A uniform rod of length 200 cm and mass 500 g is balanced on a . . .

Question : A uniform rod of length 200 cm and mass 500 g is balanced on a wedge placed at 40 cm mark. A mass of 2 kg is suspended from the rod at 20 cm and another unknown mass 'm' is suspended from the rod at 160 cm mark as shown in the figure. Find the value of 'm' such that the rod is in equilibrium. (g=10 m/s²)

(1) 1/12 kg 

(2) 1/2 kg 

(3) 1/3 kg 

(4) 1/6 kg

Doubt by Yana

Solution : 

Mass of the rod (M)
= 500 g
= 500/1000
=1/2 kg
=0.5 kg

Applying the principle of moment along the wedge

Anticlockwise Moment = Clockwise Moment

m1g×r1=Mg×R + m2g×r2
m1×r1=MR+m2×r2
2×(40-20)=0.5×(100-40) + m×(160-40)
2×20=0.5×60+m×120
40=30+m×120
40-30=m×120
10=m×120
10/120=m
m=10/120
m=1/12 kg 

Hence, (1) 1/12 kg, would be the correct option.