Question
A rectangular tank is $80\ m$ long and $25\ m$ broad. Water flows into it through a pipe whose cross-section is $25\ cm^2$, at the rate of $16\ km$ per hour. How much the level of the water rises in the tank in $45$ minutes?
✓
Answer
Consider $'h'$ be the rise in water level.
Volume of water in rectangular tank $= 8000 \times 2500 \times h cm^2$
Cross-sectional area of the pipe $= 25\ cm^2$
Water coming out of the pipe forms a cuboid of base area $25\ cm^2$ and length equal to the distance travelled in $45$ minutes with the speed $16\ km/hour$ i.e.,
length = Length $=16000\times100\times\frac{45}{60}\text{cm}$
Therefore, The Volume of water coming out pipe in $45$ minutes $=25\times16000\times100\times\Big(\frac{45}{60}\Big)$
Now, volume of water in the tank = Volume of water coming out of the pipe in $45$ minutes $\Rightarrow8000\times2500\times\text{h}=16000\times100\times\frac{45}{60}\times25$
$\Rightarrow\text{h}=\frac{25\times16000\times100\times45}{60\times8000\times2500}=1.5\text{cm}$
Volume of water in rectangular tank $= 8000 \times 2500 \times h cm^2$
Cross-sectional area of the pipe $= 25\ cm^2$
Water coming out of the pipe forms a cuboid of base area $25\ cm^2$ and length equal to the distance travelled in $45$ minutes with the speed $16\ km/hour$ i.e.,
length = Length $=16000\times100\times\frac{45}{60}\text{cm}$
Therefore, The Volume of water coming out pipe in $45$ minutes $=25\times16000\times100\times\Big(\frac{45}{60}\Big)$
Now, volume of water in the tank = Volume of water coming out of the pipe in $45$ minutes $\Rightarrow8000\times2500\times\text{h}=16000\times100\times\frac{45}{60}\times25$
$\Rightarrow\text{h}=\frac{25\times16000\times100\times45}{60\times8000\times2500}=1.5\text{cm}$
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