Question

The frequency (\( \nu \)) of an oscillating liquid drop may depend upon the radius (\( r \)) of the drop, density (\( \rho \)) of the liquid, and the surface tension (\( s \)) of the liquid as: \[ \nu = k r^a \rho^b s^c \]
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. \(\left(\frac{3}{2},-\frac{1}{2}, \frac{1}{2}\right)\)

Hint:

In dimensional analysis, equate the powers of mass (M), length (L), and time (T) systematically to find the unknown constants.

Answer 1

The Correct Option is B

The dimensional formula of frequency is:

\[ [\nu] = [T^{-1}] \]

The given equation is:

\[ \nu = r^a \rho^b s^c \]

Substituting the dimensional formulas:

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

Simplifying:

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

Equating dimensions of \( M \), \( L \), and \( T \):

\[ b + c = 0 \quad \text{(1)} \]

\[ a - 3b = 0 \quad \text{(2)} \]

\[ -2c = -1 \quad \text{(3)} \]

From equation (3):

\[ c = \frac{1}{2} \]

Substitute \( c = \frac{1}{2} \) into equation (1):

\[ b + \frac{1}{2} = 0 \implies b = -\frac{1}{2} \]

Substitute \( b = -\frac{1}{2} \) into equation (2):

\[ a - 3\left(-\frac{1}{2}\right) = 0 \implies a + \frac{3}{2} = 0 \implies a = -\frac{3}{2} \]

Thus, the values are:

\[ a = -\frac{3}{2}, \quad b = -\frac{1}{2}, \quad c = \frac{1}{2} \]

Answer 2

The correct answer is (B) : \(\left(-\frac{3}{2},-\frac{1}{2}, \frac{1}{2}\right)\)
[T−1]=[L1]a[M1L−3]b[LMLT−2​]c 
⇒T−1=Mb+c⋅La−3b⋅T−2c 
c=21​,b=−21​,a−3b=0 
a+23​=0⇒a=−23​