Statement-1: When a charged particle is placed in the cavity in a conducting sphere, the induced charge on the outer surface of the sphere is found to be uniformly distributed.
Statement-2 : Conducting surface is equipotential surface.

Options

A. Statement-1 is True, Statement-2 is True, Statement-2 is a cor.
B. Statement-1 is True, Statement-2 is True, Statement-2 is not a
C. Statement-1 is False, Statement-2 is True
D. Statement-1 is True, Statement-2 is False.

Answer

To analyze the statements provided in the question, we can break down the solution into clear steps:

### Step-by-Step Solution:

**Step 1: Understand the Setup**
- We have a conducting spherical shell with a cavity inside it. A charged particle (let's say charge \( Q \)) is placed in this cavity.

**Hint 1:** Visualize the conducting sphere and the charge inside it. Remember that conductors allow free movement of charges.

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**Step 2: Induction of Charges**
- The presence of charge \( Q \) in the cavity induces a charge of \(-Q\) on the inner surface of the conducting shell. This is due to electrostatic induction, where the conductor rearranges its charges to maintain electrostatic equilibrium.

**Hint 2:** Recall that the total induced charge on the inner surface must equal the charge placed inside the cavity.

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**Step 3: Charge Distribution on the Outer Surface**
- Since the inner surface has a charge of \(-Q\), to maintain overall neutrality, a charge of \( +Q \) will appear on the outer surface of the conducting shell.

**Hint 3:** Remember that the total charge on the conductor must remain constant and equal to the initial charge placed inside.

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**Step 4: Repulsion Among Charges**
- The positive charges on the outer surface will repel each other due to electrostatic forces. In a conductor, charges are free to move, and they will distribute themselves to minimize repulsion.

**Hint 4:** Think about how charges behave in a conductor and how they try to maximize their distance from each other.

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**Step 5: Uniform Distribution of Charge**
- As the positive charges repel each other, they will move to the outermost surface of the conductor. Eventually, they will distribute themselves uniformly across the outer surface of the sphere, achieving a constant surface charge density.

**Hint 5:** Consider the geometry of the sphere and how charges would naturally spread out to maintain equilibrium.

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**Step 6: Equipotential Surface**
- Since the charges on the outer surface are uniformly distributed, the electric potential at every point on this surface is the same. Therefore, the conducting surface acts as an equipotential surface.

**Hint 6:** Recall that in electrostatics, conductors in equilibrium have the same potential throughout their surface.

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**Step 7: Conclusion**
- Both statements are correct:
- Statement 1: The induced charge on the outer surface of the sphere is uniformly distributed.
- Statement 2: The conducting surface is an equipotential surface.
- Statement 2 provides the correct explanation for Statement 1.

**Final Answer:** Both statements are true, and Statement 2 is the correct explanation for Statement 1.

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