
In a cylindrical vessel of diameter 24 cm filled up with a sufficient quantity of water, a solid spherical ball of radius 6 cm is completely immersed. Then the increase in the height of water level is:
(a) 1.5 cm
(b) 2 cm
(c) 3 cm
(d) 1 cm
Answer
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Hint: We are asked to find an increase in the length of water if the spherical ball is immersed in it. We start by finding the volume of the ball using the formula, \[\text{Volume}=\dfrac{4}{3}\pi {{r}^{3}},\] where r is the radius. Then we will find the volume of the cylinder using the formula \[\pi {{r}^{2}}H.\] As we know that volume of ball = volume of water. We will compare these two to get our solution.
Complete step by step answer:
We are given that we have a cylindrical vessel of diameter 24 cm filled up with water. A spherical ball of radius 6 cm is completely immersed into it.
Now, first of all, we will find the volume of the spherical ball. We know that the volume of the sphere is given as,
\[\text{Volume}=\dfrac{4}{3}\pi {{r}^{3}}\]
As r = 6, we get,
\[\text{Volume of ball}=\dfrac{4}{3}\times \pi \times {{\left( 6 \right)}^{3}}\]
By the Archimedes principle, we know that the amount of object immersed in water is the amount that will move up. So, water up to a height say “h” will move up such that the volume is the same as the volume of the ball.
Now, the volume of the cylinder is given as \[\pi {{r}^{2}}H.\] As we have diameter = 24 cm, so,
\[\text{Radius}=\dfrac{\text{Diameter}}{2}\]
\[\Rightarrow \text{Radius}=\dfrac{24}{2}\]
\[\Rightarrow \text{Radius}=12cm\]
Now, the volume of the cylinder will be
\[\text{Volume of cylinder}=\pi \times {{\left( 12 \right)}^{2}}\times h\]
Now by Archimedes principle, the volume of the cylinder = volume of the ball
\[\Rightarrow \pi {{\left( 12 \right)}^{2}}h=\dfrac{4}{3}\pi {{\left( 6 \right)}^{3}}\]
Cancelling the like terms, we get,
\[\Rightarrow h=\dfrac{6\times 6\times 6\times 4}{3\times 12\times 12}\]
Simplifying, we get,
\[\Rightarrow h=2cm\]
Therefore, the increase in the length of water is 2 cm.
So, the correct answer is “Option B”.
Note: In order to find the volume of the cylinder, we will first have to find the radius using \[r=\dfrac{d}{2}\] and then use the formula for the volume of the cylinder. Students shouldn’t solve all the calculations at the starting because as we compare, the last term gets cancelled out and we get the answer easily. The early calculation makes the solution more complex.
Complete step by step answer:
We are given that we have a cylindrical vessel of diameter 24 cm filled up with water. A spherical ball of radius 6 cm is completely immersed into it.
Now, first of all, we will find the volume of the spherical ball. We know that the volume of the sphere is given as,
\[\text{Volume}=\dfrac{4}{3}\pi {{r}^{3}}\]
As r = 6, we get,
\[\text{Volume of ball}=\dfrac{4}{3}\times \pi \times {{\left( 6 \right)}^{3}}\]
By the Archimedes principle, we know that the amount of object immersed in water is the amount that will move up. So, water up to a height say “h” will move up such that the volume is the same as the volume of the ball.
Now, the volume of the cylinder is given as \[\pi {{r}^{2}}H.\] As we have diameter = 24 cm, so,
\[\text{Radius}=\dfrac{\text{Diameter}}{2}\]
\[\Rightarrow \text{Radius}=\dfrac{24}{2}\]
\[\Rightarrow \text{Radius}=12cm\]
Now, the volume of the cylinder will be
\[\text{Volume of cylinder}=\pi \times {{\left( 12 \right)}^{2}}\times h\]
Now by Archimedes principle, the volume of the cylinder = volume of the ball
\[\Rightarrow \pi {{\left( 12 \right)}^{2}}h=\dfrac{4}{3}\pi {{\left( 6 \right)}^{3}}\]
Cancelling the like terms, we get,
\[\Rightarrow h=\dfrac{6\times 6\times 6\times 4}{3\times 12\times 12}\]
Simplifying, we get,
\[\Rightarrow h=2cm\]
Therefore, the increase in the length of water is 2 cm.
So, the correct answer is “Option B”.
Note: In order to find the volume of the cylinder, we will first have to find the radius using \[r=\dfrac{d}{2}\] and then use the formula for the volume of the cylinder. Students shouldn’t solve all the calculations at the starting because as we compare, the last term gets cancelled out and we get the answer easily. The early calculation makes the solution more complex.
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