The reduced mass of two particles having masses $m$ and $2 m$ is

2 m

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3 m

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\frac{2 m}{3}

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\frac{m}{2}

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Solution

The correct option is C $\frac{2 m}{3}$
Reduced mass of particle $= μ = \frac{m_{1} m_{2}}{m_{1} + m_{2}}$
In this condition $m_{1} = m$
and $m_{2} = 2 m$
Putting the value in the equation-
$μ = \frac{m \times 2 m}{m + 2 m}$
$= \frac{2 m^{2}}{3 m}$
$= \frac{2 m}{3}$

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