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# A U tube is rotated about one of it’s limbs with an angular velocity ω. Find the  difference in height H of the liquid (density ρ) level, where diameter of the tube d

Kevin Nash
6 years ago
Hello Student,
Weight of liquid of height H
= πd2 / 4 x H x ρ x g …(i)
Let us consider a mass dm situated at a distance x from A as shown in the figure. The centripetal force required for the mass to rotate = (dm) xω2
∴ The total centripetal force required for the mass of length L to rotate
= $\int_{0}^{L} (dm)$2 where dm = ρ x πd2 / 4 x dx
∴ Total centripetal force
= $\int_{0}^{L} (\rho x \tfrac{\pi d^2}{4} x dx)$x (xω2)
= ρ x πd2 / 4 x ω2 x L2 / 2 …(ii)
This centripetal force is provided by the weight of liquid of height H.
From (i) and (ii)
πd2 / 4 x H x ρ x g = ρ x πd2 / 4 x ω2 x L2 / 2
H = ω2 L2 / 2g
Thanks
Kevin Nash
9 months ago
Dear student,

Weight of liquid of height H
= πd2 / 4 x H x ρ x g …(i)
Let us consider a mass dm situated at a distance x from A as shown
in the figure. The centripetal force required for the mass to rotate = (dm) x ω2

∴ The total centripetal force required for the mass of length L to rotate
= ∫(dm) x ω2 where dm = ρ x πd2 / 4 x dx             [limit 0 à L]

∴ Total centripetal force
= ∫(ρx πd2/4 xdx)x(xω2)               [limit 0 à L]
= ρ x πd2 / 4 x ω2 x L2 / 2 …(ii)

This centripetal force is provided by the weight of liquid of height H.
From (i) and (ii)
πd2 / 4 x H x ρ x g = ρ x πd2 / 4 x ω2 x L2 / 2
H = ω2 L2 / 2g

Thanks and regards,
Kushagra