uniform rod of length b capable of turning about its end which is out of water, rests inclined to the vertical. If its specific gravity is 5/9, find the length immersed in water
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length of rod immersed in water is 13b/9
it is given that uniform rod of length b capable of turning about its end which is out of water, rests inclined to the vertical. If its specific gravity is 5/9.
we have to find the length of rod immersed in water.
see figure,
net torque = 0
⇒mg(b/2sinθ) (↓) + Vρg(b - x/2)sinθ (↑) = 0
⇒mg(b/2sinθ) - Vρg(b - x/2)sinθ = 0
⇒mb/2 = Vρ(b - x/2)
given, specific gravity of rod is 5/9
so, m = 5/9Vρ
then, 5/9 × b/2 = (b - x/2)
⇒5b/18 = b - x/2
⇒13b/18 = x/2
⇒x = 13b/9
therefore length of rod immersed in water is 13b/9
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Answer:
⇒mg(b/2sinθ) (↓) + Vρg(b - x/2)sinθ (↑) = 0
⇒mg(b/2sinθ) - Vρg(b - x/2)sinθ = 0
⇒mb/2 = Vρ(b - x/2)
given, specific gravity of rod is 5/9
so, m = 5/9Vρ
then, 5/9 × b/2 = (b - x/2)
⇒5b/18 = b - x/2
⇒13b/18 = x/2
⇒x = 13b/9
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