Q166 / 200 · MHT CET - Medical 2026 (26 April Shift 1) · Physics
The excess pressure inside a spherical drop of water is four times the other drop of water. Their respective mass ratio is
- A$1 : 4$
- B$8 : 1$
- Correct answerC$1 : 64$
- D$1 : 16$
Solution
The excess pressure inside a spherical drop of water is given by $\Delta P = \dfrac{2T}{R}$
Given $\Delta P_1 = 4 \Delta P_2$
$\dfrac{2T}{R_1} = 4 \left( \dfrac{2T}{R_2} \right)$
$\dfrac{R_1}{R_2} = \dfrac{1}{4}$
The mass of a spherical drop is $m = \dfrac{4}{3} \pi R^3 \rho$
The ratio of their masses is $\dfrac{m_1}{m_2} = \left( \dfrac{R_1}{R_2} \right)^3$
$\dfrac{m_1}{m_2} = \left( \dfrac{1}{4} \right)^3 = \dfrac{1}{64}$
The respective mass ratio is $1 : 64$
Answer: $1 : 64$
Given $\Delta P_1 = 4 \Delta P_2$
$\dfrac{2T}{R_1} = 4 \left( \dfrac{2T}{R_2} \right)$
$\dfrac{R_1}{R_2} = \dfrac{1}{4}$
The mass of a spherical drop is $m = \dfrac{4}{3} \pi R^3 \rho$
The ratio of their masses is $\dfrac{m_1}{m_2} = \left( \dfrac{R_1}{R_2} \right)^3$
$\dfrac{m_1}{m_2} = \left( \dfrac{1}{4} \right)^3 = \dfrac{1}{64}$
The respective mass ratio is $1 : 64$
Answer: $1 : 64$
Hence, the correct answer is option: 'C'.