1U are almost der.
look,sinθ=h/l where h=L/2
so sinθ=L/2l=1/2√(Ï0/Ï)
A uniform rod of density ∂ is placed in a wide tank containing a liquid of density ∂o(∂o>∂). The depth of liquid in the tank is half the length of the rod. The rod is in equilibrium , with its lower end resting on the bottom of the tank. In this postion the rod makes an angle @(theta) with the horizontal..
a) sin @ = √(∂o/∂)2
b)sin @ = (∂o/∂)2
c)sin @ = √(∂o/∂)
d) sin @ = (∂o/∂)
the ans given is a
i've done till here...
after this i'm not getting any relation with sin@
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12 Answers
1GUYS YOU HAVE GONE WRONG !!!
THE PART OF ROD IMMERSED UNDER WATER IS NOT L/2 !!!
11ya dats where the problem....
u r telling
sin @ = h/l where h = L/2
but from where are you getting h=L/2...
(you are assuming h to be the height of water in the tank rite?? then how is that h = L/2 = dist of com from either end of the rod????)
(i've assumed these...
L = length of rod
l = length of rod inside water
from pt where rod touch the surface tank to where W acts is L/2
from pt where rod touch the surface tank to where B acts is l/2..)
[12]
1
Taking moments about pt. A (to eliminate this hell of Normal Reaction),
moment due to upthrust = moment due to weight as rod is in equilibrium
AL g Ï cosθ . L/2 = A x Ï0 g cosθ . x/2
so L2 . Ï = x2 . Ï0
so x = L √(ÏÏ0)
now take the triangle involving the part of submerged rod, the depth of the liquid and the base of the tank.
sinθ = (depth of the tank) / AB
or sinθ = (L/2) / x
so sinθ = √(Ï0/Ï)2
Ans A.
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