The excess pressure inside a spherical drop of water A is four times that of another drop B. Then the ratio…

The excess pressure inside a spherical drop of water A is four times that of another drop B. Then the ratio of mass of drop $A$ to that of drop $B$ is
  1. $1: 4$
  2. $1: 8$
  3. $1: 16$
  4. $1: 64$

Solution

Excess pressure inside $1^{\text {st }}$ spherical drop (A) is given by, $\dot{P}_A=\frac{2 T}{r_1}$
For $2^{\text {nd }} \operatorname{drop}(B), P_B=\frac{2 T}{r_2}$ $\therefore \quad-\frac{2 \mathrm{~T}}{\mathrm{r}_1}=4\left(\frac{2 \mathrm{~T}}{\mathrm{r}_2}\right) \ldots$ (given, $\mathrm{P}_{\mathrm{A}}=4 \mathrm{P}_{\mathrm{B}}$ ) $\therefore \quad \frac{r_1}{r_2}=\frac{1}{4}$ Now, $\frac{m_1}{m_2}=\frac{V_1 \rho_1}{V_2 \rho_2}$ As both are drops of water, $\rho_1=\rho_2$ $\therefore \quad \frac{\mathrm{m}_1}{\mathrm{~m}_2}=\frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{\frac{4}{3} \pi \mathrm{r}_1^3}{\frac{4}{3} \pi \pi_2^3}=\left(\frac{1}{4}\right)^3=\frac{1}{64}$

Asked in: MHT CET 2024 (04 May Shift 1)

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