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

We apply the condition of rotational equilibrium (balance of torques) about the pivot at the 40 cm mark.
Torque $$\tau$$ due to a force is given by $$\tau = F \times d$$, where $$F$$ is the weight and $$d$$ is the perpendicular distance from the pivot.

Identify the torques:

Rod’s own weight: mass = 250 g acts at the centre (50 cm).
Lever arm = $$(50 - 40) = 10\text{ cm}$$.
This produces a clockwise torque of magnitude $$250 \times 10\text{ (g·cm)}$$.

Suspended mass 400 g at 10 cm: lever arm = $$(40 - 10) = 30\text{ cm}$$.
This produces an anticlockwise torque of magnitude $$400 \times 30\text{ (g·cm)}$$.

Unknown mass $$m$$ at 90 cm: lever arm = $$(90 - 40) = 50\text{ cm}$$.
This produces a clockwise torque of magnitude $$m \times 50\text{ (g·cm)}$$.

For equilibrium, sum of anticlockwise torques = sum of clockwise torques:

$$400 \times 30 \;=\; 250 \times 10 \;+\; m \times 50 \quad\,-(1)$$

Compute the numerical values in $$(1)$$:

$$12000 \;=\; 2500 \;+\; 50m \quad\,-(2)$$

Rearrange $$(2)$$ to solve for $$m$$:

$$50m \;=\; 12000 - 2500 = 9500$$
$$m = \frac{9500}{50} = 190$$

Therefore, the required mass to be suspended at the 90 cm mark is 190 g. This corresponds to Option A.