Q1) a uniform meter scale of weight 50gram force is balanced at 40cm mark when a weight of 100 gram force is suspended at the 5cm mark were must the weight of 80 gram force be suspended to balance the meter scale.
Q2) a uniform wooden beam AB cm long and weighing 250 gram force is supported on the wedge calculate the greatest weight which can be placed at A without causing the beam to tilt
29In ans 1
Balance the torque abt the 40cm mark...
anticlockwise torque = 0.1*0.35*g
clockwise torque = 0.08*x*g (let the 80g block be at x distance from 40cm mark) + 0.05*0.5*g
0.1*0.35*g = 0.08*x*g +0.05*0.5*g
so x=0.125
so the block with mass 80g shud be at 52.5 cm(edited)
In qn.2 please specify the lenght of rod..
