This question is based on Kepler's Third Law of Planetary Motion, which states that the square of the orbital period (T) of a planet is directly proportional to the cube of the semi-major axis of its orbit (r). For elliptical orbits, the semi-major axis is the average distance from the sun. The relationship can be expressed as T² ∝ r³. Let T₁ be the present duration of the year and r₁ be the present distance of the Earth from the Sun. So, T₁² = k r₁³, where k is a constant. If the new distance is r₂ = r₁/ 4, let the new duration of the year be T₂. Then T₂² = k r₂³ = k (r₁/4)³. Dividing the two equations: (T₂² / T₁²) = (k (r₁/4)³) / (k r₁³) (T₂ / T₁)² = (r₁³/64) / r₁³ (T₂ / T₁)² = 1/64 Taking the square root of both sides: T₂ / T₁ = √(1/64) T₂ / T₁ = 1/8 T₂ = T₁/8. Thus, the duration of the year will be one-eighth of the present year.