A) ring
B) solid sphere
C) hollow sphere
D) disc
Correct Answer: D
Solution :
As \[v=\sqrt{\frac{2gh}{1+\frac{{{k}^{2}}}{{{r}^{2}}}}}\] |
Given \[h=\frac{3{{v}^{2}}}{4g}\] |
\[{{v}^{2}}=\frac{2gh}{1+\frac{{{k}^{2}}}{{{r}^{2}}}}=\frac{2g3{{v}^{2}}}{4g\left( 1+\frac{{{k}^{2}}}{{{r}^{2}}} \right)}=\frac{6g{{v}^{2}}}{4g\left( 1+\frac{{{u}^{2}}}{{{v}^{2}}} \right)}\] \[1=\frac{3}{2\left( 1+\frac{{{k}^{2}}}{{{v}^{2}}} \right)}\] |
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. |
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