A tank with a small hole at the bottom has been filled with water and kerosene (specific gravity 0.8 ). The height of water is 3 ~m and that of kerosene 2 ~m . When the hole is opened the velocity of fluid coming out

Correct answer: B. 9.6 ~ms^ -1 aligned &h_1 _1 g+h_2 _2 g=(1)/(2) _1 v^2 & 1000 ×10 ×3+800 ×10 ×2 &=(1)/(2) ×1000 × v^2 & v^2=2 × (46000)/(1000) & v= 92 =9.6 ~m/ s aligned

JEE Main 201411 AprilOnline

Exam: JEE Main
Year: 2014
Date: 11 April
Mode: Online
Subject: Physics
Chapter: Mechanical Properties of Fluids

JEE Main 2014 · 11 April · Online · Physics · Mechanical Properties of Fluids

A tank with a small hole at the bottom has been filled with water...

A tank with a small hole at the bottom has been filled with water and kerosene (specific gravity

0.8

). The height of water is

3 \mathrm{~m}

and that of kerosene

2 \mathrm{~m}

. When the hole is opened the velocity of fluid coming out from it is nearly: (take

\mathrm{g}=10 \mathrm{~ms}^{-2}

and density of water

\left.=10^3 \mathrm{~kg} \mathrm{~m}^{-3}\right)

Options

A $10.7 \mathrm{~ms}^{-1}$
B $9.6 \mathrm{~ms}^{-1}$
C $8.5 \mathrm{~ms}^{-1}$
D $7.6 \mathrm{~ms}^{-1}$

Correct answer

B. $9.6 \mathrm{~ms}^{-1}$

Step-by-step solution

\begin{aligned}&h_1 \rho_1 g+h_2 \rho_2 g=\frac{1}{2} \rho_1 \mathrm{v}^2 \\ &\Rightarrow 1000 \times10 \times3+800 \times10 \times2 \\ &=\frac{1}{2} \times1000 \times \mathrm{v}^2 \\ &\Rightarrow \mathrm{v}^2=2 \times \frac{46000}{1000}\\ &\Rightarrow \mathrm{v}=\sqrt{92}=9.6 \mathrm{~m}/ \mathrm{s} \end{aligned}

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