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
- 290 g
- 200 g
- 190 g
- 300 g
Answer
(3) 190 g
Solution
Take moments about the knife-edge (40 cm mark); g cancels.
The anticlockwise moment is due to 400 g at the 10 cm mark (arm = 30 cm), while the clockwise moments are due to the rod's weight at the 50 cm mark (arm 10 cm) and m at the 90 cm mark (arm 50 cm), so
400 30 = 250 10 + m 50
12000 = 2500 + 50m
m = 950050 = 190 g
Answer: (3)