A uniform rod of mass 250 g having length 100 cm is balanced on a sharp edge at 40 cm mark. A mass of 400 g is suspended at 10 cm mark. To maintain the balance of the rod, the mass to be suspended at 90 cm mark, is

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300 g

190 g

200 g

290 g

answer is B.

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Detailed Solution

$\begin{array}{l} τ_{Net} = 0 \Rightarrow (400 g \times 30) = (250 g \times 10) (mg \times 50) \\ m = \frac{12000 - 2500}{50} = \frac{9500}{50} \\ M = 190 g \end{array}$

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