Question

Please answer it

Question no 16

Answer 1

HERE, Ur answer dear .....⬇

since , uniform metre rule of mass 100 g is balanced on the fulcrum at mark 40 cm.

so distance of c.o.m. of rule from balanced point is 10 cm.

Net moment at balanced point should be zero.

so, m× 20 = 100 × 10
m = 50 g

if m is moved to mark of 10 cm than rule will tilt to the side where m is suspended.

Now, for balance this
50 × 30 = 100 × 10 + 50× (x-40)
(x-40) = (1500-1000)/50 = 10 cm
X = 50 cm

so, another 50 g mass will be suspended at 50 cm Marks.

Hope it's helps you.
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