Question

The gravitational potential of two homogeneous spherical shells A and B (separated by large distance) of same surface mass density at their respective centres are in the ratio 3 : 4. If the two shells coalesce into single one such that surface mass density remains same, then the ratio of potential at an internal point of the new shell to shell A is equal to :

• A. 3 : 2
• B. 4 : 3
• C. 5 : 3
• D. 3 : 5

Answer 1

Answer: (C)

5 : 3

Answer 2

The problem involves two spherical shells coalescing into a single one while maintaining the surface mass density. We need to find the ratio of gravitational potentials.

1. Initial State of Shells A and B:

Let $R_A$ and $R_B$ be the radii of shell A and shell B, respectively.
Let $\sigma$ be the constant surface mass density.

The mass of a spherical shell is given by $M = \sigma \times (\text{surface area}) = \sigma (4\pi R^2)$.
So, $M_A = \sigma (4\pi R_A^2)$ and $M_B = \sigma (4\pi R_B^2)$.

The gravitational potential at any internal point (including the center) of a homogeneous spherical shell of mass $M$ and radius $R$ is constant and given by $V = -\frac{GM}{R}$.

For shell A, the potential at its center is $V_A = -\frac{GM_A}{R_A}$.
Substitute $M_A$:
$V_A = -\frac{G(\sigma 4\pi R_A^2)}{R_A} = -G\sigma 4\pi R_A$.

For shell B, the potential at its center is $V_B = -\frac{GM_B}{R_B}$.
Substitute $M_B$:
$V_B = -\frac{G(\sigma 4\pi R_B^2)}{R_B} = -G\sigma 4\pi R_B$.

Given that the ratio of their potentials at their respective centers is 3:4:

$$\frac{V_A}{V_B} = \frac{-G\sigma 4\pi R_A}{-G\sigma 4\pi R_B} = \frac{R_A}{R_B}$$

So,

$$\frac{R_A}{R_B} = \frac{3}{4} \implies R_B = \frac{4}{3} R_A$$

2. Coalescence into a New Shell:

Let the new shell be C, with radius $R_C$ and mass $M_C$.
The surface mass density remains the same, $\sigma$.

The total mass of the new shell is the sum of the masses of shells A and B:

$$M_C = M_A + M_B = \sigma (4\pi R_A^2) + \sigma (4\pi R_B^2) = \sigma 4\pi (R_A^2 + R_B^2)$$

Since the new shell also has surface mass density $\sigma$, its mass can also be expressed as:

$$M_C = \sigma (4\pi R_C^2)$$

Equating the two expressions for $M_C$:

$$\sigma 4\pi R_C^2 = \sigma 4\pi (R_A^2 + R_B^2)$$ $$R_C^2 = R_A^2 + R_B^2$$

Now, substitute $R_B = \frac{4}{3} R_A$:

$$R_C^2 = R_A^2 + \left(\frac{4}{3} R_A\right)^2 = R_A^2 + \frac{16}{9} R_A^2$$ $$R_C^2 = \frac{9R_A^2 + 16R_A^2}{9} = \frac{25R_A^2}{9}$$

Taking the square root:

$$R_C = \sqrt{\frac{25R_A^2}{9}} = \frac{5}{3} R_A$$

3. Ratio of Potential at an Internal Point of the New Shell to Shell A:

The gravitational potential at an internal point of the new shell C is $V_C = -\frac{GM_C}{R_C}$.
Substitute $M_C = \sigma 4\pi R_C^2$:

$$V_C = -\frac{G(\sigma 4\pi R_C^2)}{R_C} = -G\sigma 4\pi R_C$$

We need to find the ratio $\frac{V_C}{V_A}$:

$$\frac{V_C}{V_A} = \frac{-G\sigma 4\pi R_C}{-G\sigma 4\pi R_A} = \frac{R_C}{R_A}$$

Substitute $R_C = \frac{5}{3} R_A$:

$$\frac{V_C}{V_A} = \frac{\frac{5}{3} R_A}{R_A} = \frac{5}{3}$$

The ratio of potential at an internal point of the new shell to shell A is 5:3.

The final answer is $\boxed{\text{5 : 3}}$.

Explanation of the solution:

1. Define masses and potentials for shells A and B using constant surface mass density $\sigma$.
   $M = \sigma 4\pi R^2$, $V = -\frac{GM}{R} = -G\sigma 4\pi R$.
2. Use the given potential ratio $V_A/V_B = 3/4$ to find the ratio of radii $R_A/R_B = 3/4$, so $R_B = \frac{4}{3}R_A$.
3. For the new shell C, its total mass $M_C = M_A + M_B$. Since surface mass density is conserved, $M_C = \sigma 4\pi R_C^2$.
4. Equate the mass expressions to find $R_C^2 = R_A^2 + R_B^2$. Substitute $R_B$ to get $R_C = \frac{5}{3}R_A$.
5. The potential inside the new shell $V_C = -G\sigma 4\pi R_C$.
6. Calculate the ratio $V_C/V_A = R_C/R_A = \frac{5/3 R_A}{R_A} = 5/3$.

Answer:

The final answer is 5 : 3.
The correct option is (C).