Question

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

• A. 300 g
• B. 200 g
• C. 290 g
• D. 190 g

Hint:

In torque problems, always use the equation \( \tau_{{Net}} = 0 \) and set up the moments about a point to solve for unknowns.

Answer 1

The Correct Option is D

The torque balance equation is: \[ \tau_{{Net}} = 0 \quad \Rightarrow \quad (400g \times 30) = (250g \times 10) + (mg \times 50) \] Solving for \( m \): \[ m = \frac{12000 - 2500}{50} = 190 \, {g} \]