Units and Dimensions (JEE Main)

Question 1

Suppose the speed of light c, Planck's constant h and the universal gravitational constant G are chosen as the fundamental quantities instead of mass, length and time. The dimensional formula of mass

Accepted Answer

A. Write M = h^a c^b G^d with [h] = ML²T⁻¹, [c] = LT⁻¹, [G] = M⁻¹L³T⁻². Mass: a − d = 1. Length: 2a + b + 3d = 0. Time: −a − b − 2d = 0. Adding the length and time equations gives a + d = 0, so with a − d = 1 we get a = 1/2, d = −1/2, and then b = 1/2. Hence M = √(hc/G), the Planck mass (about 5×10⁻⁸ kg). Option B is the Planck length and option C the Planck time; the trap is to recall the familiar P

Question 2

A small liquid drop held together by surface tension can oscillate in shape. Assume its natural frequency depends only on the surface tension S, the density ρ of the liquid and the radius r of the dro

Accepted Answer

D. Key insight: only one combination of S, ρ, r has dimension T⁻¹. Let f ∝ S^a ρ^b r^c with [S] = MT⁻², [ρ] = ML⁻³. Mass: a + b = 0. Time: −2a = −1, so a = 1/2, b = −1/2. Length: −3b + c = 0, so c = −3/2. Thus f ∝ √(S/(ρ r³)). New ratio = √((1/2)/2³) = √(1/16) = 1/4. Option C comes from f ∝ √(S/(ρ r)) (wrong power of r), option B from S/(ρ r²) without the square root, and option A from changing only

Question 3

A particle of mass m moves along the x-axis under a restoring force F = −kx³ (not simple harmonic). It is released from rest at x = A. Assuming the period depends only on m, k and A, if the amplitude

Accepted Answer

B. Here k = F/x³, so [k] = MLT⁻²/L³ = ML⁻²T⁻². Let T ∝ m^a k^b A^c. Time: −2b = 1 gives b = −1/2. Mass: a + b = 0 gives a = 1/2. Length: −2b + c = 0 gives c = −1. So T ∝ √(m/k) (1/A). Doubling A halves the period. The trap is option A: isochronism holds only for the linear force F = −kx, where [k] = MT⁻² leaves no room for A. Exam pattern: JEE Main – time period from dimensional analysis with a non-s

Units and Dimensions: questions with solutions

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