A metal rod of uniform
thickness and of length 1m
is suspended at its 25 cm
division with help of a string.
The rod remains horizontally
straight when a block of
mass 2kg is suspended to
the rod at its 10cm division.
The mass of rod is:
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Answer:
Torques balance each other
⇒mg(25)=2g(15)
15cm is a distance of the suspended mass from point of suspension
⇒25m=30
m= 25/30
=1.2kg
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