- Home
- Standard 11
- Physics
5.Work, Energy, Power and Collision
medium
A body initially at rest and sliding along a frictionless track from a height $h$ (as shown in the figure) just completes a vertical circle of diameter $AB =D$ . The height $h$ is equal to
A
$\;\frac{3}{2}D$
B
$D$
C
$\frac{5}{4}D\;$
D
$\frac{7}{5}D$
(NEET-2018)
Solution
To complete a vertical circle, speed at A should be
$v_{A}=\sqrt{5 g R}$
using energy conservation $m g h=\frac{1}{2} m v_{A}^{2}$
$h=\frac{1}{2} \frac{v_{A}^{2}}{g}=\frac{1}{2} \frac{5 g}{g} \frac{D}{2} \quad\left(R=\frac{D}{2}\right)$
$h=\frac{5 D}{4}$
Standard 11
Physics
