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Difficulty: HardPhysical Quantities, Units and Dimensions

The frequency of oscillation ff of a small liquid droplet executing spherical oscillations depends on the surface tension γ\gamma of the liquid, its mass density ρ\rho, and the radius rr of the droplet according to the relationship f=kγaρbrcf = k \gamma^a \rho^b r^c, where kk is a dimensionless constant. Which of the following represents the correct value of the exponent cc?

  1. −32-\frac{3}{2}Answer
  2. B
    −12-\frac{1}{2}
  3. C
    12\frac{1}{2}
  4. D
    32\frac{3}{2}

Answer

The correct value of the exponent cc is −32-\frac{3}{2}.
By writing the dimensional equation [M0L0T−1]=[MT−2]a[ML−3]b[L]c[M^0 L^0 T^{-1}] = [M T^{-2}]^a [M L^{-3}]^b [L]^c, we solve for the exponents: −2a=−1-2a = -1 gives a=1/2a = 1/2, a+b=0a + b = 0 gives b=−1/2b = -1/2, and −3b+c=0-3b + c = 0 gives 3/2+c=0  ⟹  c=−3/23/2 + c = 0 \implies c = -3/2. Thus, the option equal to −3/2-3/2 is correct.

Step-by-Step Solution

1
Express each physical quantity in terms of its base dimensions [M][M], [L][L], and [T][T].
Frequency f=[T−1]f = [T^{-1}], Surface tension γ=ForceLength=[MT−2]\gamma = \frac{\text{Force}}{\text{Length}} = [M T^{-2}], Density ρ=[ML−3]\rho = [M L^{-3}], and Radius r=[L]r = [L].
Dimensional analysis requires reducing derived physical quantities to their base units.
2
Substitute dimensions into the given formula f=kγaρbrcf = k \gamma^a \rho^b r^c.
[M0L0T−1]=[MT−2]a[ML−3]b[L]c=Ma+bL−3b+cT−2a[M^0 L^0 T^{-1}] = [M T^{-2}]^a [M L^{-3}]^b [L]^c = M^{a+b} L^{-3b+c} T^{-2a}.
The principle of dimensional homogeneity requires both sides of a physical equation to have matching dimensions.
3
Equate exponents for each base dimension MM, LL, and TT.
For TT: −2a=−1  ⟹  a=12-2a = -1 \implies a = \frac{1}{2}. For MM: a+b=0  ⟹  b=−a=−12a + b = 0 \implies b = -a = -\frac{1}{2}. For LL: −3b+c=0  ⟹  −3(−12)+c=0-3b + c = 0 \implies -3\left(-\frac{1}{2}\right) + c = 0.
Matching powers across orthogonal base dimensions provides a system of linear equations.
4
Solve for the target exponent cc.
\frac{3}{2} + c = 0 \implies c = -\frac{3}{2}.
Subtracting 3/23/2 from both sides gives the exact value of exponent cc.

Key Concept

Dimensional Homogeneity and Dimensional Analysis
Estimated Time:2m 0s
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Topics

Physical Quantities, Units and Dimensions

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