A particle describes a horizontal circle in a conical funnel whose inner surface is smooth with speed of 0.5 m/s. What is the height of the plane of circle from vertex of the funnel?

Question

A particle describes a horizontal circle in a conical funnel whose inner surface is smooth with speed of 0.5 m/s. What is the height of the plane of circle from vertex of the funnel? (A) 0.25 cm (B) 2 cm (C) 4 cm (D) 2.5 cm

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/c7b4b225aeb85169785e0c9be0b96255-.png" /> The particle is moving in circular path From the figure, mg = R sin θ dots(i) (mv2/r) =R cos θ dots (ii) From equations (i) and (ii) we get tan θ =(rg/v2) but tan θ =(r/h) ∴ h=(v2/g) = ((0.5)2/10) =0.025 m = 2.5 cm

Q. A particle describes a horizontal circle in a conical funnel whose inner surface is smooth with speed of $0.5 m / s$ . What is the height of the plane of circle from vertex of the funnel?

$0.25 c m$

$2 c m$

$4 c m$

$2.5 c m$

The particle is moving in circular path
From the figure, $m g = R s in θ \dots (i)$
$\frac{m v ^{2}}{r} = R cos θ \dots (ii)$
From equations (i) and (ii) we get
$t an θ = \frac{r g}{v ^{2}}$ but $t an θ = \frac{r}{h}$
$∴ h = \frac{v ^{2}}{g} = \frac{( 0.5 ) ^{2}}{10}$
$= 0.025 m = 2.5 c m$

Q. A particle describes a horizontal circle in a conical funnel whose inner surface is smooth with speed of 0.5m/s. What is the height of the plane of circle from vertex of the funnel?

0.25cm

2cm

4cm

2.5cm