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A small object of uniform density rolls up a curved surface with an initial velocity 'v'. It reaches upto a maximum height of $\frac{3v^2}{4g}$ with respect to the initial position. The object is
A
Disc
B
Ring
C
Solid sphere
D
Hollow sphere
Detailed Solution
Conservation of mechanical energy: $\frac{1}{2}mv^2\left(1 + \frac{K^2}{R^2}\right) = mgh$
$\frac{1}{2}mv^2\left(1 + \frac{K^2}{R^2}\right) = mg\left(\frac{3v^2}{4g}\right) \Rightarrow \frac{K^2}{R^2} = \frac{1}{2}$
So the object is a disc.
$\frac{1}{2}mv^2\left(1 + \frac{K^2}{R^2}\right) = mg\left(\frac{3v^2}{4g}\right) \Rightarrow \frac{K^2}{R^2} = \frac{1}{2}$
So the object is a disc.