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Chapter 7: Problem 1007

A vesel whose bottom has round holes with diameter of \(0.1 \mathrm{~mm}\) is filled with water. The maximum height to which the water can be filled without leakage is (S.T. of water \(=[(75\) dyne $\left.\\} / \mathrm{cm}], \mathrm{g}=1000 \mathrm{~m} / \mathrm{s}^{2}\right)$ (A) \(100 \mathrm{~cm}\) (B) \(75 \mathrm{~cm}\) (C) \(50 \mathrm{~cm}\) (D) \(30 \mathrm{~cm}\)

Short Answer

Expert verified
The maximum height to which the water can be filled without leakage is \(75 \mathrm{~cm}\).

Step by step solution

01

Write the expression for surface tension force

The force due to surface tension (F) is given by the product of the surface tension (S) and the circumference of the hole: F = S × 2πr Where, r is the radius of the hole.
02

Calculate the radius

Calculate the radius (r) from the given diameter of the hole (0.1 mm): r = diameter/2 = 0.1/2
03

Convert units

Convert the units of r from millimeters to centimeters: r = 0.1/2 × (1/10) = 0.005 cm
04

Write the expression for gravitational force

Gravitational force (F') is given by the product of volume, density (ρ), and acceleration due to gravity (g): F' = Vρg Where, V is the volume of the water cylinder inside the hole.
05

Write the expression for volume

Write the volume (V) of the water cylinder as the product of the cross-sectional area of the hole and the maximum height (H) of the water: V = πr^2H
06

Write the conditions for balance between forces

At equilibrium, the gravitational force (F') must be equal to the surface tension force (F): S × 2πr = πr^2Hρg
07

Solve for H

Divide both sides by πr and the given values for S, ρ and g to solve for H: H = \(\frac{(2S)}{rρg}\) = \(\frac{150~\text{dyne/cm} × 2}{0.005~\text{cm} × 1~\text{g/cm}^{3} × 1000~\text{cm/s}^2}\) H = 75 cm The correct answer is (B) \(75 \mathrm{~cm}\). The maximum height to which the water can be filled without leakage is 75 cm.

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