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Two uniform rods A and B of lengths 5m and 2 m are placed end to end. If their linear densities are 4 $k g / m$ and $1 k g / m$ , the position of their centre of mass from their interface is

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$\frac{15}{9} m$ on the side of lighter rod.

$\frac{11}{24} m$ on the side of lighter rod

$\frac{20}{9} m$ on the side of heavier rod

$\frac{24}{11} m$ on the side of heavier rod

answer is A.

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Detailed Solution

$m_{1} = 5 \times 4 = 20 k g$ , $m_{2} = 2 \times 1 = 2 k g$ ,

$x_{1} = \frac{5}{2} = 2.5 m$ , $x_{2} = \frac{2}{2} = 1 m$

$X_{C M} = \frac{- 20 \times 2.5 + 2 \times 1}{22}$ $= \frac{- 50 + 2}{22} = \frac{- 48}{22} \Rightarrow \frac{- 24}{11}$

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