Problem 19 - Master Resource Book in JEE Main Physics

Question

A piece of gold weights \(50 \mathrm{~g}\) in air and \(45 \mathrm{~g}\) in water. If there is a cavity inside the piece of gold, then find its volume [Density of gold \(=19.3 \mathrm{~g} / \mathrm{cc}]\). (a) \(2.4 \mathrm{~cm}^{3}\) (b) \(2.4 \mathrm{~m}^{3}\) (c) \(4.2 \mathrm{~m}^{3}\) (d) \(4.2 \mathrm{~mm}^{7}\)

Step-by-Step Solution

Answer

The cavity's volume is approximately 2.4 cm³. Option (a) is correct.

Step 1: Understand the Problem

We have a piece of gold weighing 50 g in air and 45 g in water. The loss of weight in water is due to buoyancy, which equals the weight of the water displaced by the object. We need to find the volume of the cavity inside the gold.

Step 2: Apply Archimedes' Principle

Using Archimedes' Principle, the loss of weight in water is equal to the weight of the water displaced. Thus, the water displaced weighs 5 g (50 g - 45 g). Since the density of water is 1 g/cm³, this means the volume of water displaced, and hence the volume of the cavity, is 5 cm³.

Step 3: Check the Density of Gold

From the problem, the density of gold is given as 19.3 g/cm³. We need to ensure our calculations align with this density to ascertain if there's a cavity. A solid piece of gold should weigh less than its apparent weight in water if there are cavities.

Step 4: Calculate Expected Volume without Cavity

For a solid piece of gold without a cavity, its density and mass would give the volume: \[ \text{Volume} = \frac{50 \text{ g}}{19.3 \text{ g/cm}^3} \approx 2.59 \text{ cm}^3. \] Since the displaced water suggests a volume of 5 cm³, this indicates a discrepancy due to the cavity.

Step 5: Determine Volume of the Cavity

The gold's volume suggests it fills only 2.59 cm³ solidly. However, the piece displaces 5 cm³ of water, so the volume of the cavity is: \[ \text{Cavity Volume} = 5 \text{ cm}^3 - 2.59 \text{ cm}^3 = 2.41 \text{ cm}^3. \] We're looking for the closest option available.

Key Concepts

Density

Density is a fundamental concept that helps us understand how much mass is contained in a given volume of a substance. It is typically expressed in grams per cubic centimeter (g/cm³) or kilograms per cubic meter (kg/m³). This relationship can be represented by the equation:

• Density \( \rho = \frac{\text{Mass}}{\text{Volume}} \)

For gold in our problem, it has a density of 19.3 g/cm³. This means for every cubic centimeter of volume, the gold has 19.3 grams of mass. Determining the density of an object can help identify it, as different materials have characteristic densities.
In this exercise, the difference in mass readings (50 g in air and 45 g in water) prompted the investigation of a cavity through analyzing the density. By comparing the calculated density to the known density of gold, the problem hints at a void or cavity affecting the object’s true density."},{

Buoyancy

Buoyancy is the upward force that a fluid exerts on an object immersed in it. This principle is crucial when objects are submerged in a fluid, like water, and float or sink. The famous Archimedes' Principle quantifies buoyancy as the weight of the fluid displaced by the object. This is crucial in figuring out the apparent loss of weight an object experiences when submerged.
In our problem, the piece of gold boils down from 50 g in air to 45 g in water. The 5 g difference represents the buoyancy force acting on the gold, equating to the weight of 5 g of water displaced. This calculation prominently shows how buoyancy directly affects measurements, assisting in identifying discrepancies, such as cavities, in the object's volume.

Volume Displacement

Volume displacement refers to the space that an object occupies when submerged in a liquid, measured by the liquid level's rise. This concept underpins the calculation of buoyancy and helps us determine the actual volume of hidden or irregular shapes, such as any cavity inside solid objects when immersed in a fluid.
For the gold in this exercise, once submerged, the water it displaces weighs 5 g. This means the volume of water—or liquid—pushed aside by the gold equals 5 cm³ owing to the water’s density (1 g/cm³).
Through understanding volume displacement, we can deduce the volume of the suspected cavity. Initially, we calculated the solid volume of gold, expecting it to match the water displacement. Since there is a mismatch indicating that more volume is displaced than calculated, it points to the presence of an internal cavity measuring 2.41 cm³, offering insightful depth into the real nature of the gold.