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Let V and E be the gravitational potent...

Let `V` and `E` be the gravitational potential field. Then select the correct alternative (s)

A

the plot of `E` against `r` (distance from centre) is discontinuous for a spherical shell

B

the plot of `V` against `r` is continous for a spherical shell

C

the plot of `E` against `r` is discontinuous for a solid sphere

D

the plot of `V` against `r` is continuous for a solid sphere

Text Solution

AI Generated Solution

The correct Answer is:
To solve the question regarding the gravitational potential field \( V \) and the gravitational field \( E \), we will analyze each statement provided in the question step by step. ### Step 1: Analyze the first statement **Statement**: The plot of \( E \) against \( R \) is discontinuous for the spherical shell. - For a spherical shell, the gravitational field \( E \) inside the shell is zero, and outside the shell, it behaves like a point mass. Therefore, there is a discontinuity at the surface of the shell where the field jumps from 0 to a non-zero value. - **Conclusion**: This statement is **correct**. ### Step 2: Analyze the second statement **Statement**: The plot of \( V \) against \( R \) is continuous for the spherical shell. - The gravitational potential \( V \) inside the shell is constant (equal to the potential at the surface), and outside it decreases with \( 1/R \). Since there is no jump in potential at the surface, the graph is continuous. - **Conclusion**: This statement is **correct**. ### Step 3: Analyze the third statement **Statement**: The plot of \( E \) against \( R \) is discontinuous for the solid sphere. - For a solid sphere, the gravitational field \( E \) inside the sphere varies linearly with \( R \) (proportional to \( R \)), and outside it behaves like a point mass (inversely proportional to \( R^2 \)). There is no discontinuity; the graph is continuous. - **Conclusion**: This statement is **incorrect**. ### Step 4: Analyze the fourth statement **Statement**: The plot of \( V \) against \( R \) is continuous for a solid sphere. - The gravitational potential \( V \) for a solid sphere is continuous. Inside the sphere, it increases linearly with \( R \), and outside it decreases with \( 1/R \). There is no discontinuity at the surface. - **Conclusion**: This statement is **correct**. ### Final Conclusion Based on the analysis: - The correct statements are: 1, 2, and 4.
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