A sprinkler shown in the figure rotates about its hinge point in a horizontal plane due to water flow discharged through its two exit nozzles.

The total flow rate Q through the sprinkler is 1 litre/sec and the cross-sectional area of each exit nozzle is 1 cm^{2}. Assuming equal flow rate through both arms and a frictionless hinge, the steady state angular speed of rotation (in rad/s) of the sprinkler is ___________ (correct to two decimal places).

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PY 5: GATE ME 2018 Official Paper: Shift 1

CT 1: Ratio and Proportion

2846

10 Questions
16 Marks
30 Mins

**Concept:**

In the given situation, as there is no external torque. Then the total angular momentum change is zero. About the hinge point.

The absolute velocities (VA & VB)

VA = VA(rel) + (rAo ω) = 5 + 0.1 × ω

VB = VB(rel) – (rBo ω) = 5 – 0.2 ω

Now ΣT = 0 (about 0)

⇒ ṁA × AO × VA – ṁB × BO × VB = 0

**Calculation:**

Given Q = 1 liter/sec = 1000 cm^{3}/s

Area = 1 cm^{2}, AO = 10 cm, BO = 20 cm

Relative velocity at (A and B) w.r.t sprinkler

\({V_A} = {V_B} = \left( {\frac{{Q/2}}{A}} \right)\)

Q = discharge, A = area of nozzle

\({V_{Ar}} = {V_{Br}} = \left( {\frac{{Q/2}}{A}} \right) = \frac{Q}{{2A}} = \frac{{1000}}{{2 \times 1}} = 500\;cm/s\; = 5\;m/s\)

V_{A(rel)} = V_{B(rel)} = 5 m/s

Now the absolute velocities (V_{A} & V_{B})

V_{A} = V_{A(rel)} + (r_{Ao} ω) = 5 + 0.1 × ω

V_{B} = V_{B(rel)} – (r_{Bo} ω) = 5 – 0.2 ω

Now ΣT = 0 (about 0)

⇒ ṁ_{A} × AO × V_{A} – ṁ_{B} × BO × V_{B} = 0

Now ṁ_{A} = ṁ_{B} = (ρQ/2)

⇒ AO × V_{A} – BO × V_{B} = 0

⇒ AO × V_{A} = BO × V_{B}

0.1 (5 + 0.1 ω) = 0.2 (5 – 0.2 ω)

⇒ 5 + 0.1 ω = 5 × 2 – 0.4 ω

⇒ 0.5 ω = 5

⇒ ω = 10 rad/s

** Mistake** that can happen here is while taking the direction of rotation while calculating the absolute velocity.

V_{A} = 5 + 0.1 ω instead mistake V_{A} = 5 – 0.1 ω

Or V_{B} = 5 – 0.2 ω instead mistake V_{B} = 5 + 0.2 ω

Any one value altered will result in wrong answer