
Marbles of diameter \[1.4\] cm each are dropped into a cylindrical beaker of diameter 7 cm containing some water. Find the number of marbles that should be dropped into the beaker so that the water level rises by \[5.6\] cm.
Answer
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Hint: Here, we will find the total number of marbles which will be dropped. We will equate the total volume of the water rises and the total volume of the marbles dropped. Then by solving the obtained equation we will get the total number of marbles dropped to raise the level of the water by \[5.6\] cm.
Complete step-by-step answer:
Let \[n\] be the number of marbles dropped.
Firstly, we will calculate the volume of each marble. As we know that marble is of spherical shape.
So, volume of each marble\[ = \dfrac{{4\pi {r^3}}}{3}\] where, \[r\] is the radius of the marble and value of \[{\rm{\pi }}\] is \[3.14\].
It is given that the diameter of the marble is \[1.4\] cm. As the radius is half of the diameter. Therefore, radius of the marble is \[0.7\] cm.
Thus, Volume of the raised water \[ = \dfrac{{4 \times 3.14 \times {{0.7}^3}}}{3} = 1.44{\rm{ c}}{{\rm{m}}^3}\]
Now we have to find out the volume of the raise water.
Volume of the raised water \[ = \pi {r^2}h\] where, \[r\] is the radius of the cylindrical beaker, \[h\] is the height by which the water level rises and value of \[{\rm{\pi }}\] is \[3.14\].
It is given that the diameter of the cylindrical beaker is 7 cm. Therefore, radius of the cylindrical beaker is \[3.5\] cm.
Thus, Volume of the raised water \[ = 3.14 \times {3.5^2} \times 5.6 = 215.51{\rm{c}}{{\rm{m}}^3}\]
Now we know that the total volume of the raised water is equal to the total volume of the marble.
Volume of the raised water \[ = {\rm{ }}n \times \] Volume of each marble
\[ \Rightarrow 215.51 = n \times 1.44\]
So, by solving this equation we will get the value of \[n\].
\[ \Rightarrow n = 150\]
Hence, 150 marbles should be dropped into the beaker so that the water level rises by \[5.6\] cm.
Note: Here, we have used the formula of volume of a sphere because we have assumed that the shape of marble is spherical. Though it is not mentioned in the question that the marble is spherical, but in real life, it is spherical. When the number of marble increases, then the water level is raised. This is because the marble takes up some space in the beaker.
Complete step-by-step answer:
Let \[n\] be the number of marbles dropped.
Firstly, we will calculate the volume of each marble. As we know that marble is of spherical shape.
So, volume of each marble\[ = \dfrac{{4\pi {r^3}}}{3}\] where, \[r\] is the radius of the marble and value of \[{\rm{\pi }}\] is \[3.14\].
It is given that the diameter of the marble is \[1.4\] cm. As the radius is half of the diameter. Therefore, radius of the marble is \[0.7\] cm.
Thus, Volume of the raised water \[ = \dfrac{{4 \times 3.14 \times {{0.7}^3}}}{3} = 1.44{\rm{ c}}{{\rm{m}}^3}\]
Now we have to find out the volume of the raise water.
Volume of the raised water \[ = \pi {r^2}h\] where, \[r\] is the radius of the cylindrical beaker, \[h\] is the height by which the water level rises and value of \[{\rm{\pi }}\] is \[3.14\].
It is given that the diameter of the cylindrical beaker is 7 cm. Therefore, radius of the cylindrical beaker is \[3.5\] cm.
Thus, Volume of the raised water \[ = 3.14 \times {3.5^2} \times 5.6 = 215.51{\rm{c}}{{\rm{m}}^3}\]
Now we know that the total volume of the raised water is equal to the total volume of the marble.
Volume of the raised water \[ = {\rm{ }}n \times \] Volume of each marble
\[ \Rightarrow 215.51 = n \times 1.44\]
So, by solving this equation we will get the value of \[n\].
\[ \Rightarrow n = 150\]
Hence, 150 marbles should be dropped into the beaker so that the water level rises by \[5.6\] cm.
Note: Here, we have used the formula of volume of a sphere because we have assumed that the shape of marble is spherical. Though it is not mentioned in the question that the marble is spherical, but in real life, it is spherical. When the number of marble increases, then the water level is raised. This is because the marble takes up some space in the beaker.
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