MHT CET · Physics · Work Power Energy
A body slides down a smooth inclined plane having angle ' \(\theta\) ' and reaches the bottom with velocity 'v'. If a body is a sphere then its linear velocity at the bottom of the plane is
- A \(\sqrt{\frac{2}{7}} \mathrm{v}\)
- B \(\sqrt{\frac{3}{7}} \mathrm{v}\)
- C \(\sqrt{\frac{5}{7}} \mathrm{v}\)
- D \(\sqrt{\frac{9}{7}} \mathrm{v}\)
Answer & Solution
Correct Answer
(C) \(\sqrt{\frac{5}{7}} \mathrm{v}\)
Step-by-step Solution
Detailed explanation
The linear velocity of the body, \(v=\sqrt{2 g h}\)
The velocity of the sphere about its centre,
\(v_{C M}=\sqrt{\frac{2 g h}{1+\frac{K^{2}}{R^{2}}}}=\frac{v}{\sqrt{1+\frac{K^{2}}{R^{2}}}} \ldots(\text { i })\)
For uniform solid sphere,
\(\frac{K^{2}}{R^{2}}=\frac{2}{5}\)
Substituting value in Eq. (i), we get
\(v_{C M}=\frac{v}{\sqrt{1+\left(\frac{2}{5}\right)}}=\sqrt{\frac{5}{7} v}\)
The velocity of the sphere about its centre,
\(v_{C M}=\sqrt{\frac{2 g h}{1+\frac{K^{2}}{R^{2}}}}=\frac{v}{\sqrt{1+\frac{K^{2}}{R^{2}}}} \ldots(\text { i })\)
For uniform solid sphere,
\(\frac{K^{2}}{R^{2}}=\frac{2}{5}\)
Substituting value in Eq. (i), we get
\(v_{C M}=\frac{v}{\sqrt{1+\left(\frac{2}{5}\right)}}=\sqrt{\frac{5}{7} v}\)
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