A small sphere of mass $m$ suspended by a thread is first taken aside so that the thread forms a right angle with the vertical and then released. The total acceleration of the sphere when it makes an angle $\theta$ with the vertical is

• Correct answer
  A

  $g\sqrt{1+3{\text{cos}}^2\theta }$
• B

  $g\sqrt{1+{\text{cos}}^2\theta }$
• C

  $g\sqrt{1+{\sin }^2\theta }$
• D

  $g\sqrt{1+3{\sin }^2\theta }$

Solution

Between A and B, $\text{mgL cos }\theta =\frac{1}{2}{\text{mv}}_{\text{B}}^2$

$\therefore {\text{v}}_{\text{B}}^2=2\text{gL cos }\theta$

Now, ${\text{a}}_{\text{r}}=\frac{{\text{v}}_{\text{B}}^2}{\text{L}}=2\text{g cos }\theta$                



and ${\text{a}}_{\text{t}}=\text{g }\text{sin }\theta$

$\therefore \text{a}=\sqrt{{\text{a}}_{\text{t}}^2+{\text{a}}_{\text{r}}^2}=\text{g}\sqrt{1+3{\text{cos}}^2\theta }$

Hence, the correct answer is option: 'A'.