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The velocity at which the mass of a part...

The velocity at which the mass of a particle becomes twice of its rest mass, will be -

A

`(2c)/(3)`

B

`c/2`

C

`(c sqrt3)/(2)`

D

`(3c)/(4)`

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The correct Answer is:
To find the velocity at which the mass of a particle becomes twice its rest mass, we can use the concept of relativistic mass. Here’s a step-by-step solution: ### Step 1: Understand the relationship between moving mass and rest mass The moving mass \( m \) of a particle is related to its rest mass \( m_0 \) by the equation: \[ m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}} \] where: - \( m \) is the moving mass, - \( m_0 \) is the rest mass, - \( v \) is the velocity of the particle, - \( c \) is the speed of light in vacuum. ### Step 2: Set up the equation for the given condition According to the problem, we want to find the velocity \( v \) at which the moving mass \( m \) becomes twice the rest mass \( m_0 \): \[ m = 2m_0 \] Substituting this into the equation gives: \[ 2m_0 = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}} \] ### Step 3: Simplify the equation We can cancel \( m_0 \) from both sides (assuming \( m_0 \neq 0 \)): \[ 2 = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \] ### Step 4: Square both sides Squaring both sides of the equation results in: \[ 4 = \frac{1}{1 - \frac{v^2}{c^2}} \] ### Step 5: Rearrange the equation Rearranging gives: \[ 1 - \frac{v^2}{c^2} = \frac{1}{4} \] This simplifies to: \[ \frac{v^2}{c^2} = 1 - \frac{1}{4} = \frac{3}{4} \] ### Step 6: Solve for \( v \) Taking the square root of both sides: \[ \frac{v}{c} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \] Thus, we can express \( v \) as: \[ v = c \cdot \frac{\sqrt{3}}{2} \] ### Final Answer The velocity at which the mass of a particle becomes twice its rest mass is: \[ v = \frac{\sqrt{3}}{2}c \] ---
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