Question

In the figure shown, $A B$ is horizontal diameter of the vertical circle. Time taken by a particle to slide along chord $A C$ is $t_{1}$ and that corresponding to chord $B C$ is $t_{2}$. If $t_{1} \times t_{2}=\frac{x R}{y}$, then find $(x+y)$

Answer 1

$A C=2 R \cos \theta ; B C=2 R \sin \theta$
$2 R \cos \theta=\frac{1}{2}(g \sin \theta) t_{1}^{2}$ ...(i)
$2 R \sin \theta=\frac{1}{2}(g \cos \theta) t_{2}^{2}$....(ii)

From (i) and (ii), we get
$t_{2} / t_{1}=\tan \theta$
and $t_{1} \times t_{2}=4 R / 9$