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
Moments about the knife-edge (40 cm mark), g cancels:
Anticlockwise: 400 g at 10 cm mark, arm = 30 cm
Clockwise: rod's weight at 50 cm mark (arm 10 cm) and m at 90 cm mark (arm 50 cm)
400 30 = 250 10 + m 50
12000 = 2500 + 50m
m = 950050 = 190 g
Answer: (3)