A particle describes a horizontal circle on the smooth surface of an inverted cone. If the height of the plane of the circle above the vertex is 9.8 cm, then the speed of the particle will be (Take g=9.8 text m/s2 )

Question

A particle describes a horizontal circle on the smooth surface of an inverted cone. If the height of the plane of the circle above the vertex is 9.8 cm, then the speed of the particle will be (Take g=9.8 text m/s2 ) (A)  0.98 text ms-1  (B)  19.6 text ms-1  (C)  4.9 ms-1  (D) None of these

Tardigrade Admin · Accepted Answer

The given situation can be shown as R cos α =(mv2/r) R sin α =mg ∴ tan α =(gr/v2) Also tan α =(r/h) ∴ (r/h)=(gr/v2) ⇒ v2=gh=0.98 ms-1

Q. A particle describes a horizontal circle on the smooth surface of an inverted cone. If the height of the plane of the circle above the vertex is 9.8 cm, then the speed of the particle will be (Take $g = 9.8 m / s^{2}$ )

$0.98 m s^{- 1}$

$19.6 m s^{- 1}$

$4.9 m s^{- 1}$

None of these

The given situation can be shown as
$R cos α = \frac{m v ^{2}}{r}$
$R sin α = m g$
$∴$ $tan α = \frac{g r}{v ^{2}}$
Also $tan α = \frac{r}{h}$
$∴$ $\frac{r}{h} = \frac{g r}{v ^{2}}$
$\Rightarrow$ $v^{2} = g h = 0.98 m s^{- 1}$

Q. A particle describes a horizontal circle on the smooth surface of an inverted cone. If the height of the plane of the circle above the vertex is 9.8 cm, then the speed of the particle will be (Take g=9.8m/s2 )

0.98ms−1

19.6ms−1

4.9ms−1