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(C) The relativistic linear momentum $p$ is given by the formula: $p = \frac{m_0 v}{\sqrt{1 - v^2/c^2}}$,where $m_0$ is the rest mass,$v$ is the veloc

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Its linear momentum will be more than double

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(C) The relativistic linear momentum $p$ is given by the formula: $p = \frac{m_0 v}{\sqrt{1 - v^2/c^2}}$,where $m_0$ is the rest mass,$v$ is the velocity,and $c$ is the speed of light.When the velocity $v$ is doubled to $2v$,the new momentum $p'$ becomes $p' = \frac{m_0 (2v)}{\sqrt{1 - (2v)^2/c^2}} = \frac{2 m_0 v}{\sqrt{1 - 4v^2/c^2}}$.Comparing $p'$ with the original momentum $p$,we see that the denominator $\sqrt{1 - 4v^2/c^2}$ is smaller than $\sqrt{1 - v^2/c^2}$.Since the denominator decreases,the value of the momentum increases by a factor greater than $2$. Thus,the linear momentum will be more than double.

$List-I$	$List-II$
$A$. Michelson-Morley experiment	$I$. The existence of anti-matter
$B$. Stern-Gerlach experiment	$II$. The existence of de-Broglie matter waves
$C$. Davisson-Germer experiment	$III$. Electrons have spins
$D$. Anderson discovery of positron	$IV$. The non-existence of ether

For a body moving with relativistic speed,if the velocity is doubled,then

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