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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