Question

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.

• A. $\sqrt{2.5}$m
• B. $1$m
• C. $3$m
• D. $1.5$m

Answer 1

The correct option is A $\sqrt{2.5}$m

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