Question

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 then

• A. $1:2$
• B. $1:\sqrt{2}$
• C. $2:1$
• D. $\sqrt{2}:1$

Answer 1

The correct option is B $1:\sqrt{2}$

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

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

But,

$S_{AB}=S_{CD}$

$\therefore \frac{S_{AB}}{S_{CD}}=\frac{\frac{1}{2}g{t}_{AB}^2}{\frac{1}{2}g\cos 60t_{CD}^2}$

$1=2\frac{t_{AB}^2}{t_{CD}^2}$

$\frac{t_{AB}}{t_{CD}}=\frac{1}{\sqrt{2}}$