a uniform meter rule of mass 100g balances at the 40cm, when a massX is placed at 10cm mark,what is the value of X.
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hii mate!!
uniform metre rule of mass 100 g is balanced on a fulcrum at mark 40 cm. so, distance of c.o.m. of rule from balanced point is 10cm.
Net moment at balancing point should be zero. So,
m×20 = 100×10
m = 50g
if m is moved to mark of 10cm then 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 = 10cm
X = 50cm.
So, another 50g mass will be suspended at 50cm mark
hope it helps !! ☺☺☺
uniform metre rule of mass 100 g is balanced on a fulcrum at mark 40 cm. so, distance of c.o.m. of rule from balanced point is 10cm.
Net moment at balancing point should be zero. So,
m×20 = 100×10
m = 50g
if m is moved to mark of 10cm then 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 = 10cm
X = 50cm.
So, another 50g mass will be suspended at 50cm mark
hope it helps !! ☺☺☺
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