To solve this problem, we first need to understand the concepts of instantaneous velocity and average velocity when a particle moves along a circular path.
For a particle moving in a circular path, the instantaneous velocity is always tangent to the circle. If the particle moves with a constant speed, its speed is the same at all points on the circle, and so is its instantaneous velocity magnitude.
When the particle turns by an angle \(90\degree\), it moves along one quarter of the circle's circumference. We can calculate the arc length \(s\) that corresponds to this angle:
s = \frac{\theta}{360\degree} \times 2\pi r = \frac{90\degree}{360\degree} \times 2\pi r = \frac{\pi}{2}r
The average velocity is defined as the total displacement divided by the total time taken for the movement. The particle moves along one quarter of the circle, so the total displacement is the straight line between the start and end points of this arc, which forms the hypotenuse of an isosceles right triangle with the circle's radius as the equal sides:
Using Pythagoras' theorem, the displacement \(d\) is calculated as follows:
d = \sqrt{r^2 + r^2} = \sqrt{2}r
By definition, the average velocity \(V_{avg}\) is:
V_{avg} = \frac{d}{t} = \frac{\sqrt{2}r}{t}
Given that the ratio of the instantaneous velocity to the average velocity is \( \pi : x\sqrt{2} \), we equate this to the earlier expressions of velocity:
\frac{v}{\frac{\sqrt{2}r}{t}} = \frac{\pi}{x\sqrt{2}}
Substituting \(v = \frac{\pi r}{2t}\), the time for the arc is \( t = \frac{\pi r}{2v} \). The equation becomes:
\frac{\pi r}{2t} \times \frac{t}{\sqrt{2}r} = \frac{\pi}{x\sqrt{2}}
After canceling terms, this implies:
\frac{\pi}{\sqrt{2}} = \frac{\pi}{x\sqrt{2}}
Solving for \(x\), we see that:
x = 2
Thus, the value of \(x\) is correctly 2. Hence, the correct answer is 2.
A body of mass 1000 kg is moving horizontally with a velocity of 6 m/s. If 200 kg extra mass is added, the final velocity (in m/s) is: