
Two cylinders of the same height but of radius $r$ and \[2r\] are immersed in the same liquid, if the buoyant force exerted on the 1st cylinder is 50 units. What will be the buoyant force on the 2nd cylinder?
A) 200 units
B) 100 units
C) 50 units
D) 25 units
Answer
218.1k+ views
Hint: When an object is placed in liquid, it will float or sink depending on the density of the object as compared to the density of the liquid. The buoyant force acting on the object immersed in the liquid depends on the volume of the object.
Formula used: In this solution, we will use the following formula
1. Buoyant force: ${F_B} = \rho Vg$ where $\rho $ is the density of the liquid, $V$ is the volume of the object immersed in the liquid, and $g$ is the gravitational acceleration.
2. volume of a cylinder: $V = \pi {R^2}h$ where $R$ is the radius and \[h\] is the height of the cylinder
Complete step by step answer:
When an object is placed in water, the buoyant force that acts on it will depend on the volume of the object. Since we’ve been told that the cylinder is immersed in the object, we can assume that it is completely under the water.
Now, we know that volume of a cylinder is calculated as
$V = \pi {R^2}h$
Since the height of both the cylinders is the same, only the radius will affect the volume of the cylinder. So, we can write that
$V \propto {R^2}$
Since the radius of the second cylinder is twice that of the first, we can write
$\dfrac{{{V_1}}}{{{V_2}}} = \dfrac{{{R_1}^2}}{{{R_2}^2}}$
Which gives us
${V_2} = {V_1}\dfrac{{{R_2}^2}}{{{R_1}^2}}$
Substituting ${R_2} = 2r$ and ${R_1} = r$, we get
${V_2} = 4{V_1}$
Since the volume of the cylinder increases by four times, the buoyant force will also change by 4 times since the buoyant force also depends linearly on the volume of the object.
Hence the buoyant force will be
${F_B} = 50 \times 4 = 200\,{\text{units}}$
Hence the correct choice is option (A).
Note: In this question, we must realize that since the cylinders are immersed in the water, their entire volume will be in the water. If this were not the case, we would have to balance the buoyant force with the weight of the cylinder however since we have not been given any masses in the question that can serve as a clue that the entire cylinder is inside the water.
Formula used: In this solution, we will use the following formula
1. Buoyant force: ${F_B} = \rho Vg$ where $\rho $ is the density of the liquid, $V$ is the volume of the object immersed in the liquid, and $g$ is the gravitational acceleration.
2. volume of a cylinder: $V = \pi {R^2}h$ where $R$ is the radius and \[h\] is the height of the cylinder
Complete step by step answer:
When an object is placed in water, the buoyant force that acts on it will depend on the volume of the object. Since we’ve been told that the cylinder is immersed in the object, we can assume that it is completely under the water.
Now, we know that volume of a cylinder is calculated as
$V = \pi {R^2}h$
Since the height of both the cylinders is the same, only the radius will affect the volume of the cylinder. So, we can write that
$V \propto {R^2}$
Since the radius of the second cylinder is twice that of the first, we can write
$\dfrac{{{V_1}}}{{{V_2}}} = \dfrac{{{R_1}^2}}{{{R_2}^2}}$
Which gives us
${V_2} = {V_1}\dfrac{{{R_2}^2}}{{{R_1}^2}}$
Substituting ${R_2} = 2r$ and ${R_1} = r$, we get
${V_2} = 4{V_1}$
Since the volume of the cylinder increases by four times, the buoyant force will also change by 4 times since the buoyant force also depends linearly on the volume of the object.
Hence the buoyant force will be
${F_B} = 50 \times 4 = 200\,{\text{units}}$
Hence the correct choice is option (A).
Note: In this question, we must realize that since the cylinders are immersed in the water, their entire volume will be in the water. If this were not the case, we would have to balance the buoyant force with the weight of the cylinder however since we have not been given any masses in the question that can serve as a clue that the entire cylinder is inside the water.
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