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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 $\left.=10^3 \mathrm{~kg} \mathrm{~m}^{-3}\right)$
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2795 Upvotes
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The correct answer is:
$9.6 \mathrm{~ms}^{-1}$
$9.6 \mathrm{~ms}^{-1}$
$$
\begin{aligned}
&h_1 \rho_1 g+h_2 \rho_2 g=\frac{1}{2} \rho_1 \mathrm{v}^2 \\
&\Rightarrow 1000 \times 10 \times 3+800 \times 10 \times 2 \\
&=\frac{1}{2} \times 1000 \times \mathrm{v}^2 \\
&\Rightarrow \mathrm{v}^2=2 \times \frac{46000}{1000} \\
&\Rightarrow \mathrm{v}=\sqrt{92}=9.6 \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
\begin{aligned}
&h_1 \rho_1 g+h_2 \rho_2 g=\frac{1}{2} \rho_1 \mathrm{v}^2 \\
&\Rightarrow 1000 \times 10 \times 3+800 \times 10 \times 2 \\
&=\frac{1}{2} \times 1000 \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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