JEE Main 2025 Physics question
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
Answer: (2)
Solution
Balancing torques: 400×30 = 250×10 + m×50, m = 9500/50 = 190g