A man carrying a monkey on his shoulder does cycling smoothly on a circular track of radius $9 \mathrm{~m}$ and completes 120 resolutions in 3 minutes. The magnitude of centripetal acceleration of monkey is (in $\mathrm{m} / \mathrm{s}^2$ ) :

<p>First, let's calculate the centripetal acceleration experienced by the monkey while the man does cycling smoothly on a circular track. The formula for centripetal acceleration ($a_c$) is given by:</p> <p>$a_c = \frac{v^2}{r}$</p> <p>where <ul> <li>$v$ is the velocity of the monkey and the man cycling,</li><br> <li>$r$ is the radius of the circular track.</p></li> </ul> <p>To find $v$, we first need to find the circumference of the circle, which is given by:</p> <p>$C = 2\pi r$</p> <p>Given the radius $r = 9 \, \mathrm{m}$, the circumference $C$ is:</p> <p>$C = 2\pi \times 9 = 18\pi \, \mathrm{m}$</p> <p>Then, we determine the total distance travelled by calculating how many times they complete the circle in the given time. With 120 revolutions in 3 minutes (180 seconds), the total distance $D$ travelled is:</p> <p>$D = 120 \times C = 120 \times 18\pi = 2160\pi \, \mathrm{m}$</p> <p>To find the speed $v$, which is distance over time, we divide the total distance by the total time in seconds:</p> <p>$v = \frac{D}{t} = \frac{2160\pi}{180} = 12\pi \, \mathrm{m/s}$</p> <p>Now, using the formula for centripetal acceleration $a_c = \frac{v^2}{r}$, we can plug in the values:</p> <p>$a_c = \frac{(12\pi)^2}{9} = \frac{144\pi^2}{9} = 16\pi^2 \mathrm{~m/s}^2$</p> <p>Therefore, the magnitude of the centripetal acceleration of the monkey is $16\pi^2 \, \mathrm{m/s}^2$, which corresponds to Option B.</p>