A particle slides down on a smooth incline of inclination ${30}^o$ , fixed in an elevator going up with an acceleration $2 m/s^2.$ The box of incline has width 4m. The time taken by the particle to reach the bottom will be

• A

  $\frac{8}{9}\sqrt{3}s$
• B

  $\frac{9}{8}\sqrt{3}s$
• Correct answer
  C

  $\frac{4}{3}\sqrt{\frac{\sqrt{3}}{2}}s$
• D

  $\frac{3}{4}\sqrt{\frac{\sqrt{3}}{2}}s$

Solution

In the frame of elevator,

a = acceleration of the particle with respect to the elevator

$\therefore m\sin 30 \left(g+2\right)=ma$

$a=\left(g+2\right)\sin {30}^o$

$=\left(10+2\right) .\frac{1}{2}$

$=6 m/s^2$

$\therefore$ The distance travelled by the particle from the top to the bottom,

$d=\frac{4}{\cos {30}^o}=\frac{4}{\frac{\sqrt{3}}{2}}=\frac{8\sqrt{3}}{3}m$

Using $s=ut+\frac{1}{2} a{t}^2$ ,

$\frac{8\sqrt{3}}{3}=0\times t+\frac{1}{2}6t^2$

$\Rightarrow t^2=\frac{16\sqrt{3}}{3\times 6}$

$\therefore t=\frac{4}{3}\sqrt{\frac{\sqrt{3}}{2}} s$

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