When the speed of a particle is doubled, the ratio of its kinetic energy to its momentum

Remains the same

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Gets doubled

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Becomes half

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Becomes four times

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Solution

The correct option is C Gets doubled
Ratio of $\frac{K E}{m o m e n t u m} = \frac{\frac{1}{2} m v^{2}}{m v} \Rightarrow \frac{K E}{m o m e n t u m} \propto v$
so when the speed is doubled, $\frac{K E}{m o m e n t u m}$ is also doubled

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