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A cylindrical vessel is filled with water upto the height $1 \mathrm{~m}$ from the base. A small orifice is opened at some height in the cylinder and the water level is reduced to height of orifice in $20 \mathrm{~s}$. If the base area of the cylinder is 100 times the area of orifice, then the height of orifice from the base is (take, $g=10 \mathrm{~m} / \mathrm{s}^2$ )
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Verified Answer
The correct answer is:
$80 \mathrm{~cm}$
Height of water in vesse $=1 \mathrm{~m}$
Let orifice is at a depth $h$ from above
Time taken to reduce the water level up to orifice,
$$
t=\frac{A}{A_0} \sqrt{\frac{2 x}{g}}
$$
where, $A_0=$ area of orifice and $A=$ base area.
$$
\text { Given, } \begin{gathered}
\frac{A}{A_0}=100, t=20 \mathrm{~s} \Rightarrow 20=100 \sqrt{\frac{2 x}{10}} \\
\sqrt{\frac{x}{5}}=\frac{1}{5} \Rightarrow x=\frac{1}{5}=0.2 \mathrm{~m} \text { or } 20 \mathrm{~cm}
\end{gathered}
$$
Now, height of orifice from base,
$$
h=100-20=80 \mathrm{~cm}
$$
Let orifice is at a depth $h$ from above
Time taken to reduce the water level up to orifice,
$$
t=\frac{A}{A_0} \sqrt{\frac{2 x}{g}}
$$
where, $A_0=$ area of orifice and $A=$ base area.
$$
\text { Given, } \begin{gathered}
\frac{A}{A_0}=100, t=20 \mathrm{~s} \Rightarrow 20=100 \sqrt{\frac{2 x}{10}} \\
\sqrt{\frac{x}{5}}=\frac{1}{5} \Rightarrow x=\frac{1}{5}=0.2 \mathrm{~m} \text { or } 20 \mathrm{~cm}
\end{gathered}
$$
Now, height of orifice from base,
$$
h=100-20=80 \mathrm{~cm}
$$
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