A stone is tied to one end of a spring 50 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 10 revolutions in 20 s, what is the magnitude of acceleration of the stone.

The correct option is 493 cm/sec².

To find the magnitude of the acceleration of the stone, we can break the problem down into simple steps:

Step 1: Identify the given parameters
Radius of the circular path ($r$) = 50 cm
Number of revolutions = 10
Time taken ($t$) = 20 s

Step 2: Find the frequency of revolution ($f$)
Frequency is defined as the number of revolutions per unit time:
$f=\frac{\text{Number of revolutions}}{t}$
$f=\frac{10}{20}=0.5\text{rev/s}$

Step 3: Calculate the angular velocity ($\omega$)
The angular velocity is given by the formula:
$\omega =2\pi f$
Substituting $f=0.5\text{rev/s}$ and $\pi \approx 3.1416$:
$\omega =2\times 3.1416\times 0.5=3.1416\text{rad/s}$

Step 4: Calculate the magnitude of the centripetal acceleration ($a$)
For an object moving in a circle with constant speed, its acceleration is directed towards the center and is given by:
$a={\omega }^{2}r$
Substituting the values of $\omega$ and $r$:
$a={\left(3.1416\right)}^2\times 50$
$a=9.8696\times 50\approx 493.48{\text{cm/s}}^2$
Rounding to the nearest integer, we get:
$a\approx 493{\text{cm/s}}^2$

Thus, the magnitude of the acceleration of the stone is 493 cm/sec².