A body of mass of 100g is attached to a 1m long string and it is revolving in a vertical circle. When the string makes an angle of 60° with the vertical then its speed is 2 m /s . The tension in the string at θ = 60° will be

The correct option is 0.89 N.

Let us understand the motion of a body in a vertical circle and derive the tension step-by-step.

When a body moves in a vertical circle, the forces acting on it at any point of its path are:
1. The weight of the body ($mg$) acting vertically downwards.
2. The tension ($T$) in the string acting along the string towards the center of the circle.

If the string makes an angle $\theta$ with the vertical (specifically, measured from the lowest point or the downward vertical), the component of gravity acting along the string (away from the center) is given by:
$mg\cos \theta$

The net force towards the center provides the necessary centripetal force for the circular motion. Therefore, we can write the equation of motion as:
$T-mg\cos \theta =\frac{m{v}^2}{r}$

Rearranging the formula to solve for the tension ($T$):
$T=mg\cos \theta +\frac{m{v}^2}{r}$

Let us identify the given values from the question:
• Mass of the body, $m=100\text{ g}=0.1\text{ kg}$
• Length of the string (radius of the circle), $r=1\text{ m}$
• Angle with the vertical, $\theta =60°$
• Speed of the body, $v=2\text{ m/s}$
• Acceleration due to gravity, $g\approx 9.8{\text{ m/s}}^2$

Now, substitute these values into the tension equation:

$T=\left(0.1\right)\left(9.8\right)\cos (60°)+\frac{\left(0.1\right){\left(2\right)}^2}{1}$

We know that $\cos (60°)=0.5$. Substituting this value:
$T=\left(0.1\right)\left(9.8\right)\left(0.5\right)+\frac{0.1\times 4}{1}$

Calculating the individual terms:
• Gravity component: $0.1\times 9.8\times 0.5=0.49\text{ N}$
• Centripetal component: $0.1\times 4=0.4\text{ N}$

Adding the two components together to find the total tension:
$T=0.49+0.4=0.89\text{ N}$

Thus, the tension in the string at $\theta =60°$ is 0.89 N.