Question

plz answer it correctly​

Answer 1

Let the mass of the particle be $\sf{m.}$ Let $\theta$ be the angle made by the vertical so that its angular displacement will be $\sf{\pi-\theta.}$

At the point B, the centripetal force is given by,

$\sf{\longrightarrow \dfrac{mv^2}{L}=T_B-mg\cos(\pi-\theta)}$

$\sf{\longrightarrow \dfrac{mv^2}{L}=T_B+mg\cos\theta}$

Then, tension in the string at point B,

$\sf{\longrightarrow T_B=\dfrac{mv^2}{L}-mg\cos\theta}$

We've to find value of $\theta$ for which,

$\sf{\longrightarrow T_B=0}$

$\sf{\longrightarrow \dfrac{mv^2}{L}-mg\cos\theta=0}$

$\sf{\longrightarrow mv^2=mgL\cos\theta}$

Multiplying by $\sf{\dfrac{1}{2},}$

$\sf{\longrightarrow \dfrac{1}{2}\,mv^2=\dfrac{1}{2}\,mgL\cos\theta\quad\quad\dots(1)}$

Applying energy conservation at points A and B,

$\sf{\longrightarrow \dfrac{1}{2}\,mu^2=\dfrac{1}{2}\,mv^2+mg(L-L\cos(\pi-\theta))}$

$\sf{\longrightarrow \dfrac{3}{2}\,mgL=\dfrac{1}{2}\,mv^2+mgL(1+\cos\theta)}$

$\sf{\longrightarrow \dfrac{1}{2}\,mv^2=mgL\left(\dfrac{1}{2}-\cos\theta\right)}$

From (1),

$\sf{\longrightarrow \dfrac{1}{2}\,mgL\cos\theta=mgL\left(\dfrac{1}{2}-\cos\theta\right)}$

$\sf{\longrightarrow \cos\theta=1-2\cos\theta}$

$\sf{\longrightarrow \cos\theta=\dfrac{1}{3}}$

$\sf{\longrightarrow\underline{\underline{\theta=\cos^{-1}\left(\dfrac{1}{3}\right)}}}$

Hence 4th option is the answer.

Answer 2

hey mate ♡

d) cos^-1 (1/3)

is the answer♥