Determine the excess pressure inside a drop of mercury having a radius of 2 mm . The surface tension of mercury is given as 0.465 N/m .

Correct answer: C. 465 N/m^2 The excess pressure inside a liquid drop is given by the formula Δ P = (2T)/(R) . Given: T = 0.465 N/m R = 2 mm = 2 × 10^ -3 m Substituting the values in the formula: Δ P = 2 × 0.465 2 × 10^ -3 Δ P = 0.465 × 10^3 Δ P = 465 N/m^2

Concepts Of Physics MCQ Edition [Volume 1]

Exam: Concepts Of Physics MCQ Edition [Volume 1]
Subject: Physics
Chapter: Some Mechanical Properties of Matter

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Some Mechanical Properties of Matter

Determine the excess pressure inside a drop of mercury having a...

Determine the excess pressure inside a drop of mercury having a radius of

2 \text{ mm}

. The surface tension of mercury is given as

0.465 \text{ N/m}

Options

A $232.5 \text{ N/m}^2$
B $46.5 \text{ N/m}^2$
C $465 \text{ N/m}^2$
D $930 \text{ N/m}^2$

Correct answer

C. $465 \text{ N/m}^2$

Step-by-step solution

The excess pressure inside a liquid drop is given by the formula

\Delta P = \dfrac{2T}{R}

.

Given:

T = 0.465 \text{ N/m}

R = 2 \text{ mm} = 2 \times 10^{-3} \text{ m}

Substituting the values in the formula:

\Delta P = \dfrac{2 \times 0.465}{2 \times 10^{-3}}

\Delta P = 0.465 \times 10^3

\Delta P = 465 \text{ N/m}^2

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