Question

Three particles $A,B$ and $C$ move in a circle in anticlockwise direction with speeds $5{\text{ms}}^{-1},2.5{\text{ms}}^{-1}$ and $10{\text{ms}}^{-1}$ respectively. The initial positions of $A,B$ and $C$ are as shown in the figure. The ratio of distance travelled by $A$ and $C$ at the instant $A$, $B$ and $C$ meet for the first time is

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

Answer 1

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


When $A$, $B$ and $C$ meet for the first time, let ${\theta }_A,{\theta }_B,{\theta }_C$ be their angular displacements in time $t$.

Now,

${\theta }_A={\omega }_{A}t=\frac{V_A\times t}{R}=\frac{5t}{R}$

${\theta }_B=\frac{V_B\times t}{R}=\frac{5t}{2R}$

${\theta }_C=\frac{V_C\times t}{R}=\frac{10t}{R}$

Then, ratio of arc lengths i.e distance travelled by $A$ and $C$ is,

$\frac{S_A}{S_C}=\frac{R{\theta }_A}{R{\theta }_C}=\frac{1}{2}$

Hence, option $\left(b\right)$ is the correct answer.