Due to the excess pressure of surface tension find the change in the radius (in mathringA ) of a liquid drop. of surface tension 0.075 N / m and Bulk modulus 1.25 × 108 N / m 2.

Question

Due to the excess pressure of surface tension find the change in the radius (in mathringA ) of a liquid drop. of surface tension 0.075 N / m and Bulk modulus 1.25 × 108 N / m 2. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Δ P=B (Δ V/V) ⇒ (2 T/R)=B (Δ V/V) ; V=(4/3) π R3 ; (Δ V/V)=(3 Δ R/R) ⇒ (2 T/R)=B((3 Δ R/R)) ⇒ Δ R=(2 T/3 B)=(2 × 0.075/3 × 1.25 × 108)=4 mathringA

Q. Due to the excess pressure of surface tension find the change in the radius (in $\overset{˚}{A}$ ) of a liquid drop. of surface tension $0.075 N / m$ and Bulk modulus $1.25 \times 1 0^{8} N / m^{2}$ .

$Δ P = B \frac{Δ V}{V}$
$\Rightarrow \frac{2 T}{R} = B \frac{Δ V}{V}; V = \frac{4}{3} π R^{3}; \frac{Δ V}{V} = \frac{3Δ R}{R}$
$\Rightarrow \frac{2 T}{R} = B (\frac{3Δ R}{R})$
$\Rightarrow Δ R = \frac{2 T}{3 B} = \frac{2 \times 0.075}{3 \times 1.25 \times 1 0 ^{8}} = 4 \overset{˚}{A}$

Q. Due to the excess pressure of surface tension find the change in the radius (in A˚ ) of a liquid drop. of surface tension 0.075N/m and Bulk modulus 1.25×108N/m2.

ΔP=BVΔV​ ⇒R2T​=BVΔV​;V=34​πR3;VΔV​=R3ΔR​ ⇒R2T​=B(R3ΔR​) ⇒ΔR=3B2T​=3×1.25×1082×0.075​=4A˚

Q. Due to the excess pressure of surface tension find the change in the radius (in $\overset{˚}{A}$ ) of a liquid drop. of surface tension $0.075 N / m$ and Bulk modulus $1.25 \times 1 0^{8} N / m^{2}$ .

$Δ P = B \frac{Δ V}{V}$
$\Rightarrow \frac{2 T}{R} = B \frac{Δ V}{V}; V = \frac{4}{3} π R^{3}; \frac{Δ V}{V} = \frac{3Δ R}{R}$
$\Rightarrow \frac{2 T}{R} = B (\frac{3Δ R}{R})$
$\Rightarrow Δ R = \frac{2 T}{3 B} = \frac{2 \times 0.075}{3 \times 1.25 \times 1 0 ^{8}} = 4 \overset{˚}{A}$