Question

Two particles $1$ and $2$ are allowed to descend on two frictionless chords $OP$ and $OQ$. The ratio of the speeds of the particles $1$ and $2$ respectively when they reach on the circumference is:

• A. $\frac{1}{4}$
• B. $\frac{1}{2}$
• C. $1$
• D. $\frac{1}{2\sqrt{2}}$

Answer 1

The correct option is B $\frac{1}{2}$

Let radius of circle = $R$

$Oq=Pq=R$, So $△OPq$ is a equilateral triangle

So Particle on chord $OP$ will travel vertical distance = $Rcos60=\frac{R}{2}$

$\therefore$ From work-energy theorem $mg\left(R/2\right)=\frac{1}{2}m{v}_1^2⟹v_1=\sqrt{Rg}$

Particle on chord $OQ$ will travel vertical distance = $2R$

Similarly for particle on $OQ$- $v_2=\sqrt{4Rg}=2\sqrt{Rg}$

Ratio of their speed = $\frac{v_1}{v_2}=\frac{1}{2}$