MCQ
A disc arranged in a vertical plane has two groves of same length directed along the vertical chord $AB$ and $CD$ as shown in the fig. The same particles slide down along $AB$ and $CD$. The ratio of the time $t_{AB}/t_{CD}$ is
  • A
    $1 : 2$
  • $1 :\sqrt 2 $
  • C
    $2 : 1$
  • D
    $\sqrt 2 :1$

Answer

Correct option: B.
$1 :\sqrt 2 $
b
$S_{A B}=\frac{1}{2} g t_{A B}^{2}$

$S_{C D}=\frac{1}{2} g \cos 60 t_{C D}^{2}$

But,

$S_{A B}=S_{C D}$

$\therefore \frac{S_{A B}}{S_{C D}}=\frac{\frac{1}{2} g t_{A B}^{2}}{\frac{1}{2} g \cos 60 t^{2} C D}$

$1=2 \frac{t^{2}_{AB}}{t^{2}_{CD}}$

$\frac{t_{A B}}{t_{C D}}=\frac{1}{\sqrt{2}}$

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