A length of path $A C B$ is 1500 m and the length of the path $A D B$ is 2100 m . Two particles star

Question

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

Professor PC Thomas Classes · Answer

(a): The velocity of first particle is $v_1=\frac{1500}{12} \mathrm{~m} \mathrm{~s}^{-1}$ and the velocity of second particle is $v_2=\frac{2100}{12} \mathrm{~m} \mathrm{~s}^{-1}$. Let after completing $n_1$ and $n_2$ trips, they will again meet at the point $B$.\ $	herefore \quad \frac{3600 n_1}{v_1}=\frac{3600 n_2}{v_2}=t$\ [total path length $=1500+2100=3600$ ]\ or
$
\frac{n_1}{n_2}=\frac{v_1}{v_2}=\frac{\frac{1500}{12}}{\frac{2100}{12}}=\frac{1500}{2100}=\frac{5}{7}
$\ 
 $	herefore \quad t_{\min }=\frac{3600 	imes 5}{v_1}=\frac{3600 	imes 5}{1500} 	imes 12=144 \mathrm{~s}=(12)^x \mathrm{~s}$\ $	herefore \quad x=2$