If the time period t of the oscillation of a drop of liquid of density d, radius r, vibrating under surface tension s is given by the formula t = √r2bscda/2. It is observed that the time period is directly proportional to √(d/s). The value of b should therefore be: (a) 3/4 (b) √(3) (c) 3/2 (d) 2/3

Answer: (c). Write t = rbsc/2da/4. Since t ∝ √(d/s): a/4 = 1/2 and c/2 = -1/2. Dimensions: [r] = L, [s] = MT⁻², [d] = ML⁻³. Powers of L: b - 3/4 = 0 ⇒ b - 3/2 = 0 So the answer is (c) b = 3/2.

If the time period t of the oscillation of a drop of liquid of… | JEE Main 2013 (23 Apr, Shift 1)

Q76JEE Main 2013 · 23 APR · SHIFT 1Medium

Applications of Dimensional Analysis

If the time period $t$ of the oscillation of a drop of liquid of density $d$ , radius $r$ , vibrating under surface tension $s$ is given by the formula $t = \sqrt{r^{2b}s^{c}d^{a/2}}$ . It is observed that the time period is directly proportional to $\sqrt{\frac{d}{s}}$ . The value of $b$ should therefore be:

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Answer: (c)

Write $t = r^{b}s^{c/2}d^{a/4}$ . Since $t \propto \sqrt{d/s}$ : $\frac{a}{4} = \frac{1}{2}$ and $\frac{c}{2} = -\frac{1}{2}$ .
Dimensions: $[r] = L$ , $[s] = MT^{-2}$ , $[d] = ML^{-3}$ .
Powers of $L$ : $b - 3\cdot\frac{a}{4} = 0 \Rightarrow b - \frac{3}{2} = 0$
So the answer is (c) $b = \frac{3}{2}$ .

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