Q154 / 200 · Manipal (MET) 2013 · Physics
A wet open umbrella is held vertical and it whirld about the handle at a uniform rate of 21 revolutions in 44 s . If the rim of the umbrella is 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
- Correct answerA$\sqrt{2.5} \mathrm{~m}$
- B1 m
- C3 m
- D1.5 m
Solution
From equation of motion
$h=u t+\frac{1}{2} g t^2$
where $u$ is initial velocity and $t$ is time. Since, $u=0$
$\therefore \quad t=\sqrt{\frac{2 h}{g}}=\frac{\sqrt{2 \times 4.9}}{9.8}=1 \mathrm{~s}$
The horizontal range of the drop $=x$, then
$x=\left(\frac{v_t}{o}\right) t$
Also, $\omega=\frac{\Delta \theta}{\Delta t}=\frac{21 \times 2 \pi}{44}=3 \mathrm{rad} / \mathrm{s}$
Tangential speed $0.5 \times 3 = 1.5$
$x=1.5 \times 1=1.5 \mathrm{~m}$
$\begin{aligned} \text {Locus of drop }=\sqrt{x^2+r^2} & =\sqrt{(1.5)^2+(0.5)^2} \\ & =\sqrt{2.5} \mathrm{~m}\end{aligned}$
Hence, the correct answer is option: 'A'.