Q.20 Consider a spherical particle of mass m and radius r moving in a medium. Its velocity at any
time t is given by v=𝑣0 exp(-6𝜋xrt/m) ,where 𝑣0 is initial velocity of the particle. The
dimensions of X are
(A) MLT-1 (B) M-1LT (C) ML-1T-1 (D) Dimensionless
Dimensions of X in Spherical Particle Velocity Equation | Stokes Law Dimensional Analysis
The spherical particle velocity equation \( v = v_0 \exp\left(-\frac{6\pi X r t}{m}\right) \) models drag in viscous media per Stokes’ law. Determining dimensions of X ensures the exponential is dimensionless, vital for dimensional analysis in physics exams like CSIR NET. This guide solves the problem step-by-step, explains options, and links to Stokes law derivation.
Dimensional Analysis
The velocity equation is \( v = v_0 \exp\left(-\frac{6\pi X r t}{m}\right) \). The exponential argument must be dimensionless (\([M^0 L^0 T^0]\)), so \(\left[\frac{X r t}{m}\right] = [M^0 L^0 T^0]\).
Known dimensions are: mass \([m] = [M]\), radius \([r] = [L]\), time \([t] = [T]\). Thus, \([X] = \frac{[M]}{[L][T]} = [M L^{-1} T^{-1}]\).
Physical Context
This form arises from Stokes’ law drag force \( F_d = 6\pi \eta r v \), where \(\eta\) is viscosity with \([\eta] = [M L^{-1} T^{-1}]\).
Newton’s second law gives \( m \frac{dv}{dt} = -6\pi \eta r v \), solving to \( v = v_0 e^{-\frac{6\pi \eta r t}{m}} \).
Thus, X represents viscosity \(\eta\).
Option Analysis
(C) \([M L^{-1} T^{-1}]\): Correct
\([M L^{-1} T^{-1}][L][T]/[M] = [M^0 L^0 T^0]\).
- (A) \([M L T^{-1}]\): Matches momentum; \([M L T^{-1}][L][T]/[M] = [L T^2]\) (not dimensionless).
- (B) \([M^{-1} L T]\): \([M^{-1} L T][L][T]/[M] = [L^2 T^2 M^{-2}]\) (not dimensionless).
- (D) Dimensionless: \([1][L][T]/[M] = [M^{-1} L T]\) (not dimensionless).
Why Dimensional Analysis Matters
Exponents in equations like this must yield dimensionless arguments. Here, velocity \([L T^{-1}]\) on both sides matches, but the exponent \(\frac{6\pi X r t}{m}\) requires \([X] = [M L^{-1} T^{-1}]\), matching viscosity \(\eta\) from \( F = 6\pi \eta r v \).
This confirms X as the viscosity coefficient.
Correct Answer
Option (C) \([M L^{-1} T^{-1}]\)
Detailed verification: \(\left[\frac{X r t}{m}\right] = [M L^{-1} T^{-1} \cdot L \cdot T / M] = 1\).
Other options fail dimensional balance, as analyzed above.
