Question

A solid sphere of mass $2\mathrm{kg}$ and density ${10}^5\mathrm{kg}/{\mathrm{m}}^3$ hanging from a string is lowered into a vessel of uniform cross-section area $10{\mathrm{m}}^2$ containing a liquid of density $0.5\times {10}^5\mathrm{kg}/{\mathrm{m}}^3$, until it is fully immersed. The increase in pressure of liquid at the base of the vessel is (Assume liquid does not spills out of the vessel and take $\mathrm{g}=10\mathrm{m}/{\mathrm{s}}^2)$ (1) $2\mathrm{N}/{\mathrm{m}}^2$ (2) $1\mathrm{N}/{\mathrm{m}}^2$ (3) $4\mathrm{N}/{\mathrm{m}}^2$ (4) $\frac{1}{2}\mathrm{N}/{\mathrm{m}}^2$

Short answer

Increase in pressure at the base of the vessel is 1 N/m^2, option (2).

Step-by-step solution

Understand the given data

Mass of the sphere, m = 2 kgDensity of the sphere, ${\rho }_s$ = ${10}^5$ kg/m^3Cross-sectional area of the vessel, A = 10 m^2Density of the liquid, ${\rho }_l$ = $0.5\times {10}^5$ kg/m^3Gravitational acceleration, g = 10 m/s^2

Calculate volume of the sphere

Using the formula for density: $\rho =\frac{m}{V}$Volume of the sphere, $V$$$V=\frac{m}{{\rho }_s}=\frac{2\text{ kg}}{{10}^5{\text{ kg/m}}^3}=2\times {10}^{-5}{\text{ m}}^3$$

Determine the buoyant force

Buoyant force is given by Archimedes' principle, which states it is equal to the weight of the displaced liquid.$$F_b={\rho }_l\cdot V\cdot g$$$$F_b=(0.5\times {10}^5{\text{ kg/m}}^3)\cdot (2\times {10}^{-5}{\text{ m}}^3)\cdot 10{\text{ m/s}}^2$$$$F_b=10\text{ N}$$

Find increase in pressure at the base of the vessel

Increase in pressure $\mathrm{\Delta }P$ is given by:$$\mathrm{\Delta }P=\frac{\text{Buoyant force}}{\text{cross-sectional area}}$$$$\mathrm{\Delta }P=\frac{10\text{ N}}{10{\text{ m}}^2}=1{\text{ N/m}}^2$$

Final result

The increase in pressure of the liquid at the base of the vessel is 1 N/m^2, which corresponds to option (2).

Key concepts

Buoyant Force

The buoyant force is an upward force exerted by a fluid that opposes the weight of an immersed object. Imagine dropping an object into water. The water pushes the object up. This push is the buoyant force. It happens because the fluid pressure at the bottom of the object is greater than at the top.

How Does It Work?

• When an object is submerged, the fluid exerts pressures on all surfaces of the object.
• Pressure increases with depth, leading to a higher force at the bottom.
• This difference in pressure causes an upward net force.

The magnitude of the buoyant force can be found using Archimedes' principle. It's equal to the weight of the fluid displaced by the object.

For our exercise:
$$F_b=10\text{ N}$$

Pressure Increase

When an object is submerged in a fluid, it displaces some of the fluid. This displacement causes the pressure in the fluid to change. The increase in pressure at a specific point is due to the weight of the fluid column above that point.

Why Does It Happen?

• An increase in the amount of fluid leads to an increase in weight.
• This extra weight results in higher pressure at deeper points.

To calculate this pressure increase at the base of the vessel, we use the formula: $$\frac{\text{Buoyant force}}{\text{cross-sectional area}}$$ Here, it's essential to consider the area of the container. A pressure increase means more force spread over a specific area. In our case,
$$\frac{10\text{ N}}{10{\text{ m}}^2}\to 1{\text{ N/m}}^2$$

Archimedes' Principle

Archimedes' principle is a fundamental law of physics that explains why objects seem lighter in water. It states that any object, wholly or partially submerged in a fluid, is buoyed up by a force equal to the weight of the fluid displaced by the object.

The Principle in Action

• Submerge an object in water.
• Measure the volume of water displaced.
• Calculate the weight of this water.
• This weight is the buoyant force acting on the object.

For our sphere in the exercise:
$${\rho }_l\times V\times g\to \text{0.5}\times {10}^5{\text{ kg/m}}^3\times 2\times {10}^{-5}{\text{ m}}^3\times 10{\text{ m/s}}^2\to 10\text{ N}$$