The frequency ($v$) of an oscillating liquid drop may depend upon radius (r) of the drop, density ($\rho$) of liquid and the surface tension (s) of the liquid as : $v=r^{a} \rho^{b} s^{c}$. The values of $a, b$ and $c$ respectively are

(A) $\left(-\frac{3}{2},-\frac{1}{2}, \frac{1}{2}\right)$

(B) $\left(\frac{3}{2},-\frac{1}{2}, \frac{1}{2}\right)$

(C) $\left(\frac{3}{2}, \frac{1}{2},-\frac{1}{2}\right)$

(D) $\left(-\frac{3}{2}, \frac{1}{2}, \frac{1}{2}\right)$

JEE Main · 2023 · 24 Jan 2023 · Shift 2 · Units & Measurement

The frequency (

v

) of an oscillating liquid drop may depend upon radius (r) of the drop, density (

ρ

) of liquid and the surface tension (s) of the liquid as :

v = r^{a} ρ^{b} s^{c}

. The values of

a, b

and

c

respectively are

Answer: (A)

Comparing dimensions,

[T^{- 1}] = [L]^{a} [M L^{- 3}]^{b} [M T^{- 2}]^{c} = [M^{b + c} L^{a - 3 b} T^{- 2 c}]

. Matching the powers of mass gives

b + c = 0

; matching the powers of time gives

- 2 c = - 1

, so

c = \frac{1}{2}

and therefore

b = - \frac{1}{2}

; matching the powers of length gives

a - 3 b = 0

, so

a = - \frac{3}{2}

. The exponents are

(- \frac{3}{2}, - \frac{1}{2}, \frac{1}{2})

JEE Main · 2023 · 24 Jan 2023 · Shift 2 · Single correct option