MCQ
A small object of uniform density rolls up a curved surface with an initial velocity $v$. It reaches upto a maximum height of $3v^2/4g$ with respect to the initial position. The object is
- ARing
- BSolid sphere
- CHollow sphere
- ✓Disc
✓
Answer
Correct option: D.
Disc
d
By the concept of energy conservation
By the concept of energy conservation
$\frac{1}{2} m v^2+\frac{1}{2} I \omega^2=M g h$
$\frac{1}{2} M v^2+\frac{1}{2} I\left(\frac{v}{R}\right)^2=M g\left(\frac{3 v^2}{4 g}\right)\left(\because \omega=\frac{v}{R}\right)$
$\frac{1}{2} I \frac{v^2}{R^2}=\frac{3}{4} M v^2-\frac{1}{2} M v^2=\frac{1}{4} M v^2$
$I=\frac{1}{2} M R^2$
The object is Disc
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