Question

Two concentric metal shells of radii $r_A = 1$ cm and $r_B = 2$ cm are given charges and action performed as shown.

• A. 1 | 4 | 3 | 2
• B. 1 | 2 | 3 | 2
• C. 3 | 2 | 3 | 1
• D. 4 | 4 | 3 | 2

Answer 1

Answer: (B)

Option (B)

Answer 2

We must decide what happens when one of the shells is “grounded” or when the shells are connected. In our problem the two shells (inner A of radius 1 cm and outer B of radius 2 cm) carry the initial charges given in List–I. Then the following takes place:

1. For (P):
   • Initial: q₍A₎ = 2 µC, q₍B₎ = 4 µC.
   • Process: Shell A is grounded. Since A’s potential becomes zero, using

   V₍A₎ = k (q₍A′₎/r₍A₎) + k (q₍B₎/r₍B₎) = 0

   we get

   q₍A′₎ = –(r₍A₎/r₍B₎) q₍B₎ = –(1/2)×4 = –2 µC.

   Thus A’s charge changes while B remains unchanged. This is outcome type (1).

2. For (Q):
   • Initial: q₍A₎ = 4 µC, q₍B₎ = –4 µC.
   • Process: Shell B is grounded. For B we have

   V₍B₎ = k [ (q₍A₎ + q₍B′₎)/r₍B₎ ] = 0 ⟹ q₍B′₎ = –q₍A₎ = –4 µC.

   Because A is isolated it stays at 4 µC. So the net charges do not change; both shells remain with the same charges. This is outcome type (2).

3. For (R):
   • Initial: q₍A₎ = –2 µC, q₍B₎ = 4 µC ⟹ Total Q = 2 µC.
   • Process: A and B are connected by a conducting wire so they come to a common potential. For concentric spheres the potentials are given by

   q₍A′₎/r₍A₎ = q₍B′₎/r₍B₎

   and

   q₍A′₎ + q₍B′₎ = 2 µC.

   Solving, we get

   q₍A′₎ = 2/3 µC and q₍B′₎ = 4/3 µC.

   Thus both charges change. This is outcome type (3).

4. For (S):
   • Initial: q₍A₎ = –1 µC, q₍B₎ = 2 µC.
   • Process: Shell A is grounded. Again using

   V₍A₎ = k (q₍A′₎/r₍A₎) + k (q₍B₎/r₍B₎) = 0 ⟹ q₍A′₎ = –(r₍A₎/r₍B₎) q₍B₎ = –(1/2)×2 = –1 µC.

   So A remains at –1 µC while B, being isolated, also keeps its net charge (2 µC). In other words the net change is nil; this outcome is like (2).

We then obtain the following matching: (P) → (1)
(Q) → (2)
(R) → (3)
(S) → (2)

Checking the options given in List–II we see that Option (B) has the mapping:
P: 1, Q: 2, R: 3, S: 2.

Thus the correct answer is Option (B).