A spherical drop of water has 1 mm radius. If the surface tension of water is 75 × 10-3 N / m, then difference of pressure between inside and outside of the drop is

Question

A spherical drop of water has 1 mm radius. If the surface tension of water is 75 × 10-3 N / m, then difference of pressure between inside and outside of the drop is (A) 35 N / m 2 (B) 70 N / m 2 (C) 140 N / m 2 (D) 150 N / m 2

Tardigrade Admin · Accepted Answer

Excess pressure =(2 T/R) =(2 × 75 × 10-3/1 × 10-3) =150 N / m 2 T= surface tension R= radius

Q. A spherical drop of water has $1 mm$ radius. If the surface tension of water is $75 \times 1 0^{- 3} N / m$ , then difference of pressure between inside and outside of the drop is

$35 N / m^{2}$

$70 N / m^{2}$

$140 N / m^{2}$

$150 N / m^{2}$

Excess pressure $= \frac{2 T}{R}$
$= \frac{2 \times 75 \times 1 0 ^{- 3}}{1 \times 1 0 ^{- 3}}$
$= 150 N / m^{2}$
$T =$ surface tension
$R =$ radius

Q. A spherical drop of water has 1mm radius. If the surface tension of water is 75×10−3N/m, then difference of pressure between inside and outside of the drop is

35N/m2

70N/m2

140N/m2

150N/m2