Question

Three point particles P, Q, R move in a circle of radius 'r' with different but constant speeds. They start
moving at t = 0 from their initial positions as shown in the figure. The angular velocities (in rad/sec) of P,Q and R are 5pi, 2pi & 3pi respectively, in the same sense. The time interval after which they all meet is:-
(1) 2/3 sec
(2) 1/6 sec
(3) 1/2 sec
(4) 3/2 sec

pls explain with correct explanation and answer...
(irrelevants and incorrect will get report)​

Answer 1

Angular velocity of P, $\sf W_P$ = 5π rad/s

Angular velocity of Q, $\sf W_Q$ = 2π rad/s

Angular velocity of R, $\sf W_R$ = 3π rad/s

Now, let's assume P as the origin. So,

• Angle travelled by P, θ = $\sf W_P$ × t = 5πt - (i)

• Angle travelled by Q, θ = $\sf W_Q$ × t = 2πt - (ii)

• Angle travelled by R, θ = $\sf W_R$ × t = 3πt - (iii)

Let's use the rule of elimination, in equation (i), (ii) and (iii). So,

• 5πt = 5π × $\dfrac{2}{3}$ = $\dfrac{10π}{3}$

• 2πt = 2π × $\dfrac{2}{3}$ + $\dfrac{π}{2}$ = $\dfrac{4π}{3}$

• 3πt = 3π × $\dfrac{2}{3}$ + π = 2π

Similarly, when t was either $\dfrac{1}{6}$ or $\dfrac{1}{2}$, the value of θ was different. The value of θ should be between 0 and 2π. Now, we put the value of t = $\dfrac{3}{2}$ . Now,

• 5πt = 5π × $\dfrac{3}{2}$ = $\dfrac{15π}{2}$ - (a)

• 2πt = 2π × $\dfrac{3}{2}$ + $\dfrac{π}{2}$= $\dfrac{7π}{2}$ - (b)

• 3πt = 3π × $\dfrac{3}{2}$ + π = $\dfrac{11π}{2}$ - (c)

But the value of θ is different in the above equations. As we know,

• $\dfrac{15π}{2}$ = 6π + $\dfrac{3π}{2}$

• $\dfrac{7π}{2}$ = 2π + $\dfrac{3π}{2}$

• $\dfrac{11π}{2}$ = 4π + $\dfrac{3π}{2}$

$\bold{Hope\;it \; helps\;!}$