Engineering

Physics

Physical Quantities

Question

A liquid of density r is filled in tube having three vertical arms each of length L as shown. Initially height of liquid in each arm is L/2. If left vertical arm is now closed and whole system is rotated with angular velocity ω about vertical axis passing through middle arm, the height of liquid in left arm becomes $\frac{5 L}{8} . Given (L = \frac{40}{3} m, ρ = 10^{3} kg / m^{3}, g = 10 m / s^{2}, P_{atm} = 10^{5} N / m^{2})$ , air is treated as ideal gas

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Linked Question 1

Find the angular velocity ω

A

$ω = \frac{\sqrt{7}}{2} rad / \sec$

B

$ω = \frac{\sqrt{3}}{4} rad / \sec$

C

$ω = \frac{\sqrt{21}}{4} rad / \sec$

D

ω = 5 rad/sec

When tube starts rotating

h₁ + h₂ + h₃ = 3 $(\frac{L}{2})$ ................(i)

Let air pressure on left arm now becomes P₂

For fluid AB

P_AA – P_BA = m ω² r_cm

(P₂ + h₁pg) – (P_am + h₂pg) = LPω² $(\frac{L}{2})$ ....(ii)

$(here h_{1} = \frac{5 L}{8})$

fw fluid BC

P_CA – P_BA = mω²r_cm

(P_atm + h₃Pg) – (P_atm + h₂Pg)

$= LP ω^{2} (\frac{L}{2})$ ..............(iii)

For air in left arm

P₂V₂ = P₁ V₁

P₂(L – h₁) = P_atm ; $P_{2} = \frac{P_{atm} \frac{L}{2}}{L - \frac{5 L}{8}}$ ..................(iv)

From (i), (ii), (iii) and (iv) $ω = \frac{\sqrt{21}}{8}$ h₂ = 0

Linked Question 2

The height of liquid in the middle arm

A

$\frac{40}{3} m$

B

$\frac{10}{3} m$

C

zero m

D

$\frac{20}{3} m$

when tube starts rotating

h₁ + h₂ + h₃ = 3 $(\frac{L}{2})$ .........(i)

Let air pressure on left arm now becomes P₂

For fluid AB

P_AA – P_BA = m ω² r_cm

(P₂ + h₁pg) – (P_am + h₂pg) = LPω² $(\frac{L}{2})$ ....(ii)

$(here h_{1} = \frac{5 L}{8})$

fw fluid BC

P_CA – P_BA = mω²r_cm

(P_atm + h₃Pg) – (P_atm + h₂Pg)

$= LP ω^{2} (\frac{L}{2})$ .........(iii)

For air in left arm

P₂V₂ = P₁ V₁

P₂(L – h₁) = P_atm ; $P_{2} = \frac{P_{atm} \frac{L}{2}}{L - \frac{5 L}{8}}$ .........(iv)

From (i), (ii), (iii) and (iv) $ω = \frac{\sqrt{21}}{8}$ h₂ = 0