A simple pendulum of length 1 m has a bob of mass 100 g . If the...

Given:
Length of the pendulum, $l = 1 \text{ m}$
Mass of the bob, $m = 100 \text{ g} = 0.1 \text{ kg}$
Speed of the bob, $v = 1.4 \text{ m/s}$
Angle with the vertical, $\theta = 0.20 \text{ rad}$
Acceleration due to gravity, $g = 9.8 \text{ m/s}^2$

Using the small angle approximation for $\cos\theta$:
$\cos\theta \approx 1 - \dfrac{\theta^2}{2} = 1 - \dfrac{(0.20)^2}{2} = 1 - 0.02 = 0.98$

The forces acting on the bob along the radial direction are the tension $T$ towards the pivot and the component of weight $mg \cos\theta$ away from the pivot. The net force provides the necessary centripetal acceleration:
$T - mg \cos\theta = \dfrac{mv^2}{l}$

Rearranging for tension $T$:
$T = mg \cos\theta + \dfrac{mv^2}{l}$

Substituting the values into the equation:
$T = (0.1 \times 9.8 \times 0.98) + \dfrac{0.1 \times (1.4)^2}{1}$

$T = 0.9604 + 0.196$

$T = 1.1564 \text{ N}$

Rounding off to two decimal places, we get:
$T \approx 1.16 \text{ N}$

Answer: $1.16 \text{ N}$