Question
Water flows through a cylindrical pipe, whose inner radius is 1cm, at the rate of 80cm/ sec in an empty cylindrical tank, the radius of whose base is 40cm. What is the rise of water level in tank in half an hour?
✓
Answer
Main concept: Volume of flowing water $\text{A}.\upsilon.\text{t}$ Area of base = A = Area of cross-section of flowing water height = distance = speed × time $\upsilon.\text{t}$ Flowing water is filled in cylindrical tank. Hence, the volume of flowing water is equal to volume of water in cylindrcical tank.Flowing water:
$\text{A}=\pi\text{r}^2(\text{cylinder})$ $\upsilon=80\text{cm/s}$Cylinder:
R = 40cm H = x $\text{r}=1\text{cm and t}=\frac{1}{2}\text{hr}=\frac{1}{2}\times3600\sec=1800\sec$ $\therefore$ Volume of water in cylindrical tank = Volume of flowing water $\Rightarrow\ \ \pi\text{R}^2\text{H}=\text{A}.\upsilon.\text{t}$ $\Rightarrow\ \ \pi\text{R}^2\text{H}=\pi\text{r}^2\upsilon.\text{t}$ $\Rightarrow\ \ 40\times40\times\text{x}=1\times1\times80\times1800$ $\Rightarrow\ \ \text{x}=\frac{80\times1800}{40\times40}=5\times18=90\text{cm}$ Hence, the rise of water level in cylindrical tank is 90cm.
$\text{A}=\pi\text{r}^2(\text{cylinder})$ $\upsilon=80\text{cm/s}$Cylinder:
R = 40cm H = x $\text{r}=1\text{cm and t}=\frac{1}{2}\text{hr}=\frac{1}{2}\times3600\sec=1800\sec$ $\therefore$ Volume of water in cylindrical tank = Volume of flowing water $\Rightarrow\ \ \pi\text{R}^2\text{H}=\text{A}.\upsilon.\text{t}$ $\Rightarrow\ \ \pi\text{R}^2\text{H}=\pi\text{r}^2\upsilon.\text{t}$ $\Rightarrow\ \ 40\times40\times\text{x}=1\times1\times80\times1800$ $\Rightarrow\ \ \text{x}=\frac{80\times1800}{40\times40}=5\times18=90\text{cm}$ Hence, the rise of water level in cylindrical tank is 90cm.
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