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Question

If a sphere is rolling, the ratio of its rational energy to total kinetic energy is given by:-


A
7:10
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B
2:5
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C
10:5
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D
2:7
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Solution

The correct option is D $$2 : 7$$
$$\begin{array}{l} For\, \, a\, \, rolling\, \, sphere,\, \, the\, \, rotation\, \, kinetic\, \, energy\, \, and\, \, the\, \, total\, \, kinetic\, \, energy\, \, respectively\, \, aare \\ { K_{ r } }=\frac { 1 }{ 2 } I{ \omega ^{ 2 } }\, \, and\, \, { K_{ t } }=\frac { 1 }{ 2 } m{ v^{ 2 } }+\frac { 1 }{ 2 } I{ \omega ^{ 2 } } \\ The\, \, moment\, \, of\, \, inertia\, \, of\, \, a\, \, sphere\, \, of\, \, mass\, \, m\, \, and\, \, radius\, \, r\, \, about\, \, its\, \, centre\, \, is \\ I=\frac { 2 }{ 5 } m{ r^{ 2 } } \\ also,\, \omega =\frac { v }{ r }  \\ therefore,\, I{ \omega ^{ 2 } }=\frac { 2 }{ 5 } m{ r^{ 2 } }\times \frac { { { v^{ 2 } } } }{ { { r^{ 2 } } } } =\frac { 2 }{ 5 } m{ v^{ 2 } } \\ Thus,\, \, { K_{ r } }=\frac { 1 }{ 5 } m{ v^{ 2 } }\, \, and\, \, { K_{ t } }=\frac { 1 }{ 2 } m{ v^{ 2 } }+\frac { 1 }{ 5 } m{ v^{ 2 } }=\frac { 7 }{ { 10 } } m{ v^{ 2 } } \\ Hence,\, \, \frac { { { K_{ r } } } }{ { { K_{ t } } } } =\frac { { \frac { 1 }{ 5 } m{ v^{ 2 } } } }{ { \frac { 7 }{ { 10 } } m{ v^{ 2 } } } } =\frac { 2 }{ 7 } =2:7 \end{array}$$
Hence,
option $$(D)$$ is correct answer.

Physics

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