Problem 76 - A Complete Resource Book in Physics for JEE Main

Question

This question contains Statement 1 and Statement 2 of the four choices given after the statements, choose the one that best describes the two statements. [2008] Statement 1: For a mass \(M\) kept at the centre of a cube of side \(a\) the flux of gravitational field passing through its sides \(4 \pi G M\). Statement 2: If the direction of a field due to a point source is radial and its dependence on the distance \(r\) from the source is given as \(\frac{1}{r^{2}}\), its flux through a closed surface depends only on the strength of the source enclosed by the surface and not on the size or shape of the surface. (A) Statement 1 is false, Statement 2 is true. (B) Statement 1 is true, Statement 2 is true; Statement 2 is a correct explanation for Statement 1 . (C) Statement 1 is false, Statement 2 is true; Statement 2 is not a correct explanation for Statement \(1 .\) (D) Statement 1 is true, Statement 2 is false.

Step-by-Step Solution

Answer

(B) Statement 1 is true, Statement 2 is true; Statement 2 is a correct explanation for Statement 1.

Step 1: Gravitational field passing through the sides of a cube with mass M kept at its center can be found using Gauss's law for gravity. Gauss's law states that the gravitational flux through any closed surface is proportional to the mass enclosed by that surface. We can use this law and integrate the gravitational field over the entire surface(area) of the cube to find the total flux. Let's derive and evaluate the gravitational flux for this given scenario. #Step 2: Deriving Gravitational flux using Gauss's Law for gravity#:

The gravitational field due to mass \(M\) at a distance \(r\) is given by: \(E = -\frac{GMm}{r^2} \hat{r}\) The infinitesimal electric flux \(d \phi\) through an infinitesimal area \(dA\) is given by: \(d\phi = E \cdot dA = \left(-\frac{GMm}{r^2} \hat{r}\right) \cdot (dA \hat{r})\) \(d\phi = -\frac{GMm}{r^2} dA\) The total gravitational flux through the cube is given by the integral of the electric flux over the entire surface of the cube. We can do this integration for all six faces of the cube. \(\phi = \int d\phi = -GMm \int_{cube} \frac{dA}{r^2} \) #Step 3: Calculate gravitational flux for the given problem scenario#:

Step 2: The Gravitational flux over any surface enclosing mass M equals: \(\phi = -4\pi GMm\) As per the Gauss' Law for gravity, the flux through any closed surface is independent of the size of the surface and only depends on the mass enclosed by the surface. So, the statement 1 can be rephrased, however, its sense is still correct. Hence, Statement 1 is technically true. #Step 4: Analyzing Statement 2#:

As we already derived the gravitational flux using Gauss's law and saw that the flux depends only on the strength of the source enclosed by the surface and not on the size or shape of the surface, we conclude that Statement 2 is true. #Step 5: Check if Statement 2 is a correct explanation for Statement 1#:

Step 3: Statement 2 states the principle of Gauss's Law applied in this context. Essentially, this was used to derive the result of Statement 1. So, Statement 2 is a correct explanation for Statement 1. #Final Answer#:

Based on our analysis of the given statements, we can conclude that the correct choice is: (B) Statement 1 is true, Statement 2 is true; Statement 2 is a correct explanation for Statement 1.

Key Concepts

Gravitational Field Flux

Gravitational field flux is a measure of the amount of gravitational field passing through a given area. If we consider the Earth, which has a mass that creates a gravitational field around it, this field can be visualized as lines spreading outwards. These lines represent the direction in which a small mass would be attracted if placed within the field. Now imagine a surface, such as the side of a cube, positioned in this field. The gravitational field flux through this surface would be the number of lines passing perpendicularly through it.

To quantify this concept, we use Gauss's Law for Gravity, which parallels the similar electrical concept. Gauss's Law for Gravity states that the gravitational flux through a closed surface is equal to \( -4\pi G M \) where \( G \) is the gravitational constant and \( M \) is the mass enclosed by the surface. Notice the negative sign, which is a reflection of the fact that gravity is an attractive force, pulling masses together, unlike electric charges which can also repel. Therefore, the more mass there is inside the surface, the higher the gravitational field flux through that surface.

Closed Surface Flux

Understanding the concept of closed surface flux is crucial in physics, especially when applying Gauss's Law. A closed surface can be thought of as a balloon that encompasses a given volume of space with no openings. For Gauss's Law, the flux through this closed surface is what we're interested in. Importantly, it doesn't matter what shape the surface is—as long as it is entirely closed.

When calculating closed surface flux for gravity, we find that the total flux depends solely on the enclosed mass and not on the specific shape or size of the surface. This is why, in the exercise, the cubic shape does not impact the calculation of flux. Whether it's a sphere, a cube, or any other closed shape, the gravitational flux will be the same if the mass inside is the same. The flux through a closed surface is like a 'net' amount; it considers all the field lines going in and out. This profound principle means that for a planet like Earth, the gravitational flux through a hypothetical closed surface surrounding it will be constant regardless of the surface's shape.

Integration in Physics

Integration is a mathematical operation used extensively in physics to sum up small pieces of quantities to get a total. It is vital in calculating various physical properties, like the gravitational field flux discussed previously. When we integrate, we essentially add up an 'infinitesimal' quantity—like the tiny bits of flux through each part of a surface—to find a whole. It's like determining how much paint is needed for a wall by adding up the paint coverage for each square inch.

Integration allows us to move from the specific (field at a point) to the general (total flux through a surface). This is key in our understanding of gravitation, as it permits the evaluation of gravitational forces and fluxes over complex volumes and surfaces. For example, when we speak of the gravitational field flux through a cube, we use integration to sum the effects at every point on the cube’s surface. This inclusion of integration in our problem-solving toolkit is essential to produce results that reflect the complexities of physical systems.