Question 47 of 130

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

A$190\mathrm {~g}$B$200\mathrm {~g}$C$300\mathrm {~g}$D$290\mathrm {~g}$

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Answer: (A) $190\mathrm {~g}$

$\begin{aligned} & \tau_{\text {Net }}=0 \Rightarrow(400 \mathrm{~g} \times 30)=(250 \mathrm{~g} \times 10)(\mathrm{mg} \times 50) \\ & \mathrm{m}=\frac{12000-2500}{50}=\frac{9500}{50} \\ & \mathrm{M}=190 \mathrm{~g}\end{aligned}$

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