If it takes 5 minutes to fill a 15 litre bucket from a water tap of diameter 2 π cm then the Reynolds number for the flow is (density of water = 10 ^ 3 kg / m ^ 3 andviscosityofwater = 10 ^ - 3 Pa . s ) close to:

Correct answer: A. 5500 Reynolds number R = (ρvD)/(η) D = Diameter of litre Also rate of flow = (volume)/(time) = Av (V)/(t) = π D ^ 2 4 × v ⇒ v = 4V π D ^ 2 t ∴ R = (ρD)/(η) × 4V π D ^ 2 t = (4ρV)/(πηDt) = 4 × 10 ^ 3 × 15 × 10 ^ - 3 π × 10 ^ - 3 × 2 × 5 × 60 π × 10 ^ 2 = 10000 π 5500

JEE Main 201510 AprilOnline

Exam: JEE Main
Year: 2015
Date: 10 April
Mode: Online
Subject: Physics
Chapter: Mechanical Properties of Fluids

JEE Main 2015 · 10 April · Online · Physics · Mechanical Properties of Fluids

If it takes 5 minutes to fill a 15 litre bucket from a water tap of...

If it takes 5 minutes to fill a 15 litre bucket from a water tap of diameter

\dfrac{2}{\sqrt{\pi}}cm

then the Reynolds number for the flow is (density of water =

{10}^{3}kg / {m}^{3}andviscosityofwater = {10}^{- 3}Pa . s

) close to:

Options

A $5500$
B $550$
C $1100$
D $11,000$

Correct answer

A. $5500$

Step-by-step solution

Reynolds number

R = \dfrac{ρvD}{η}

D =

Diameter of litre

Also rate of flow

= \dfrac{volume}{time} = Av

\dfrac{V}{t} = \dfrac{\pi{D}^{2}}{4} \times v ⇒ v = \dfrac{4V}{\pi{D}^{2}t}

∴ R = \dfrac{ρD}{η} \times \dfrac{4V}{\pi{D}^{2}t}

= \dfrac{4ρV}{\piηDt}

= \dfrac{4 \times {10}^{3} \times 15 \times {10}^{- 3}}{\pi \times {10}^{- 3} \times 2 \times 5 \times 60}\sqrt{\pi} \times {10}^{2}

= \dfrac{10000}{\sqrt{\pi}} \approx 5500

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