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…

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
  1. $190\mathrm {~g}$
  2. $200\mathrm {~g}$
  3. $300\mathrm {~g}$
  4. $290\mathrm {~g}$

Solution


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

Asked in: JEE Main 2025 (28 Jan Shift 2)

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