Answer
Verified
413.7k+ views
Hint: Find the volume of water that flows through the pipe in one hour and equate it to the formula of volume of the cylindrical tank, that is, \[V = \pi {r^2}h\] where ‘r’ is the radius of the base and ‘h’ is the height of the cylindrical tank.
Complete step-by-step answer:
It is given that the water flows out through a circular pipe, whose internal diameter is \[1\dfrac{1}{3}\] cm. Converting mixed fraction \[1\dfrac{1}{3}\] into an improper fraction, we have:
\[1\dfrac{1}{3} = \dfrac{{1 \times 3 + 1}}{3}\]
\[1\dfrac{1}{3} = \dfrac{4}{3}\]
The radius of the circular pipe is half of the diameter, that is, half of \[\dfrac{4}{3}\] cm.
\[{r_{pipe}} = \dfrac{1}{2} \times \dfrac{4}{3}\]
\[{r_{pipe}} = \dfrac{2}{3}cm\]
The area of cross-section of the pipe is given as \[\pi {r_{pipe}}^2\], hence, we have:
\[A = \pi {r_{pipe}}^2\]
\[A = \pi {\left( {\dfrac{2}{3}} \right)^2}\]
\[A = \dfrac{{4\pi }}{9}c{m^2}...........(1)\]
The water flows at a rate of 0.63 m per second into the cylindrical tank.
\[0.63m/s = 63cm/s\]
The volume of water flowing into the cylinder per second is equal to the product of the area of cross-section and the flow rate.
\[v = \dfrac{{4\pi }}{9} \times 63\]
\[v = \dfrac{{252\pi }}{9}c{m^3}/s\]
The volume of water that flows into the tank in one hour is given by:
\[V = \dfrac{{252\pi }}{9} \times 60 \times 60\]
\[V = 100800\pi c{m^3}\]
Converting volume into cubic meters by dividing by \[{10^6}\].
\[V = \dfrac{{100800\pi }}{{{{10}^6}}}{m^3}\]
\[V = 0.1008\pi {m^3}...........(2)\]
We know that the volume of a cylinder with radius r and height h is given as follows:
\[V = \pi {r^2}h\]
The radius of the base of the cylindrical tank is 0.2 m.
\[V = \pi {(0.2)^2}h\]
\[V = 0.04\pi h...........(3)\]
We know that equation (2) and equation (3) are equal. Then, we have:
\[0.04\pi h = 0.1008\pi \]
Solving for h, we have:
\[h = \dfrac{{0.1008}}{{0.04}}\]
\[h = 2.52m\]
Hence, the water fills up to a height of 2.52 m.
Hence, the correct answer is option (b).
Note: Convert all quantities into the same units before multiplying or dividing them, otherwise, the answer will differ by powers of 10, which is wrong.
Complete step-by-step answer:
It is given that the water flows out through a circular pipe, whose internal diameter is \[1\dfrac{1}{3}\] cm. Converting mixed fraction \[1\dfrac{1}{3}\] into an improper fraction, we have:
\[1\dfrac{1}{3} = \dfrac{{1 \times 3 + 1}}{3}\]
\[1\dfrac{1}{3} = \dfrac{4}{3}\]
The radius of the circular pipe is half of the diameter, that is, half of \[\dfrac{4}{3}\] cm.
\[{r_{pipe}} = \dfrac{1}{2} \times \dfrac{4}{3}\]
\[{r_{pipe}} = \dfrac{2}{3}cm\]
The area of cross-section of the pipe is given as \[\pi {r_{pipe}}^2\], hence, we have:
\[A = \pi {r_{pipe}}^2\]
\[A = \pi {\left( {\dfrac{2}{3}} \right)^2}\]
\[A = \dfrac{{4\pi }}{9}c{m^2}...........(1)\]
The water flows at a rate of 0.63 m per second into the cylindrical tank.
\[0.63m/s = 63cm/s\]
The volume of water flowing into the cylinder per second is equal to the product of the area of cross-section and the flow rate.
\[v = \dfrac{{4\pi }}{9} \times 63\]
\[v = \dfrac{{252\pi }}{9}c{m^3}/s\]
The volume of water that flows into the tank in one hour is given by:
\[V = \dfrac{{252\pi }}{9} \times 60 \times 60\]
\[V = 100800\pi c{m^3}\]
Converting volume into cubic meters by dividing by \[{10^6}\].
\[V = \dfrac{{100800\pi }}{{{{10}^6}}}{m^3}\]
\[V = 0.1008\pi {m^3}...........(2)\]
We know that the volume of a cylinder with radius r and height h is given as follows:
\[V = \pi {r^2}h\]
The radius of the base of the cylindrical tank is 0.2 m.
\[V = \pi {(0.2)^2}h\]
\[V = 0.04\pi h...........(3)\]
We know that equation (2) and equation (3) are equal. Then, we have:
\[0.04\pi h = 0.1008\pi \]
Solving for h, we have:
\[h = \dfrac{{0.1008}}{{0.04}}\]
\[h = 2.52m\]
Hence, the water fills up to a height of 2.52 m.
Hence, the correct answer is option (b).
Note: Convert all quantities into the same units before multiplying or dividing them, otherwise, the answer will differ by powers of 10, which is wrong.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
The branch of science which deals with nature and natural class 10 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Define absolute refractive index of a medium
Find out what do the algal bloom and redtides sign class 10 biology CBSE
Prove that the function fleft x right xn is continuous class 12 maths CBSE
Trending doubts
State the differences between manure and fertilize class 8 biology CBSE
Why are xylem and phloem called complex tissues aBoth class 11 biology CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
What would happen if plasma membrane ruptures or breaks class 11 biology CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
What precautions do you take while observing the nucleus class 11 biology CBSE
What would happen to the life of a cell if there was class 11 biology CBSE
Change the following sentences into negative and interrogative class 10 english CBSE