JEE Main 2023 · 24 Jan · Shift 2

JEE Main 2023 Physics question on Units and Measurements

The frequency (vv) of an oscillating liquid drop may depend upon radius (rr) of the drop, density (ρ\rho) of liquid and the surface tension (ss) of the liquid as : v=raρbscv = r^a \rho^b s^c. The values of a, b and c respectively are
  1. A.
    (32,12,−12)\left(\dfrac{3}{2}, \dfrac{1}{2}, -\dfrac{1}{2}\right)
  2. B.
    (−32,12,12)\left(-\dfrac{3}{2}, \dfrac{1}{2}, \dfrac{1}{2}\right)
  3. C.
    (−32,−12,12)\left(-\dfrac{3}{2}, -\dfrac{1}{2}, \dfrac{1}{2}\right)
  4. D.
    (32,−12,12)\left(\dfrac{3}{2}, -\dfrac{1}{2}, \dfrac{1}{2}\right)
Show answer and solution

Correct option: C

Solution

Using dimensional analysis, write dimensions of each quantity:
- v: T⁻¹
- r: L
- ρ: ML⁻³
- s (surface tension): MT⁻²

So: T⁻¹ = Lᵃ · (ML⁻³)ᵇ · (MT⁻²)꜀

Matching dimensions:

M: 0 = b + c → c = -b

T: -1 = -2c → c = 1/2, so b = -1/2

L: 0 = a - 3b → a = 3b = -3/2

Therefore a = -3/2, b = -1/2, c = 1/2, which matches option 3.

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