A hollow vertical cylinder of radius R and height h has a smooth internal surface. A small particle is placed in contact with the inner side of the upper rim at a point P . It is given a horizontal speed v0 tangential to the rim. It leaves the lower rim at point Q , vertically below P . The number of revolutions made by the particle will be

Question

A hollow vertical cylinder of radius R and height h has a smooth internal surface. A small particle is placed in contact with the inner side of the upper rim at a point P . It is given a horizontal speed v0 tangential to the rim. It leaves the lower rim at point Q , vertically below P . The number of revolutions made by the particle will be (A) (h/2 π R) (B) (v0/√2 g h) (C) (2 π R/h) (D) (v0/2 π R)(√(2 h/g))

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-sqipulgmhzpyvzbz.jpg" /> Time to reach the bottom, ∵ t=√(2 h/g) ∴ (2 π R/v0)× n=√(2 h/g) â n=(v0/2 π R)√(2 h/g)

$\frac{h}{2 π R}$

$\frac{v _{0}}{2 g h}$

$\frac{2 π R}{h}$

$\frac{v _{0}}{2 π R} (\frac{2 h}{g})$

Time to reach the bottom,
$∵ t = \frac{2 h}{g}$
$∴ \frac{2 π R}{v _{0}} \times n = \frac{2 h}{g}$
$\Rightarrow n = \frac{v _{0}}{2 π R} \frac{2 h}{g}$

2πRh​

2gh​v0​​

h2πR​

2πRv0​​(g2h​​)

Time to reach the bottom, ∵t=g2h​​ ∴v0​2πR​×n=g2h​​ ⇒n=2πRv0​​g2h​​

$\frac{h}{2 π R}$

$\frac{v _{0}}{2 g h}$

$\frac{2 π R}{h}$

$\frac{v _{0}}{2 π R} (\frac{2 h}{g})$

Time to reach the bottom,
$∵ t = \frac{2 h}{g}$
$∴ \frac{2 π R}{v _{0}} \times n = \frac{2 h}{g}$
$\Rightarrow n = \frac{v _{0}}{2 π R} \frac{2 h}{g}$