A simple pendulum of length l and mass m is initially at its lowest position . It is given the minimum horizontal speed necessary to move in a circular path about the point of suspension. The tension in the string at the lowest position of the bob is:

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3 mg

4 mg

5 mg

6 mg

answer is D.

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Detailed Solution

$\begin{array}{rcl} T - mg cosθ & = & \frac{{mu}^{2}}{R} \end{array}$

Here, T is tension in string, m is mass attached,

g is acceleration due to gravity, $\begin{array}{rcl} θ \end{array}$ is angle made by string with vertical,

u is velocity of mass m and R is radius of the circle.

$\begin{array}{rcl} At lowest position θ & = & 0, \cos 0 = 1, then : \end{array}$

$\begin{array}{rcl} T - mg & = & \frac{m {(\sqrt{5 gR})}^{2}}{R} \\ \Rightarrow & T = 6 mg \end{array}$

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