Question

A particle describes a horizontal circle on the smooth surface of an inverted cone. the plane of that circle is at a height of 9.8 cm above the vertex. Then the speed of the particle is :

• A. 0.49 $m{s}^{-1}$
• B. 0.98 $m{s}^{-1}$
• C. 1.96 $m{s}^{-1}$
• D. 3.92 $m{s}^{-1}$

Answer 1

Answer: (B)

0.98 $m{s}^{-1}$

Acc to Newton law of motion

$N\cos \alpha =\frac{m{v}^2}{r}$

$N\sin \alpha =mg$

$r=h\tan \theta =$Radius of motion

$r=h\tan \theta =\frac{mg.h}{\frac{m{v}^2}{r}}$

$r=\frac{rg.h}{v^2}$

$v=\sqrt{gh}=\sqrt{9.8\times .098}$

$v=.98m{s}^{-1}$