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

Options

A. `(2c)/(3)`
B. `c/2`
C. `(c sqrt3)/(2)`
D. `(3c)/(4)`

Answer

To find the velocity at which the mass of a particle becomes twice its rest mass, we can follow these steps:

### 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 \( v \) is the velocity of the particle and \( c \) is the speed of light in vacuum.

### Step 2: Set up the equation for the condition given in the problem
According to the problem, the moving mass \( m \) is 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 divide both sides by \( m_0 \) (assuming \( m_0 \neq 0 \)):
\[ 2 = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \]

### Step 4: Square both sides
Squaring both sides to eliminate the square root gives:
\[ 4 = \frac{1}{1 - \frac{v^2}{c^2}} \]

### Step 5: Rearrange the equation
Rearranging the equation leads to:
\[ 1 - \frac{v^2}{c^2} = \frac{1}{4} \]

### Step 6: Solve for \( v^2 \)
Now, we can isolate \( v^2 \):
\[ \frac{v^2}{c^2} = 1 - \frac{1}{4} = \frac{3}{4} \]
Thus,
\[ v^2 = \frac{3}{4}c^2 \]

### Step 7: Take the square root to find \( v \)
Taking the square root of both sides gives:
\[ v = c \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}c \]

### 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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