Question

A sphere of radius \[3{\text{ }}cm\] is dropped into a cylindrical vessel of radius \[{\text{4 }}cm\]. If the sphere is submerged completely, then the height (in cm) to which the water rises is?
\[
  \left( A \right)\;2.35\;\;\;\;\;\;\;\;\;\;\;\;\; \\
  \left( B \right){\text{ }}2.30\;\;\;\;\;\;\;\; \\
  \left( C \right){\text{ }}2.25\;\;\;\;\;\;\;\;\; \\
  \left( D \right){\text{ }}2.15 \\
 \]

Answer 1

Hint: We will have to know the formula of volume of sphere and cylinder to solve this problem. Height of water rise will occur due to complete submergence of the sphere in the cylindrical vessel. Hence the volume of raised water in the cylindrical vessel will be equal to the total volume of the sphere. So first of all, find the volume of the sphere & then equalize it to the volume of water raised.

Complete step-by-step answer:
According to the question as the sphere dropped in the cylindrical vessel so there is a rise in the height of the water. So, let the rise in height in water be h cm
Given, radius of sphere = \[3{\text{ }}cm\], Radius of cylindrical vessel = \[{\text{4 }}cm\]
 As we know that the $Volume{\text{ }}of{\text{ }}the{\text{ }}sphere = \dfrac{4}{3}\pi {r^3}$
Now, we put the value of radius in the formula we get,
$ = \dfrac{4}{3} \times \pi \times {3^3}$
As we know that the $Volume{\text{ }}of{\text{ }}cylindrical{\text{ }}vessel = \pi {r^2}h$
Now, we put the value of radius in the formula we get,
$ = \pi \times {4^2} \times h$
As the volume of the water displaced by the sphere is equal to the volume of the sphere.
$\therefore Volume{\text{ }}of{\text{ }}cylindrical{\text{ }}vessel = Volume{\text{ }}of{\text{ }}the{\text{ }}sphere$
Thus, equating both sides volume of raised water in cylindrical vessel, we get
$ \Rightarrow \pi {r^2}h = \dfrac{4}{3}\pi \times {r^3}$
Now we put all the values in the equation we get,
$
   \Rightarrow \pi \times {4^2} \times h = \dfrac{4}{3} \times \pi \times {3^3} \\
    \\
 $
Now we solve the equation to find the value of h, we get
$ \Rightarrow h = \dfrac{{4 \times \pi \times 3 \times 3 \times 3}}{{3 \times \pi \times 4 \times 4}}$
Now we eliminate the similar terms we get
$
   \Rightarrow h = \dfrac{{3 \times 3}}{4} \\
   \Rightarrow h = \dfrac{9}{4} \\
   \Rightarrow h = 2.25 \\
 $
So the rise in the height in water is 2.25 cm

Hence , the correct option is C.

Note: Key formula needed here - volume of the sphere = $\dfrac{4}{3}\pi {r^3}$
Key concept used is that the volume of the sphere should be equal to the volume of water replaced by it .
In this type of problem, mistakes may be done while equating the total volume of the cylindrical vessel to that of the volume of the sphere. Hence it should be done carefully.