A uniform metre rod is balanced at the 70 cm mark by suspending a weight of 50 gf at the 40 cm mark and 200 gf at the 95 cm mark. Draw a diagram of the arrangement and calculate the weight of the metre rod.
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Answer:
Explanation:7. Total clockwise motion =
200*25 = 5000gf ............ Eq 1
Total anti-clockwise motion =
50*30+ W*20 = 1500 + W20........... Eq 2
According to
the law of motion
We know that total clockwise motion = total anti-clockwise motion,
Therefore-
Equating Eq1 and Eq2
= 1500 + W20 = 5000
W20 = 5000-1500
W20 = 3500
W = 3500/20
W = 175 gf (Ans)
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