Question

A length of path $ACB$ is $1500m$ and the length of the path $ADB$ is $2100m$. Two particles start from point $A$ simultaneously around the track $ACBDA$. One of them travels the track in clockwise sense and other in anticlockwise sense with their respective constant speeds. After $12s$ from the start, the first time they meet at the point $B$. After minimum time(ins) in which they meet first at point $B$, will they again meet at the point $B$ is time $t_{\min }=(12{)}^{x}s$. The value of $x$ is______.

Answer 1

Answer: 2

The velocity of first particle is $v_1=\frac{1500}{12}m{s}^{-1}$ and the velocity of second particle is $v_2=\frac{2100}{12}m{s}^{-1}$.
Let after completing $n_1$ and $n_2$ trips, they will again meet at the point $B$.
$\therefore \frac{3600n_1}{v_1}=\frac{3600n_2}{v_2}=t$
Or $\frac{n_1}{n_2}=\frac{v_1}{v_2}=\frac{\frac{1500}{\frac{12}{2100}}}{\frac{12}{2}}$
$=\frac{1500}{2100}=\frac{5}{7}$
$\therefore t_{\min }=\frac{3600\times 5}{v_1}$
$=\frac{3600\times 5}{1500}\times 12=144s=(12{)}^x$
$\therefore x=2$