Sometimes it is convenient to construct a system of units so that all quantities can be expressed in terms of only one physical quantity. In one such system, dimensions of different quantities are given in terms of a quantity X as follows: [position] ; [speed] ; [acceleration] ; [linear momentum] ; [force] . Then -
- (A)✓ Correct
- (B)✓ Correct
- (C)
- (D)
Answer: (A), (B)
Solution
Given: L = X^α, LT⁻¹ = X^β, LT⁻² = X^p, MLT⁻¹ = X^q, MLT⁻² = X^r. From (1)/(2): T = X^(α-β). From acceleration: L/T² = X^p gives X^α/X^(2(α-β)) = X^p, so α + p = 2β ✓. From momentum: M = X^(q-β). From force: X^(q-β) · X^α · X^(-(α-β)) = X^r, giving α + r - q = β, or p + q - r = β ✓. Options (C) and (D) do not follow from these relations.
JEE Advanced · 2020 · 27 Sep 2020 · Paper 1 · One or more correct options