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Physics · UNITS AND MEASUREMENT

हिन्दी

The period of oscillation \( T \) of a spherical liquid drop under the action of surface tension depends on the density of the liquid \( \rho \), radius of the drop \( r \), and surface tension \( S \). Using the method of dimensional analysis as described in section 1.6.2, if \( T = k \rho^a r^b S^c \) (where \( k \) is a dimensionless constant), then the values of exponents \( a, b, c \) are respectively:

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

Correct answer · Option 1

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

Explanation

Equating dimensions: \( [T] = [M L^{-3}]^a [L]^b [M T^{-2}]^c = [M^{a+c} L^{-3a+b} T^{-2c}] \). Equating exponents gives \( a + c = 0 \), \( -3a + b = 0 \), and \( -2c = 1 \). Solving yields \( c = -1/2 \), \( a = 1/2 \), and \( b = 3/2 \).
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