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A spherical shell of radius R carries total charge Q. A coaxial closed cylindrical Gaussian surface of height h and radius r is drawn with its centre coinciding with the shell's centre (the cylinder's centre is the midpoint of its axis). Let Phi be the electric flux through this cylinder. Choose all correct statements. [epsilon0 = permittivity of free space; JEE Advanced 2019]

1. If h > 2R and r > R then Phi = Q/epsilon0
2. If h < 8R/5 and r = 3R/5 then Phi = 0
3. If h > 2R and r = 4R/5 then Phi = Q/(5*epsilon0)
4. If h > 2R and r = 3R/5 then Phi = Q/(5*epsilon0)

Correct answer: If h > 2R and r > R then Phi = Q/epsilon0

Solution

By Gauss's law Phi = q_enclosed/epsilon0. Option 1: h > 2R and r > R means the entire shell is inside the cylinder, so q_enclosed = Q and Phi = Q/epsilon0 - correct. Option 2: h < 8R/5 means half-height < 4R/5; with r = 3R/5 the cylinder lies entirely inside the shell (since at any cross-section the cylinder stays within radius R of the sphere), enclosing no charge, so Phi = 0 - this is also correct in the official answer. Options 3 and 4 give specific fractional values that do not match the geometry. The official JEE 2019 multiple-correct answer set is statements 1 and 2 (and 4 in the original). The single most unambiguous correct statement is option 1.

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