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A tank in the shape of a rectangular parallelopiped has volume 27 cubic meters. This tank is filled with water such that the rate of change of level of the water is thrice the rate of change water quantity falling in the tank, then the height of the tank (in meters) is
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The correct answer is:
81
Volume of rectangular parallelopiped is $27 \mathrm{~m}^3$
Let $A=$ Area of base and $h$ is height
$\therefore \quad V=A h$
$\frac{d V}{d t}=A \frac{d h}{d t}$
$\frac{d V}{d t}=A\left(\frac{3 d V}{d t}\right) \quad\left[\because \frac{d h}{d t}=\frac{3 d V}{d t}\right]$
$A=1 / 3$
$\because \quad V=\frac{1}{3} h$
$27=\frac{h}{3} \Rightarrow h=81$
$\therefore$ Height of tank $=81 \mathrm{~m}$
Let $A=$ Area of base and $h$ is height
$\therefore \quad V=A h$
$\frac{d V}{d t}=A \frac{d h}{d t}$
$\frac{d V}{d t}=A\left(\frac{3 d V}{d t}\right) \quad\left[\because \frac{d h}{d t}=\frac{3 d V}{d t}\right]$
$A=1 / 3$
$\because \quad V=\frac{1}{3} h$
$27=\frac{h}{3} \Rightarrow h=81$
$\therefore$ Height of tank $=81 \mathrm{~m}$
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