TS EAMCET 19-July-2022 Shift 2 Solved Paper

The given situation is shown below.Mass of each coin, m=2g
Mass of 4 coin =4×m
M′=4×2⇒M′=8g
Let mass of the metre stick be m.
The centre of mass lies at its mid-point i.e at the mark of 50‌cm
When the coins are placed 10‌cm away from the end B, the centre of mass gets shifted by 4‌cm from point G towards the coins side.
Now, using the principle of moment about point P, we get
8g(46−10)−mg(50−46)‌=0
8×36−m×4‌=0
m=‌

8×36

4

‌=72g