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Standard XII

Physics

Why Com Is an Important Point

The reduced m...

Question

The reduced mass of two particles having masses $4 m$ and $2 m$ is

2 m

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3 m

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\frac{4 m}{3}

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\frac{m}{2}

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Solution

The correct option is C $\frac{4 m}{3}$
Given that,
$m_{1} = 4 m$
$m_{2} = 2 m$
Reduced mass of the two particles $= μ = \frac{m_{1} m_{2}}{m_{1} + m_{2}}$
Substitute the value in the equation -
$μ = \frac{4 m \times 2 m}{4 m + 2 m}$
$= \frac{8 m^{2}}{6 m}$
$= \frac{4 m}{3}$

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The reduced mass of two particles having masses 4m and 2m is

The correct option is C 4m3Given that, m1=4m m2=2m Reduced mass of the two particles =μ=m1m2m1+m2 Substitute the value in the equation - μ=4m×2m4m+2m =8m26m =4m3

The reduced mass of two particles having masses $4 m$ and $2 m$ is

The correct option is C $\frac{4 m}{3}$
Given that,
$m_{1} = 4 m$
$m_{2} = 2 m$
Reduced mass of the two particles $= μ = \frac{m_{1} m_{2}}{m_{1} + m_{2}}$
Substitute the value in the equation -
$μ = \frac{4 m \times 2 m}{4 m + 2 m}$
$= \frac{8 m^{2}}{6 m}$
$= \frac{4 m}{3}$