^2}{r}$

Rearranging the equation to solve for the tension $T$:
$T=mg+\frac{m{v}^2}{r}$

Substituting the values of weight and centripetal force:
$T=(2\times 10)+50$
$T=20+50=70\text{N}$

Thus, the tension in the string is exactly 70 N when the mass is at the bottom of its path.