Q102 / 180 · NEET 2013 (All India) · Physics

A small object of uniform density rolls up a curved surface with an initial velocity $v^{\prime}$. It reaches up to a maximum height of $\frac{3 v^2}{4 g}$ with respect to the initial position. The object is

  • A
    ring
  • B
    solid sphere
  • C
    hollow sphere
  • Correct answer
    D
    disc

Solution

As $v=\sqrt{\frac{2 g h}{1+\frac{k^2}{r^2}}}$ Given $h=\frac{3 v^2}{4 g}$ $v^2=\frac{2 g h}{1+\frac{k^2}{r^2}}=\frac{2 g 3 v^2}{4 g\left(1+\frac{k^2}{r^2}\right)}=\frac{6 g v^2}{4 g\left(1+\frac{u^2}{v^2}\right)}$ $1=\frac{3}{2\left(1+\frac{k^2}{v^2}\right)}$ or $1+\frac{k^2}{r^2}=\frac{3}{2}$ or $\frac{k^2}{r^2}=\frac{3}{2}-1=\frac{1}{2}$ $k^2=\frac{1}{2} r^2$ (Equation of disc) Hence, the object is disc.

Hence, the correct answer is option: 'D'.

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