Question

A small sphere is given vertical velocity of magnitude $v_o=5m{s}^{-1}$ and it swings in a vertical plane about the end of a massless string. The angle $\theta$ with the vertical at which string will break, knowing that it can withstand a maximum tension equal to twice the weight of the sphere, is

• A. ${\cos }^{-1}\frac{2}{3}$
• B. ${\cos }^{-1}\left(\frac{1}{4}\right)$
• C. ${60}^o$
• D. ${30}^o$

Answer 1

The correct option is B ${\cos }^{-1}\left(\frac{1}{4}\right)$

$\Rightarrow T-ma\cos \theta =\frac{m{v}^2}{R}$

$\Rightarrow 2mg-ma\cos \theta =\frac{m{v}^2}{R}$

$\Rightarrow 2g-g\cos \theta =\frac{v^2}{R}$

$\Rightarrow 20-10\cos \theta =\frac{25}{2}$

$\Rightarrow 20-\frac{25}{2}=10\cos \theta$

$\Rightarrow \frac{40-25}{2}=10\cos \theta$

$\Rightarrow \frac{15}{20}=\cos \theta$

$\Rightarrow \theta ={\cos }^{-1}\left(\frac{3}{15}\right)$

$\therefore$ For vertical angle ${}^{\prime }{\theta }^{\prime }$ ${\cos }^{-1}\left(\frac{1}{4}\right)$

Hence, the answer is ${\cos }^{-1}\left(\frac{1}{4}\right).$