question_answer

A wet open umbrella is held vertical and it whirled about the handle at a uniform rate of 21 revolutions in 44 s. If the rim of the umbrella is a circle of 1 m in diameter and the height of the rim above the flour is 4.9 m, the locus of the drop is a circle of radius:

A)  \[\sqrt{2.5}\,m\]                            

B)  \[1\,m\]

C)  \[3\,\,m\]                                          

D)  \[1.5\,\,m\]

Correct Answer: A

Solution :

From equation of motion                                 \[h=ut+\frac{1}{2}g{{t}^{2}}\] Where \[u\] is initial velocity and \[t\] is time. Since, \[u=0\]                 \[\therefore \]  \[t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\times 4.9}{9.8}}=1s\] The horizontal range of the drop \[=x\], then                                     \[x=\left( \frac{{{v}_{1}}}{0} \right)t\] Also, \[\omega =\frac{\Delta \theta }{\Delta t}=\frac{21\times 2\pi }{44}=3\,rad/s\] Tangential speed \[{{v}_{t}}=r\omega =0.5\times 3\times 1.5\,m/s\]                 \[\therefore \]  \[x=1.5\times 1=1.5\,m\] Locus of drop \[=\sqrt{{{x}^{2}}+{{r}^{2}}}=\sqrt{{{\left( 1.5 \right)}^{2}}+{{\left( 0.5 \right)}^{2}}}=\sqrt{2.5}\,m\]