Correct Answer: decrease by 4 times
Let the initial radius of the spherical detector be R.
Volume of a sphere is proportional to the cube of its radius:
V ∝ R³
Given that the volume increases by 8 times:
V₂ = 8V₁
So,
R₂³ = 8R₁³
R₂ = 2R₁
Intensity of radiation from a point source follows the inverse square law:
I ∝ 1 / R²
Therefore,
I₂ / I₁ = (R₁ / R₂)²
I₂ / I₁ = (1 / 2)²
I₂ / I₁ = 1 / 4
Hence, the intensity becomes one-fourth of its original value.
So, the intensity decreases by 4 times.
Updated for JEE Main 2026: This PYQ is important for JEE Mains, JEE Advanced and other competitive exams. Practice more questions from this chapter.